Principles of Mathematical Economics II: Solutions Manual, Supplementary Materials and Supplementary Exercises 9462390878, 9789462390874, 9789462390881

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Mathematics Textbooks for Science and Engineering

Shapoor Vali

Principles of Mathematical Economics II Solutions Manual, Supplementary Materials and Supplementary Exercises

Mathematics Textbooks for Science and Engineering Volume 4

More information about this series at http://www.springer.com/series/10785

Shapoor Vali

Principles of Mathematical Economics II Solutions Manual, Supplementary Materials and Supplementary Exercises

Shapoor Vali Department of Economics Fordham University New York, NY USA

Mathematics Textbooks for Science and Engineering ISBN 978-94-6239-087-4 ISBN 978-94-6239-088-1 DOI 10.2991/978-94-6239-088-1

(eBook)

Library of Congress Control Number: 2013951796 Published by Atlantis Press, Paris, France www.atlantis-press.com © Atlantis Press and the authors 2015 This book, or any parts thereof, may not be reproduced for commercial purposes in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system known or to be invented, without prior permission from the Publisher. Printed on acid-free paper

Series Information Textbooks in the series ‘Mathematics Textbooks for Science and Engineering’ will be aimed at the broad mathematics, science and engineering undergraduate and graduate levels, covering all areas of applied and applicable mathematics, interpreted in the broadest sense. Series editor Charles K. Chui Stanford University, Stanford, CA, USA Atlantis Press 8, square des Bouleaux 75019 Paris, France For more information on this series and our other book series, please visit our website www.atlantis-press.com

Editorial

Recent years have witnessed an extraordinarily rapid advance in the direction of information technology within the scientific, engineering, and other disciplines, in which mathematics play a crucial role. To meet such urgent demands, effective mathematical models as well as innovative mathematical theory, methods, and algorithms must be developed for data information understanding and visualization. The revolution of the data information explosion as mentioned above demands early mathematical training with emphasis on data manipulation at the college level and beyond. The Atlantis book series, “Mathematics Textbooks for Science and Engineering (MTSE),” is founded to meet the needs of such mathematics textbooks that can be used for both classroom teaching and self-study. For the benefit of students and readers from the interdisciplinary areas of mathematics, computer science, physical and biological sciences, various engineering specialties, and social sciences, contributing authors are requested to keep in mind that the writings for the MTSE book series should be elementary and relatively easy to read, with sufficient examples and exercises. We welcome submission of such book manuscripts from all who agree with us on this point of view. This fourth volume consists of the solution manual and supplementary materials of the previous volume: “Principles of Mathematical Economics,” by the same author. It is divided into 13 chapters, covering such topics as: Market equilibrium model, Rates of change and the derivative, Optimal level of output and long run price, Nonlinear models, Economics Dynamics, and Mathematics of interest rates and finance. It is an important companion of the author’s previous volume. Charles K. Chui

vii

Preface

It was part of my original plan in writing my book, Principles of Mathematical Economics, Volume III of this MTSE book series, to provide answers to some of the problems in the exercise sections of the book. However, when the text grew to about 500 pages and the number of problems to over 600, I decided that simply providing numerical answers, without going through the steps of formulating and solving the problems, would not be very helpful to students or general readers. My own experience from teaching quantitative courses convinced me that providing mere answers was even harmful: in many cases chasing “the answer” leads some students to set problems up incorrectly, but such that its solution is the same as the given answer. This, on some occasions, as I am sure many of my academic colleagues have experienced, leads to debate over the “proper” grade for a homework or exam problem, and periodic sermon, undoubtedly boring, about the impossibility of consistently reaching correct conclusions through wrong reasoning. I believe that the old saying “the correct formulation of a problem is 50 % of the solution” is an expression of the accumulated wisdom of scientific inquiries. For the above reasons, I decided to prepare this manual which gives a full-fledged formulation and solution to each of the problems in the text. In some cases I go beyond setting up and solving a problem and include additional materials relevant to the subject of the chapter. This book naturally accompanies Principles of Mathematical Economics, but it can also be used independently from the text. The book can be treated as a standalone collection of solved problems in different areas of mathematical economics and as additional sets of exercises, over 500, that can be used to sharpen students’ skill and depth of understandings of many economic topics. Therefore, students can benefit from this manual even if the course they take in quantitative economics uses a different textbook. The manual is organized as follows: exercises from each chapter of the text are listed, followed by their solutions. Where a problem can be solved using different methods, sketches of alternative methods are also provided. If the solution references an equation in the text, the equation number is used. But if in the process of solving a problem a new equation is derived its number is tagged by “SM” ix

x

Preface

(for “Solution Manual”) to distinguish it from the text equation. For example, the first equation derived in Chap. 4 is labeled (4.1 SM). If a modified version of a text formula is introduced, it is tagged “MOD” to indicate the modification. Finally, each set of solutions to chapter exercises is followed by an additional set of unsolved problems under the heading “Supplementary Exercises”. In the process of preparing this manual, I discovered a number of typographical errors in the text. These errors are all mine and escaped me in the process of proofreading. I have corrected the errors if they appear in the Exercise sections of the chapters, which are repeated here in the manual. I hope there are no errors in this manual, but given the sheer volume of numbers and mathematical expressions, it is still likely that you encounter some errors. In case you do, I would appreciate it if you would let me know by sending an email to [email protected]. As it always the case, writing a book of this nature requires help from many individuals. In particular, I would like to thank the publishers Dr. Keith Jones and Dr. Zeger Karssen, as well as the book series editor, Professor Charles Chui, of Atlantis Press, for their support in the publication of this book. I am grateful to Dr. Michael Malenbaum, Ellen Fishbein, and Behrang Vali for their help in editing and proofreading parts of the manuscript. Some of my students also helped me by checking some of the solutions. I thank them all, especially Tyler Shegerian. Above all, I am again indebted to my wife Firoozeh for her support and encouragement. Spring 2014

Shapoor Vali

Contents

1

Household Expenditure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

2

Variables, A Short Taxonomy . . . . . . . . . . . . . . . . . . . . . . . . . . .

17

3

Sets and Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

21

4

Market Equilibrium Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

33

5

Rates of Change and the Derivative. . . . . . . . . . . . . . . . . . . . . . .

57

6

Optimal Level of Output and Long Run Price . . . . . . . . . . . . . . .

81

7

Nonlinear Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

123

8

Additional Topics in Perfect and Imperfect Competition. . . . . . . .

151

9

Logarithmic and Exponential Functions. . . . . . . . . . . . . . . . . . . .

177

10

Production Function, Least-Cost Combination of Resources, and Profit Maximizing Level of Output . . . . . . . . . . . . . . . . . . . .

205

11

Economics Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

227

12

Mathematics of Interest Rates and Finance . . . . . . . . . . . . . . . . .

239

13

Matrices and Their Applications . . . . . . . . . . . . . . . . . . . . . . . . .

263

xi

Figures

Fig. Fig. Fig. Fig.

1.1 1.2 3.1 3.2

Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig.

3.3 3.4 3.5 3.6 4.1 4.2 4.3 4.4 4.5 4.6 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10 5.11 6.1 6.2

Fig. 6.3 Fig. 7.1 Fig. 7.2

Budget Line . . . . . . . . . . . . . . . . . . . . . . . . . . Budget Line and Income-Consumption Path . . . . pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi The graph of the positive roots of w ¼ x2 þ 25 wþ2 Graph of f ðwÞ ¼ pffiffiffiffiffiffiffiffiffiffi , drawn for 20  w  w2 16 wþ2 ffiffiffiffiffiffiffiffiffiffi p Graph of f ðwÞ ¼ 2 , drawn for 6  w  15 . .

.. .. .. 6.

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. . . .

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. . . .

3 4 25 25

.... w 16 Graph of y ¼ 3x þ 2; y ¼ 10 2x; and y ¼ 5 þ 3x . Graph of f ðxÞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Graphs of f ðxÞ and gðxÞ . . . . . . . . . . . . . . . . . . . . . . Graph of supply and demand functions . . . . . . . . . . . Graph of supply and demand functions . . . . . . . . . . . Shift in the demand function. . . . . . . . . . . . . . . . . . . Supply and demand curves for part (a). . . . . . . . . . . . Supply and demand curves for part (b) . . . . . . . . . . . The IS Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . MC curve and Levels of Output where MC ¼ 50 . . . . Marginal Cost and Average Total Cost curves . . . . . . Graph of function y ¼ x3 9x2 48x þ 50 . . . . . . . . Graph of h ¼ 15t2 þ 96t þ 4 . . . . . . . . . . . . . . . . . Graph of TR ¼ 1250N 5N 2 . . . . . . . . . . . . . . . . . . Graph of y ¼ xð1 xÞ4 . . . . . . . . . . . . . . . . . . . . . . Graph of x2 þ y2 2y ¼ 2 . . . . . . . . . . . . . . . . . . . . Graph of f ðx; yÞ ¼ x2 þ y2 2y and its Contour Plot . Graph of z ¼ 3x3 þ 3x2 y 2y þ 5 . . . . . . . . . . . . . . . pffiffiffiffiffiffiffiffiffiffiffiffiffiffi Graph of z ¼ x2 þ y2 . . . . . . . . . . . . . . . . . . . . . . pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Graph of 49 x2 y2 . . . . . . . . . . . . . . . . . . . . . Graph of A ¼ x þ 2 þ 2x and M ¼ 2x þ 2 . . . . . . . . . . Graph of A ¼ 2w2 þ 5w þ 10 and M ¼ 6w2 þ 10w þ 10 . . . . . . . . . . . . . . . . . . . . . . Graph of ATC ¼ 5Q 25 þ 240 25 . Q and MC ¼ 10Q 3 2 15x 600 . . . . . . . . . . . . . . . Graph of f ðxÞ ¼ x Graph of 16 Q3 2:5Q2 100. . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . .

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. . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . .

25 27 29 29 41 41 42 43 43 50 63 65 67 67 68 70 72 73 75 75 78 108

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109 110 128 130

xiii

xiv

Figures

Fig. 7.3 Fig. 7.4 Fig. 7.5 Fig. 7.6 Fig. Fig. Fig. Fig.

Graph of Graph of Graph of Graph of z 100P Fig. 7.11 Graph of Fig. 7.12 Graph of Fig. 7.13 Graph of Fig. 8.1 Demand, Fig. 9.1 Graph of Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig.

7.7 7.8 7.9 7.10

Graph of supply and demand . . . . . . . . . . . Graph of excess demand . . . . . . . . . . . . . . 700 10 ð2P 15Þ0:7 Graph of EDðPÞ ¼ 2Pþ3   Graph of ESðPÞ ¼ 100P0:4 1000 Pþ25 P2 þ5 . . .

9.2 9.3 9.4 10.1 10.2 10.3 11.1 11.2 11.3

............. ............. .............

133 133 134

.............

135

1800Q21 2100Q1 þ 810000 ¼ 700 10Þ 2Pþ3 þ 10 . . . . . . . . . . . . . ¼ 2:43Q41 189Q21 þ 10Q1 þ 3295 . . 0:02P 0:02P 3 2 Q41 0:6

f ðQ1 Þ ð2P f ðQ1 Þ 10P þ 10Pð3Þ f ðPÞ ¼ P ð3Þ 1600 . . . . . . . . . . . . . . . . . . . . . . . f ðPÞ ¼ 5P1:1 þ 10P0:3 110 . . . . . . . . the derivative of profit function . . . . . . the Derivative of Revenue function . . . MR, and MC curves of the monopolist . P3 e0:02Pþ0:1 . . . . . . . . . . . . . . . . . . . . . . . 2

P Graph of e0:02ðPþ0:5Þ ..................... Graph of MC and MR . . . . . . . . . . . . . . . . . Graph of TC and TR curves . . . . . . . . . . . . . Isocost and isoquant . . . . . . . . . . . . . . . . . . 3 Isoquants and expansion path. . . . . . . . . . . Total, average, and marginal product of labor . Graph of yt ¼ 6:709ð 0:17t Þ 1:709 . . . . . . Graph of Kt ¼ ð0:9Þt þ 6 . . . . . . . . . . . . . .  t Graph of Pt ¼ 1 þ 0:35ð1:1Þt 100 . . . . . . .

. . . . . . . . .

. . . . . . . . .

....... ....... .......

137 138 139

. . . . . .

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. . . . . .

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143 143 146 148 161 190

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. . . . . . . . .

191 201 201 212 219 221 232 233 235

Tables

Table 1.1

Table 1.2

Table Table Table Table Table Table Table Table

5.1 9.1 9.2 9.3 11.1 11.2 13.1 13.2

Average annual expenditures and percent changes by major category of all consumer units, Consumer Expenditure Survey, 2008–2011 . . . . . . . . . . . . . . . Percent distribution of total annual expenditures by major category for all consumer units, Consumer Expenditure Survey, 2008–2011 . . . . . . . . . . . . . . . Values of x and y0 . . . . . . . . . . . . . . . . . . . . . . . . . US GDP from 1978 to 2012, revised. . . . . . . . . . . . US population from 2000 to 2012. . . . . . . . . . . . . . National health expenditure 2003–2012 . . . . . . . . . . Answers to questions number 1 and 2 . . . . . . . . . . . Actual and estimated values of M2 . . . . . . . . . . . . . Results of problem # 24 . . . . . . . . . . . . . . . . . . . . Results of part(d) problem # 25 . . . . . . . . . . . . . . .

......

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. . . . . . . . .

7

9 70 186 200 202 230 236 280 281

xv

Chapter 1

Household Expenditure

Chapter 1, 1.2 Exercises 1. A household allocates its $2,000 monthly income to the purchase of three goods. Prices of these goods are $30, $40, and $20 per unit. (a) Write the household monthly budget constraint. (b) If this household purchases 40 units of good three each month, write its budget equation and graph it. What is the slope of the budget line? 2. Assume a household with a monthly income of $5,000. This household allocates its income to the purchasing of food and nonfood products. If the average price of food products is $20 per unit and nonfood items cost $150 per unit (a) Write the household’s budget equation. (b) If this household consumes 100 units of food products, how many units of nonfood items it can buy? (c) Assume that the price of nonfood products increases to $160. Write the new budget equation. (d) If this household wants to purchase food and nonfood items in the same proportion as in part (b), what is the household’s new bundle in part (c)? 3. Assume the household’s income in Problem 2 increases by 5 %. Repeat parts (a)–(d) of problem (2). 4. Assume that due to competition prices of food and nonfood products in Problem 2 decline by 5 % while the household’s income remains the same. Repeat parts (a) and (b) of problem (2). Compare your result with problem (3). 5. A household splits its $4,000 monthly income between necessity and luxury goods. The average price of necessities is $30 per unit and that of luxuries is $100 per unit.

© Atlantis Press and the authors 2015 S. Vali, Principles of Mathematical Economics II, Mathematics Textbooks for Science and Engineering 4, DOI 10.2991/978-94-6239-088-1_1

1

2

1 Household Expenditure

(a) Write the household budget constraint. (b) Determine the household equilibrium bundle if its proportion of necessity and luxury goods purchases is 10 to 1. (c) What is the equation of the household’s income-consumption path? (d) Assume the household income declines by 10 %. What is the household’s new bundle? (e) Assume no loss of income but an inflation rate of 10 %. What is the household’s new bundle? (f) Compare your answers in parts (d) and (e). (g) Assume the original household income increases by 10 % to $4,400. What is the household’s new bundle? (h) Assume no change in income but the price level declines by 10 %. What is the household’s new bundle? (i) Compare your answers in parts (g) and (h). (j) Compare your answers to parts (f) and (i). Are you surprised? Answers to Chapter 1, 1.2 Exercises #1 (a) Denoting the quantities of goods 1, 2, and 3 by Q 1 , Q 2 , and Q 3 , the household budget constraint is 30Q 1 + 40Q 2 + 20Q 3 = 2000 (b) If this household purchases 40 units of good 3 each month, then its budget constraint will be 30Q 1 + 40Q 2 = 2000 − 40 ∗ 20 = 1200 and the budget equation is 40Q 2 = 1200 − 30Q 1

−→

Q 2 = 30 − 0.75Q 1

Figure 1.1 is the graph of the budget equation. The slope of the budget line is −0.75. #2 (a) Denoting the quantities of food and nonfood goods by F and NF, we have 20F + 150NF = 5000 and the equation of the budget line is F = 250 − 7.5NF.

1 Household Expenditure

3

Fig. 1.1 Budget Line

(b) 20 ∗ 100 + 150NF = 5000

−→

150NF = 3000

Then we have NF = 3000/150 = 20 units. The household bundle is then (100F, 20NF). (c) The new budget constraint is 20F + 160NF = 5000 leading to a new equation for the budget line as F = 250 − 8NF (d) From part (b), we have the number of food items that this household purchases as 5 times the number of nonfood items. So the equation of the income-consumption path is F = 5NF. Therefore, we must solve the following system of two equations with two unknowns for the new bundle.  F = 250 − 8NF F = 5NF 250 − 8NF = 5NF −→ 13NF = 250 and NF = 19.23 F = 250 − 8 ∗ 19.23 = 96.15. The household new bundle is (96.15F, 19.23NF) Figure 1.2 shows the budget line and the income-consumption path.

4

1 Household Expenditure

Fig. 1.2 Budget Line and Income-Consumption Path

#3 (a) After 5 % increase in income, the new budget constraint is 20F + 150NF = 5250 leading to the budget equation F = 262.5 − 7.5NF (b) The household bundle is (100F, 21.67NF) (c) The new equation of the budget line is F = 262.5 − 8NF (d) Based on the bundle in part (b), the proportion is now 100 to 21.67, leading to F = 4.615NF. The household new bundle is then (96.02F, 20.81NF). #4 The new prices are $19 for food and $142.5 for nonfood, thus (a) 19F + 142.5NF = 5000 and the equation of the budget line is F = 263.16 − 7.5NF Notice no change in the slope of the budget equation. (b) 19 ∗ 100 + 142.5NF = 5000 −→ 142.5NF = 3100 then we have 3100 = 21.75 units. The household bundle is (100F, 21.75NF). NF = 142.5 Note that this bundle is an improvement over the bundle in part (b) of problem 3.

1 Household Expenditure

5

#5 Let denote quantity of necessities by N and luxuries by L. (a) The budget constraint is 30N + 100L = 4000 leading to the budget equation L = 40 − 0.3N (b) If N = 10L then L = 40 − 0.3(10L)

−→

L = 10 and N = 100 units

(c) The equation of the income-consumption path is L = 0.1N . (d) The new budget equation is L = 36 − 0.3N and after substituting 10L for N in this equation, we find the new bundle as (L = 9, N = 90) units. (e) With a 10 % inflation rate, the prices of N and L would be $33 and $110, respectively. The new budget constraint is 33N + 110L = 4000 leading to the budget equation L = 36.364 − 0.3N Solving this equation with N = 10L leads to the new bundle (L = 9.091, N = 90.91). (f) Comparing this bundle with the bundle in part (d) indicates that if the incomeconsumption path is linear, the negative impact of inflation on the real standard of living of the household is less than that of a comparable loss of income. (g) Now the budget constraint is 30N + 100L = 4400

−→

L = 44 − 0.3N

Assuming N = 10L, the household optimal bundle is (L = 11, N = 110) (h) This time the budget constraint is 27N + 90L = 4000

−→

L = 44.444 − 0.3N

Again, with N = 10L the household optimal bundle is (L = 11.11, N = 111.1) units.

6

1 Household Expenditure

(i) Clearly the bundle in part (h) is better than the bundle in part (g). The implication is that under the assumption of a linear income-consumption path, the favorable impact of a p% price decline is greater than that of a p% rise in income. (j) The results indicate that when the income-consumption path is linear, inflation is preferable to loss of income and deflation to rise in income. In short, change in prices is preferable to change in income. Supplementary Discussion At the time of this writing (Dec. 2013), the latest available Consumer Expenditure Survey Report is for year 2011.1 Table 1.1 of the survey is reproduced on the next page. As the table shows since the great recession of 2008 the housing expenditure (by far the largest item on an average household’s budget) among households who own their homes has declined, and instead, rental expenditure as part of total housing expenses has grown. The impact of large scale foreclosure and high unemployment finally reversed the upward trend of homeownership in the United State. From the peak of 67 % in 2007, the proportion of homeowners dropped to 66 % in 2008 and further to 64 % in 2012. This process naturally resulted in higher demand for rental housing, leading to higher rents. According to a report released (December 2013) by the Joint Center for Housing Studies of Harvard University, “in 1960, about one in four renters paid more than 30 % of income for housing. Today, one in two are cost-burdened”. “Cost-burdened” refers to the households that pay more than 30 % of their income for housing. Those who pay more than half of their incomes for housing are “severely cost-burdened”. The Study reports that “by 2011, 28 % of renters paid more than half of their incomes for housing, bringing the number with severe cost burdens up by 2.5 million in just four years, to 11.3 million”. The next table (Table 1.2) shows the proportion of household income allocated to 14 major categories of expenditure. As this table indicates, over at least a short period of time, the households allocation of income to the major categories at aggregate level are relatively stable. This, to some extent, justifies the assumption made in this chapter that households, on the average, consume goods and services in the same proportions. Chapter 1 Supplementary Exercises 1. A household allocates its $3,000 monthly income to the purchase of 4 goods with prices $25, $50, $20, and $35 per unit. (a) Write the household monthly budget constraint. (b) If this household purchases 40 units of good 3 and 20 units of good 1 each month, write its budget equation and graph it. What is the slope of the budget line? 1

Tables for 2012 survey are available, but the annual report has not yet been issued.

1 Household Expenditure

7

Table 1.1 Average annual expenditures and percent changes by major category of all consumer units, Consumer Expenditure Survey, 2008–2011 Item 2008 2009 2010 2011 (%) Change 2008–09 2009–10 2010–11 Number of consumer 120770 units (in thousands) Consumer unit characteristics Income before taxes $63,563 Age of reference person 49.1 Average number in consumer unit Persons 2.5 Children under 18 0.6 Persons 65 or older 0.3 Earners 1.3 Vehicles 2 Percent homeowner 66 Spending categories Average annual $50,486 expenditures Food 6443 Food at home 3744 Cereals and bakery 507 products Meats, poultry, 846 fish, and eggs Dairy products 430 Fruits and vegetables 657 Other food at home 1305 Food away from home 2698 Alcoholic beverages 444 Housing 17109 Shelter 10183 Owned dwellings 6760 Rented dwellings 2724 Other lodging 698 Utilities, fuels, and 3649 public services Household operations 998 Housekeeping supplies 654 Household furnishings 1624 and equipment Apparel and services 1801

120847 121107 122287 · · ·

···

···

$62,857 $62,481 $63,685 −1.1 49.4 49.4 49.7 ···

−0.6 ···

1.9 ···

2.5 0.6 0.3 1.3 2 66

··· ··· ··· ··· ··· ···

··· ··· ··· ··· ··· ···

2.5 0.6 0.3 1.3 1.9 66

2.5 0.6 0.3 1.3 1.9 65

··· ··· ··· ··· ··· ···

$49,067 $48,109 $49,705 −2.8

−2.0

3.3

6372 3753 506

6129 3624 502

6458 3838 531

−1.1 0.2 −0.2

−3.8 −3.4 −0.8

5.4 5.9 5.8

841

784

832

−0.6

−6.8

6.1

406 656 1343 2619 435 16895 10075 6543 2860 672 3645

380 679 1278 2505 412 16557 9812 6277 2900 635 3660

407 715 1353 2620 456 16803 9825 6148 3029 648 3727

−5.6 −0.2 2.9 −2.9 −2.0 −1.3 −1.1 −3.2 5.0 −3.7 −0.1

−6.4 3.5 −4.8 −4.4 −5.3 −2.0 −2.6 −4.1 1.4 −5.5 0.4

7.1 5.3 5.9 4.6 10.7 1.5 0.1 −2.1 4.4 2.0 1.8

1011 659 1506

1007 612 1467

1122 615 1514

1.3 0.8 −7.3

−0.4 −7.1 −2.6

11.4 0.5 3.2

1725

1700

1740

−4.2

−1.4

2.4 (continued)

8

1 Household Expenditure

Table 1.1 (continued) Item Transportation Vehicle purchases (net) outlay) Gasoline and motor oil Other vehicle expenses Public and other transportation Healthcare Entertainment Personal care products and services Reading Education Tobacco products and smoking supplies Miscellaneous Cash contributions Personal insurance and pensions Life and other personal insurance Pensions and social security Source U.S, Bureau of Labor Statistics

2008 2009 2010 2011 (%) Change 2008–09 2009–10 2010–11 8604 7658 7677 8293 −11.0 2755 2657 2588 2669 −3.6

0.2 −2.6

8.0 3.1

2715 2621 513 2976 2835 616 116 1046 317

1986 2536 479 3126 2693 596 110 1068 380

2132 2464 493 3157 2504 582 100 1074 362

2655 −26.9 2454 −3.2 516 −6.6 3313 5.0 2572 −5.0 634 −3.2 115 −5.2 1051 2.1 351 19.9

7.4 −2.8 2.9 1.0 −7.0 −2.3 −9.1 0.6 −4.7

24.5 −0.4 4.7 4.9 2.7 8.9 15 −2.1 −3.0

840 1737 5605 317 5288

816 1723 5471 309 5162

849 1633 5373 318 5054

775 1721 5424 317 5106

4.0 −5.2 −1.8 2.9 −2.1

−8.7 5.4 0.9 −0.3 1.0

−2.9 −0.8 −2.4 −2.5 −2.4

2. Assume a 5-member household with a monthly income of $6,000. This household allocates its income to the purchasing of food and nonfood products. If the average price of food products is $30 per unit and nonfood items cost $120 per unit (a) Write the households budget equation. (b) If this household purchases 100 units of food products each month, how many units of nonfood items it can buy? (c) Assume that the household’s allocation of its monthly budget to nonfood products is 5 times that of its budget allocation for food. What is this household optimal bundle? [Note that here the assumption is not that the household consumption of nonfood products is 5 times of that of food products, but rather the allocation of household income to nonfood items is 5 times of its allocation to food products]. (d) Assume the price of food products increases to $35 per unit. Write the household’s new budget equation. If this household wants to purchase food and nonfood items in the same proportion as in part (c), what is the household’s new bundle? (e) Assume the price of food products increases to $35 per unit. If this household wants to maintain its allocation of budget to purchase of food and nonfood

1 Household Expenditure

9

Table 1.2 Percent distribution of total annual expenditures by major category for all consumer units, Consumer Expenditure Survey, 2008–2011 Spending category 2008 2009 2010 2011 Average annual expenditures Food Food at home Food away from home Alcoholic beverages Housing Shelter Utilities, fuels, and public services Household operations Housekeeping supplies Household furnishings and equipment Apparel and services Transportation Vehicle purchases (net outlay) Gasoline and motor oil Other vehicle expenses Public transportation Healthcare Entertainment Personal care products and services Reading Education Tobacco products and smoking supplies Miscellaneous Cash contributions Personal insurance and pensions Life and other personal insurance Pensions and social security

100.00 12.8 7.4 5.3 0.9 33.9 20.2 7.2 2.0 1.3 3.2 3.6 17.0 5.5 5.4 5.2 1.0 5.9 5.6 1.2 0.2 2.1 0.6 1.7 3.4 11.1 0.6 10.5

100.0 13.0 7.6 5.3 0.9 34.4 20.5 7.4 2.1 1.3 3.1 3.5 15.6 5.4 4.0 5.2 1.0 6.4 5.5 1.2 0.2 2.2 0.8 1.7 3.5 11.2 0.6 10.5

100.0 12.7 7.5 5.2 0.9 34.4 20.4 7.6 2.1 1.3 3.0 3.5 16.0 5.4 4.4 5.1 1.0 6.6 5.2 1.2 0.2 2.2 0.8 1.8 3.4 11.2 0.7 10.5

100.0 13.0 7.7 5.3 0.9 33.8 19.8 7.5 2.3 1.2 3.0 3.5 16.7 5.4 5.3 4.9 1.0 6.7 5.2 1.3 0.2 2.1 0.7 1.6 3.5 10.9 0.6 10.3

Source U.S Bureau of Labor Statistics

items in the same proportion as in part (c), what is the household’s new bundle? 3. Assume the household’s income in Problem 2 increases by 5 %. Redo parts (a)–(e) of problem (2). 4. Assume that due to inflation, prices of food and nonfood products in Problem 2 increase by 5 % while the household’s income remains the same. Repeat parts (a), (b), and (c) of problem (2). Compare your answer to part (c) of this exercise to that of exercise 3.

10

1 Household Expenditure

5. A household allocates its $5,000 monthly income to the purchase of necessity and luxury goods. The average price of necessities is $25 per unit and that of luxuries is $125 per unit. (a) Write the household budget constraint. (b) Determine the household equilibrium bundle if the household’s allocation of its monthly budget to luxuries is 3 times that of its budget allocation for necessities. (c) What is the equation of household income-consumption path? (d) Assume the household’s income declines by 10 % and the household decides to reduce its allocation to luxuries by 10 %. What is the household’s new bundle? (e) Assume no loss of income but an inflation rate of 10 %. What is the household’s new bundle if the household decides to cut its budget allocation to luxuries by 10 % ? (f) Compare your answers in parts (d) and (e). (g) Assume the original household income increases by 10 % to $5,500 and the household decides to increase its allocation to luxuries by 10 %. What is the household’s new bundle? (h) Assume no change in income but the price level declines by 10 %. What is the household’s new bundle? (i) Compare your answers in parts (g) and (h). 6. A household allocates its $3,500 monthly income to food, housing, and medical care. Assume the household’s budget allocation to housing is twice the sum of its allocation to food and medical care. Also assume that the household budget allocation to food is twice its allocation to medical care. If the unit prices of food, housing, and medical care are $20, $30, and $50 respectively, determine the household optimal bundle. Exercises (Appendix A) 1. Write the following sums in sigma notation. (a) (b) (c) (d) (e) (f) (g)

1 + 2 + 3 + 4 + · · · + 20 a + a2 + a3 + · · · + am 1/2 1/4 +√ 1/8 + · · · +√1/64 √ +√ 5 + 6 + 7 + · · · + 25 1 + 1/3 + 1/5 + 1/7 + 1/9 + · · · + 1/31 2 + 4 + 6 + 8 + · · · + 100 (the sum of even numbers from 2 to 100) 1 + 4 + 9 + 16 + 25 + · · · + 100

1 Household Expenditure

11

2. Expand and find the sum.

(a)

5  j=2

( j + 1)2

3 

(c)

5 

(b)

i2

i=1

2 j−1

(d)

3  3k − 2 k=0

j=−2

3k + 2

3. Show that (a)

n  i=1

(b)

(ai + bi )2 =

n  i=1

(xi − yi )2 =

n 

ai2 +

n 

bi2

n 

xi2 −

n 

yi2

i=1

i=1

i=1

i=1

4. Assume n numbers x1 , x2 , ..., xn have mean x¯ and variance s 2 . Show that if we subtract a constant c from each number, the mean changes to x¯ − c but the variance stays the same. 5. Assume that in problem 4 instead of adding a constant c to each number, we multiply the numbers by c. What would be the new mean and variance? n  n(n + 1) . i is 6. Show that the sum of the first n positive integers 2 i=1

[Hint: use the trick that Karl Gauss, a 19th century German mathematician, used to solve this problem when he was about 9 years old. Write the sum twice, first in the usual order and second in the reverse order, and then add both sides of the sums.] Use the result and find (a)

100 

j

(b)

j=1

j=1

7. Find the number n such that

n  i=1

8. True or false? n  j=1

100  ( j + 2)

a j + an+1 + an+2 =

i = 20100

n+2  j=1

aj

12

1 Household Expenditure

9. Evaluate (a)

5 

(−1)k k

4  ( j + 2) j

(b)

j=1

k=1

Answers to problems in Chapter 1, Appendix A #1 (a) (b)

1 + 2 + 3 + 4 + · · · + 20 = a + a2 + a3 + · · · + am =

20 

k

k=1 m 

ai

i=1

(c)

1/2 + 1/4 + 1/8 + · · · + 1/64 =

(d)

√

(e) (f) (g)

5+

√

32  1 2i i=1

25 √  √ √ 6 + 7 + · · · + 25 = k k=5

1 + 1/3 + 1/5 + 1/7 + 1/9 + 1/11 + · · · + 1/31 = 2 + 4 + 6 + 8 + · · · + 100 =

50 

15  i=0

1 2i + 1

2i

i=1

1 + 4 + 9 + 16 + 25 + · · · + 100 =

10 

j2

j=1

#2 (a) 5  ( j +1)2 = (2+1)2 +(3+1)2 +(4+1)2 +(5+1)2 = 9 + 16 + 25 + 36 = 86 j=2

(b) 5  i=1

i 2 = (1)2 + (2)2 + (3)2 + (4)2 + (5)2 = 1 + 4 + 9 + 16 + 25 = 55

1 Household Expenditure

13

(c) 3 

j=−2

2 j−1 = 2−2−1 + 2−1−1 + 20−1 + 21−1 + 22−1 + 23−1 = 2−3 + 2−2 + 2−1 + 20 + 21 + 22 =

1 1 1 + + +0+2+4 8 4 2

= 6.875 (d) 3  3k − 2 k=0

=

3k + 2

−2 1 4 7 + + + = −1 + 0.2 + 0.5 + 0.6364 = 0.3364 2 5 8 11

#3 (a) n  i=1

(ai +bi )2 =

n n n n n n       bi2 ai2 + ai bi  = bi2 +2 ai2 + (ai2 +bi2 +2ai bi ) = i=1

i=1

i=1

i=1

i=1

i=1

(b) n n n n     (xi2 + yi2 − 2xi yi ) = (xi − yi )2 = xi2 − yi2 i=1

i=1

i=1

i=1

#4 After subtracting c from each observation, the new set is x1 − c, x2 − c, . . . , xn − c. The mean of this new set, x¯new , is

x¯new =

=

n  (xi − c) i=1

n

n  i=1

n

=

n  i=1

xi −

xi −

cn = x¯ − c n

n

n  i=1

c =

n  i=1

xi − nc n

14

1 Household Expenditure

The new variance is

2 snew =

n  [(xi − c) − (x¯ − c)]2 i=1

n−1

=

n  (xi − c − x¯ + c)2 i=1

n−1

= s2

There is no change in the variance. #5 After multiplying each observation by a constant c, the new set is cx1 , cx2 , . . . , cxn . The mean of this new set, x¯new , is

x¯new =

n  (cxi ) i=1

n

c =

n 

xi

i=1

= c x¯

n

The new mean is c times the original mean. The new variance is

2 snew =

n  [(cxi ) − (c x)] ¯ 2 i=1

n−1

=

n  [c(xi − x)] ¯ 2 i=1

n−1

= c2 s 2

The new variance is c2 times the original variance. n #6 Let’s denote the sum of first n integers i=1 i by S, that is S = 1 + 2 + 3 + · · · + (n − 2) + (n − 1) + n

(1.1SM)

We get exactly the same sum by adding the numbers in the reverse order S = n + (n − 1) + (n − 2) + · · · + 3 + 2 + 1

(1.2SM)

By adding both sides of (1.1SM) and (1.2SM) we get 2S = 1+n +[2+(n −1)]+[3+(n −2)]+· · ·+[(n −2)+3]+[(n −1)+2]+n +1 On the right hand side of this equation we have n terms which are all n + 1, thus 2S = n(n + 1)

−→

S=

n(n + 1) 2

1 Household Expenditure

15

Using this result, we have (a) 100  j=1

j=

100(100 + 1) = 5050 2

(b) 100 100 100    2 = 5050 + 200 = 5250 j+ ( j + 2) = j=1

j=1

j=1

#7 Using result from Exercise #6, we have n  i=1

n(n + 1) = 20100 2

i=

−→

n 2 + n = 40200

By solving the quadratic equation n 2 + n − 40200 = 0 we determine n = 201. #8 True #9 (a) 5  (−1)k k = (−1)1 1 + (−1)2 2 + (−1)3 3 + (−1)4 4 + (−1)5 5 k=1

= −1 + 2 − 3 + 4 − 5 = −3

(b) 4  j=1

( j + 2) j = (1 + 2)1 + (2 + 2)2 + (3 + 2)3 + (4 + 2)4 = 3 + 16 + 125 + 1296 = 1440

Chapter 1, Appendix A : Supplementary Exercises 1. Write the sum in sigma notation. (a) (b) (c) (d) (e) (f)

−1 + 2 − 3 + 4 − 5 + 6 − 7 + 8 − 9 + 10 −1 + 2 − 3 + 4 − 5 + · · · + (−1)n n 1 − 2 + 3 − 4 + 5 − 6 + 7 − 8 + 9 − 10 1 − 2 + 3 − 4 + 5 − · · · + (−1)n−1 n 1 2 3 14 + + + ··· 2 3 4 15 n 1 2 + + ··· + 2 3 n+1

16

1 Household Expenditure

2. Find the value of following sums:

(a)

5 

(2k + 2k)

(b)

i(4i 2 − 3)

(e)

k=1

(d)

4  i=1

(g)

n   1 i=1

(j)

i

−

1  i +1

n  (4k − 4k + 1 ) k=1

2 

j=−2 n  i=1

(h)

(ai − ai−1 )

5  k=1

(k)

(2 + 2 j)2

(4k − 4k−1 )

n  i=1

(n i − n i−1 )

(c)

50 

(−1) j

j=1

(f)

5   1

m=1

(i)

m

−

1  m +1

n  (4k − 4k−1 ) k=1

(l)

4  4 2i−1 i=1

[Series (e), (f), (g), (h), (i), (j), and (k) are called telescoping series, since the terms of the series, other than the first and last terms, cancel out in pairs.]

Chapter 2

Variables, A Short Taxonomy

Chapter 2, 2.1 Exercises 1. Classify the following as quantitative or qualitative variables. (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l) (m) (n) (o) (p)

Height Ethnicity Number of passengers on a bus Weight Language Religion Party affiliation US government budget deficit since 2002 US trade deficit with China since 2002 Time takes to run 100 meters Record of the top 100 runners in 2010 New York Marathon Consumer credit outstanding Last quarter corporate profits Closing price of stocks listed on NYSE on January 28, 2008 Population of major cities in the US based on the 2000 census Unemployment rate of 27 members of the European Union in the last 10 years (q) Unemployment rate of 27 members of European Union in 2007 (r) Closing price of stocks listed on NYSE from Jan. 2, 1990 to Jan. 2, 2008 (s) Record of the top 100 runners in New York Marathon in the last 5 years

2. Classify the qualitative variables in Exercise 1 as dichotomous or multinomial. 3. Classify the quantitative variables in Exercise 1 as discrete or continuous. 4. Classify the economic variables in Exercise 1 as time-series, cross-section, or mixed. 5. Classify k and s in Problem 1 as time-series, cross-section, or mixed. © Atlantis Press and the authors 2015 S. Vali, Principles of Mathematical Economics II, Mathematics Textbooks for Science and Engineering 4, DOI 10.2991/978-94-6239-088-1_2

17

18

2 Variables, A Short Taxonomy

6. Classify the economic variables in Exercise 1 as stock or flow. 7. Determine the size of the table—the number of rows, columns, and cells—you would need in order to present each of the variables that you identified as mixed in Exercise 1 (for (r) assume 3500 stocks listed on NYSE and 252 trading days). 8. How would you present the unemployment rate of the white and non-white populations in the last 20 years for 50 states of the United States? Answers to Chapter 2, Exercises 2.1 #1 Variables in (b), (e), (f), and (g) are qualitative. The rest are quantitative. #2 All qualitative variables in exercise #1 are multinomial. #3 The quantitative variables in (c) and (o) are discrete. The rest are continuous. #4 Variables in (h), (i), and (l) are time series. Variables in (n), (o), and (q) are cross section. Variables in (p), and (r) are mixed. Depending how one interprets the question, the variable “corporate profit” in (m) could be cross section or time series. #5 The variable in (k) is cross section and in (s) is mixed. #6 Variables in (h), (i), and (m) are flows. Variables in (l), (n), (o), (p), (q), and (r) are stocks. #7 For (p) we need a table of 27 by 10: 27 rows for 27 members of the EU and 10 columns for 10 years. This table has 27 * 10 = 270 cells. If we designate the variable EU Unemployment rate as EUU, it can be expressed as EUUi j

i = 1, 2, . . . , 27; j = 1, 2, . . . , 10

For (r) we need a table with 3500 rows for 3500 corporations listed on NYSE and 252 * 18 = 4,536 columns for 252 trading days over an 18-year period. This table has 3500 * 4536 = 15,876,000 cells. The variable Closing Price of Stock CPS can be expressed as CPSi j

i = 1, 2, . . . , 3500; j = 1, 2, . . . , 4536

For part (s) we need a table of 100 rows and 5 columns. This table has 500 cells. #8 Here the unemployment rate can be expressed as Uratei jk

i = 1, 2, . . . , 50; j = 1, 2, . . . , 20; k = 1, 2

where i subscript refers to “state”, j to “time”, and k to “race”, classified as White and non-White. To present this data, we need a 3-dimensional table. Those familiar with a high-level computer programming language, such as FORTRAN and C, know how to use a 3-dimensional “array” to store the data in a computer. For a paper presentation, however, we need two separate tables, one for White and one for non-White. Each table has 50 rows for States and 20 columns for years, a total of 1,000 cells.

2 Variables, A Short Taxonomy

19

Chapter 2 Supplementary Exercises 1. Classify the following as quantitative or qualitative variables. (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l) (m) (n) (o) (p) (q) (r) (s) (t) (u) (v) (w)

Gender Brand of soft drink Race (classified as white or non-white) US trade deficit with China Price index Price of Gold Exchange rate between $ and e Military rank The US money supply Energy use per e1 of GDP of EU members Farm productivity in developing countries Average manufacturing unit labor costs of G7 countries Federal grant to 50 states in 2012 Real private fixed investment Number of spectators in a 90,000 seats stadium Population of NAFTA countries in the last 20 years Money supply of OECD countries in 2012 Inflation rate of Latin American countries for the last 10 years Closing prices of stocks of companies in S&P 500 index Employment status of a person in the work force Religion (classified as Christian or non-Christian) Ethnicity (classified as Italian or non-Italian) Inflation rate of US major metropolitan areas in 2012

2. Classify the qualitative variables in Exercise 1 as dichotomous or multinomial. 3. Classify the quantitative variables in Exercise 1 as discrete or continuous. 4. Classify the economic variables in Exercise 1 as time-series, cross-section, or mixed. 5. Classify the economic variables in Exercise 1 as stock or flow. 6. Determine the size of the table—number of rows, columns, and cells—you would need in order to present each of the variables that you identified as mixed in Exercise 1. 7. How would you present the age distribution of the white and non-white populations in the last 20 years for 50 states of the United States?

Chapter 3

Sets and Functions

Chapter 3, 3.7 Exercises 1. Consider the sets of real numbers A = {1, 5, 6, 11} B = {−3, 5, 7, 11, 18, 25} C = {−4, 1, 15, 19} Find the following: (a) A ∪ B (e) A ∪ B ∪ C

(b) A ∩ B (f) A ∩ B ∩ C

(c) A ∪ C (g) (A ∪ B) ∩ C

(d) C ∪ B (h) A ∩ C

2. Write the complement of A from Exercise 1 using specification. 3. Write the set of all subsets of C in Exercise 1. 4. Given A = {a, b, c, d}, B = {c, d}, and whether the following statements are valid:

C = {a, b, e, f, c, d},

verify

(a) B is a subset of A. (b) A is a subset of C. (c) Since B is a subset of A and A is a of subset C, then B must be a subset of C. (d) C ∩ A = B (e) B ∪ A = A (f) B ∩ A = B (g) C ⊂ A (h) C ∪ A = B 5. Assume x and y are real numbers. Which of the following is not a function? 

x2 + 1

(a) y = x + 2

(b) y =

(c) y = −x 2 + 10

(d) y = x + 1

© Atlantis Press and the authors 2015 S. Vali, Principles of Mathematical Economics II, Mathematics Textbooks for Science and Engineering 4, DOI 10.2991/978-94-6239-088-1_3

21

22

3 Sets and Functions

6. In general, A × B = B × A. Under what condition they are equal? 7. Find the Cartesian product of X = {3, 4, 9} and Y = {9, 16, 81}. Write the function that maps X to Y . 8. Assume A is the set of names of 100 faculty members at a small college and B is the set of academic ranks a faculty member could have: B = {Professor, Associate Professor, Assistant Professor, Lecturer} Is a mapping from A to B a function? 9. Determine the range of the following functions for the given domain: (a) y = x + 2

D = {x ∈ R | x = 2}

(b) y = x + 3x + 5

D = {x ∈ R | 5 ≤ x ≤ 10}

(c) y = x + 5  (d) w = x 2 + 25 √ (e) f (z) = z + 5 w+2 (f) f (w) = √ w 2 − 16

D = {x ∈ R | − 4 ≤ x ≤ 10}

2

2

D = {x ∈ R | − 4 ≤ x ≤ 4} D = {z ∈ R | z > −3} D = {w ∈ R | w ≤ −6 or w ≥ 6}

10. Assume x, y, w, and z are real numbers. Determine the domain of the following functions: (a) y = (c) y =

√ 

x +5

(b) y = √

x 2 − 25

(d) y = √

1 2

(e) z = (x − 3) w+2 (g) f (w) = √ w 2 − 16

1 x +5 1

x 2 − 25  (f) f (x) = x 2 − 9x + 20 (h) y = (x 2 − 2x + 10)

−1 2

11. Given f (x) = x 2 − 2 and g(x) =

2x + 1 3x − 5

find the following: (a) f (3)

(b) g(3)

(e) f (x + 2)

(f) g(x + 2)

(i) f [ f (x)]

(j) g[g(x)]

(c) f (−3) √ (g) f ( x) (k) f [g(x)]

(d) g(−3) (h) g(x 2 + 1) (l) g[ f (x)]

3 Sets and Functions

23

12. Is g( 53 ) in Exercise 11 defined? 13. Enumerate the elements of the following sets: (a) (b) (c) (d) (e)

{x {a {y {x {x

∈ ∈ ∈ ∈ ∈

Z | x < 5} ∩ {x ∈ Z | x > 0} Z | a ≥ 10} ∩ {a ∈ Z | a < 15} N | 1 < y < 9} Z | |x| √ < 5} N | x < 10}

14. Describe the following sets: (a) (b) (c) (d)

{y ∈ N | 1 < y < 9} {y ∈ R | − 4 ≤ y < 9} {w ∈ R | w > 0} {x ∈ R | |x| > 10}

15. Write the following sets using specification: (a) (b) (c) (d) (e) (f)

Set of integers less than 10 Set of positive integer numbers Set of real numbers between −15 and 20 Set of real numbers square of which is greater than 50 Set of real numbers with square roots less than 200 Set of natural numbers with square roots less than 10

16. Graph the functions (a) y = 3x + 2

(b) y = 10 − 2x

(c) y = −5 + 3x

Find the coordinate of point of intersection between (a) and (b). Why can’t we do the same for (a) and (c)? Answers to Chapter 3, 3.7 Exercises #1 Given A = {1, 5, 6, 11} B = {−3, 5, 7, 11, 18, 25} C = {−4, 1, 15, 19} (a) A ∪ B = {−3, 1, 5, 6, 7, 11, 18, 25}

(c) A ∪ C = {−4, 1, 5, 6, 11, 15, 19}

(b) A ∩ B = {5, 11}

(d) C ∪ B = {−4, −3, 1, 5, 7, 11, 15, 18, 19, 25}

(e) A ∪ B ∪ C = {−4, −3, 1, 5, 6, 7, 11, 15, 18, 19, 25} (g) (A ∪ B) ∩ C = {1}

(h) A ∩ C = {1}

(f) A ∩ B ∩ C = ∅

24

3 Sets and Functions

#2 A¯ = {a ∈ R | a ∈ / A} #3 Denoting the power set of C by P, the set of all possible subsets of C is P = {∅, {−4}, {1}, {15}, {19}, {−4, 1}, {−4, 15}, {−4, 19}, {1, 15}, {1, 19}, {15, 19}, {−4, 1, 15}, {−4, 1, 19}, {−4, 15, 19}, {1, 15, 19}, {−4, 1, 15, 19}} #4 Parts (d) and (h) are false, and the rest are true. √ #5 Only (c) y = x 2 + 1 is not a function; for each value of x there are two values for y. If, for example, x = 3 then y = ±3.162. #6 The Cartesian product of two sets A and B is equal to the Cartesian product of B and A if and only if A and B are equal. #7 The Cartesian product of X and Y is X × Y = {3, 4, 9} × {9, 16, 81}

= {(3, 9), (3, 16), (3, 81), (4, 9), (4, 16), (4, 81), (9, 9), (9, 16), (9, 81)}

A subset of X × Y is {(3, 9), (4, 16), (9, 81)} and a function that maps X to Y is y = x 2. #8 A mapping from A to B is a function. #9 To avoid confusion between R as range and R as the set of real numbers, I will use R A to denote range (a) R A = {y ∈ R | y = 4}

(b) R A = {y ∈ R | 45 ≤ y ≤ 135} (c) R A = {y ∈ R | 5 ≤ y ≤ 105}

(d) R A = {w ∈ R | 5 ≤ w ≤ 6.403} considering only positive roots Figure 3.1 is the graph of the function w = (e) R A = { f (z) ∈ R | f (z) >

√

√

x 2 + 25 in part (d).

2} considering only positive roots

Considering only the positive square roots of real numbers, we have the following range for part (f) (f) R A = { f (w) ∈ R | − 1 ≤ f (w) ≤ −0.866 or 1 ≤ f (w) ≤ 1.7889} Figures 3.2 and 3.3 depict the behavior of the function for values of −20 ≤ w ≤ −6 and 6 ≤ w ≤ 15.

3 Sets and Functions

25

Fig. 3.1 The graph of the positive √ roots of w = x 2 + 25

Fig. 3.2 Graph of w+2 f (w) = √ , drawn w 2 − 16 for −20 ≤ w ≤ −6

Fig. 3.3 Graph of w+2 f (w) = √ , drawn w 2 − 16 for 6 ≤ w ≤ 15

#10 (a) D = {x ∈ R | x ≥ −5}

(c) D = {x ∈ R | x ≥ 5 or x ≤ −5}

(d) D = {x ∈ R | x > 5 or x < −5} (e) D = {x ∈ R | x ≥ 3}

(b) D = {x ∈ R | x > −5}

26

3 Sets and Functions

To determine the domain of function in part (f), we first solve the quadratic equation x 2 − 9x + 20 = 0. This equation has two roots, namely x = 4 and x = 5. The values of x 2 − 9x + 20 for x between 4 and 5 are negative and the function f (x) has no real value. Therefore, the domain of f (x) must be expressed as (f) D = {x ∈ R | x ≤ 4 or x ≥ 5}

(g) D = {w ∈ R | w > 4 or w < −4}

In part (h), since x 2 − 2x + 10 for all values of x is positive, then the domain of the 1 is simply function y = √ 2 x − 2x + 10 (h) D = {x ∈ R} #11 (a) f (3) = 7

(b) g(3) = 1.75

(e) f (x + 2) = (x 2 + 2)2 − 2 = x 2 + 4x + 2

(d) g(−3) = 0.357

2(x + 2) + 1 2x + 5 = 3(x + 2) − 5 3x + 1

(f) g(x + 2) =

(c) f (−3) = 7

√ √ (g) f ( x) = ( x)2 − 2 = x − 2

2x 2 + 3 2(x 2 + 1) + 1 = 3(x 2 + 1) − 5 3x 2 − 2

(h) g(x 2 + 1) =

(i) f [ f (x)] = (x 2 − 2)2 − 2 = x 4 + 4 − 4x 2 − 2 = x 4 − 4x 2 + 2 2 (j) g[g(x)] =

3





2x+1 3x−5 2x+1 3x−5





+1 −5 2

=

7x − 3 −9x + 28 4x 2 + 1 + 4x −14x 2 + 64x − 49 −2= 2 9x + 25 − 30x 9x 2 + 25 − 30x

(k) f [g(x)] =



(l) g[ f (x)] =

2x 2 − 3 2(x 2 − 2) + 1 = 3(x 2 − 2) − 5 3x 2 − 11

2x + 1 3x − 5

−2=

3 #12 In exercise 11, g( ) is not defined, since the denominator of the function 5 becomes 0.

3 Sets and Functions

27

#13 (a) X = {1, 2, 3, 4}

(b) A = {10, 11, 12, 13, 14}

(d) X = {−4, −3, −2, −1, 0, 1, 2, 3, 4}

(c) Y = {2, 3, 4, 5, 6, 7, 8}

(e) X = {1, 4, 9, 16, 25, 36, 49, 64, 81}

#14 (a) (b) (c) (d)

Counting numbers greater than 1 and less than 9. Real numbers greater than or equal to −4 and less than 9. Real numbers greater than 0, or positive real numbers. Real numbers greater than −10 and less than 10.

#15 (a) X = {x ∈ Z | x < 10}

(b) Y = {y ∈ Z | y > 0}

(c) A = {a ∈ R | − 15 < a < 20}

(d) B = {b ∈ R | b2 > 50}

(e) X = {x ∈ R |

√

x < 200}

(f) C = {c ∈ N |

√

c < 10}

#16 Figure 3.4 shows the graph of 3 functions. We find the coordinates of the point of intersection between (a) and (b) by solving 3x + 2 = 10 − 2x

−→

5x = 12 and x = 2.4

y = 3(2.4) + 2 = 9.2 We can’t do the same for (a) and (c) because these two lines are parallel. They have the same slope, 3.

Fig. 3.4 Graph of y = 3x + 2, y = 10 − 2x, and y = −5 + 3x

28

3 Sets and Functions

Chapter 3 Supplementary Exercises 1. Consider the following subsets of the lower-case English alphabet A = {a, b, d}

B = {c, d, e, h}

C = {d, h, v, w, z}

(b) A ∩ B

(c) A ∪ C

Find the following: (a) A ∪ B (e) A ∪ B ∪ C

(f) A ∩ B ∩ C

(g) (A ∪ B) ∩ C

(d) C ∪ B (h) A ∩ C

2. Find the cardinality of the sets in Exercise 1. 3. Use A, B and C in Exercise 1 and find the following Cartesian products: (a) A × B

(b) A × C

(c) B × C

(d) B × A

(e) C × A

(f) C × B

4. Use A, B and C in Exercise 1 and show that (A ∪ B) × C = (A × C) ∪ (B × C) (A ∩ B) × C = (A × C) ∩ (B × C) 5. Write the power set of B in Exercise 1. What is its cardinality? 6. Express the following set using enumeration (a) (b) (c) (d)

Set of positive even numbers less than 20 Set of prime numbers between 10 and 50 Set of companies on the Dow Jones Utility Index (Google the Dow Jones Index) Set of US branches of government

7. Express the following sets using specification: (a) (b) (c) (d) (e)

Set of real numbers between and including −20 and 100. Set of real negative numbers with absolute values greater than 200. Set of integer numbers whose squares are greater than 100 and less than 5000. Set of real numbers with inverse greater than 10. Set of real numbers with cubic roots less than 250.

8. Describe the following sets: (a) {a ∈ R | − 5 < a < 20} (b) {z ∈ R | 5 ≤ z < 10} (c) {w ∈ Z | w > 0}

3 Sets and Functions

29

Fig. 3.5 Graph of f (x)

Fig. 3.6 Graphs of f (x) and g(x)

(d) {x ∈ R | |x| > 25} (e) {y ∈ R | 5 < y 2 < 250} 9. Figure 3.5 depicts the graphs of a function f (x). (a) (b) (c) (d)

What is, approximately, the value of f (−2)? What is your estimate of the value of f (2)? What is your estimate of the value of f (0)? Approximately, at what value of x does this function reaches its local maximum? What is the maximum value of the function?

30

3 Sets and Functions

(e) Approximately, at what value of x, f (x) is minimum? (f) Approximately, at what value of x this function is zero? (g) State the domain and range of the function f (x). 10. Figure 3.6 depicts the graphs of f (x) and g(x). (a) (b) (c) (d) (e)

What are, approximately, the values of f (2) and g(2)? For what values of x, f (x) = g(x)? State the domain and range of f (x) and g(x). At what values of x is f (x) less than g(x)? At what values of x is f (x) more than g(x)?

11. State the domain of the following functions: 1

(a) y = (x − 10) 2

(b) y = √



(d) y = √

(c) y =

x 2 − 64 1

1 x − 10

x +2

x 2 − 64 

(e) z = (y − 3)− 2

(f) f (x) =

w+2 (g) f (w) = √ w 2 − 25

(h) y = (3x 2 + 26x − 40)− 2

x 2 − 6x − 40 1

12. Given f (x) = 2x 2 + 2 and g(x) =

2x − 1 3x + 5

find the following: (a) f (−2)

(b) g(−3)

(e) f (x − 2)

(f) g(x − 2)

(i) f [ f (1)]

(n) f [ f (x + 1)]

(j) g[g(2)]

(o) g[ f (1)]

(c) f (5) √ (g) f ( x) (k) f [g(−1)]

13. If f (x) = 4x − 2 and f [g(x)] = x + 3 what is g(x)? 14. Assume f (x) = 2x 2 − 3x + 1

g(x) =

x −1 x +1

(d) g(2) (h) g(x 2 − 11)

(l) g[ f (x)]

3 Sets and Functions

31

If h  = 0 find the following: (a) f (x + h) (c)

(b) g(x + h)

f (x + h) − f (x) h

(d)

g(x + h) − g(x) h

15. Graph the following functions: (a) y = 2x + 10

(b) y = −2x + 2

(c) y = 2x − 10

(a) Find the coordinates of the intersection point between (a) and (b). (b) Write the equation of a line parallel to (a) and with x-intercept of 6. (c) Write equation of a line perpendicular to (c) at point x = 2.

Chapter 4

Market Equilibrium Model

Chapter 4, 4.4 Exercises 1. A market consists of 5,000 identical households and 100 identical producers. The demand equation for a typical household over a week is given by Q id = 30 − 2P + 0.001PCDI i − 0.028Pc

i = 1, 2, 3, . . . , 5000

And the supply equation for a typical firm over a week is given by Q sj = −50 + 10P − 0.5PL − 0.1PE

j = 1, 2, 3, . . . , 100

(a) Write the market demand and supply equations. (b) Assume a households’ per capita disposable income PCDI is $8,000. Further assume that Pc , PL , and PE are $20, $100, and $80 respectively. Determine the market equilibrium price and quantity. 2. What would be the equilibrium price and quantity in problem (1) if the households’ per capita income increased to $8,500, ceteris paribus? 3. How would the equilibrium solution in problem (1) change if you doubled the number of households and producers? 4. In problem (1) assume that due to inflation, the cost of labor increases by 30 % and price of energy by 40 %. What is the new market equilibrium price and quantity? Measure the impact of the change in prices of labor and energy by comparing the new equilibrium values with values in problem (1) (You will be doing comparative static analysis) 5. In problem (1) (a) What is the weekly consumption of a household? (b) What is the weekly budget allocation of a household for this good? © Atlantis Press and the authors 2015 S. Vali, Principles of Mathematical Economics II, Mathematics Textbooks for Science and Engineering 4, DOI 10.2991/978-94-6239-088-1_4

33

34

4 Market Equilibrium Model

6. In problem (1) assume that producers collectively employ 1000 units of labor and use 500 units of energy per week. Assuming no other production costs, what is the weekly profit of a producer? 7. Consider the following market demand and supply functions for a commodity Q d = 20000 − 400P Q s = −4000 + 800P Determine the equilibrium price and quantity. 8. Assume the demand and supply functions for a good are Q d = 200 − 5P + 0.002 INC 0 Q s = −100 + 8P where INC 0 is the exogenously determined average income. (a) Find the equilibrium price and quantity if the average income is 45,000 dollars. (b) Draw the demand and supply curves. (c) Assume average income rises to $50,000. Write and graph the new demand function. (d) Find the new equilibrium solution. What is the impact of rise in income? 9. Find the equilibrium price and quantity for the following market model (a) Q d = 200 − 40P Q s = −40 + 80P

(b) Q d = 1400 − 60P Q s = −400 + 30P

10. Graph the following demand and supply functions. Determine the equilibrium solution from the graph. Solve the equation to verify your answers. (a) Q d = 20 − 2P Q s = −10 + 4P

(b) Q d = 30 − 2P Q s = −50 + 3P

What is wrong with model (b)? 11. Find the equilibrium solution for the following models: (a) Q d = 30 − P 2 Q s = −2 + 4P

(b) Q d = 95 − 3P 2 Q s = −10 + 6P

{Use the quadratic formula to solve the resultant quadratic equations. This is an example of a non-linear market equilibrium models. We will discuss non-linear models in detail in Chap. 7.}

4 Market Equilibrium Model

35

12. The demand and supply functions of two related goods are given by Q d1 = 30 − 8P1 + 4P2 Q s1 = −60 + 6P1 Q d2 = 200 + 4P1 − 4P2 Q s2 = −40 + 6P2 (a) What is the relationship between good one and good two? (b) Find the equilibrium prices and quantities. 13. The demand and supply function of two related goods are given by Q d1 = 30 − 5P1 + 2P2 Q s1 = −20 + 4P1 Q d2 = 70 + 4P1 − 4P2 Q s2 = −10 + 2P2 Find the equilibrium prices and quantities. 14. Assume that the national income model is specified as: Y = C + I0 + G 0 C = C0 + bY D T = tY (a) Identify the endogenous and exogenous variables and the parameters. (b) Determine the equilibrium national income and the multiplier. (c) What restrictions on the model’s parameters are needed in order for the model to have a meaningful solution? 15. Assume that in problem (14) C0 = 30, b = 0.92, t = 0.12, I0 = 70, and G 0 = 60. (a) Find the equilibrium values of the model’s endogenous variables. (b) What would be the change in national income if I0 changes from 70 to 75? (c) What would be the change in the equilibrium values of Y, C, and T if I0 declines from 70 to 60 while G 0 increases from 60 to 70? 16. Consider an open economy with the following consumption and tax functions: C = 5 + 0.75Y D T = 0.20Y

36

4 Market Equilibrium Model

Assume I0 = 100; G 0 = 110; X 0 = 90 and M0 = 105 (a) (b) (c) (d)

Find the equilibrium output. Is the budget in this economy balanced? Is saving equal to investment? Determine the change in the equilibrium output if the autonomous imports decline by 20.

17. Consider the following model of income determination: Y = C + I + G0 C = C0 + bY D I = I0 + iY T = tY This model includes a new equation expressing investment as a function of income (induced investment) and an autonomous or exogenous component I0 . (a) Solve the model for the equilibrium income. What is the model multiplier? (b) What restrictions on the model’s parameters are needed in order for the model to have a meaningful multiplier? 18. The following is a numerical version of the model in question (17): C = 10 + 0.60Y D I = 20 + 0.20Y T = 0.10Y If G 0 = 48, solve the model for the equilibrium values of all the endogenous variables. 19. An expanded version of the national income determination model in problem (17) is formulated by adding an autonomous or exogenous tax component to the tax function. Y = C + I + G0 C = C0 + bY D I = I0 + iY

I0 > 0

T = T0 + tY

T0 > 0

0 < i < 1 and 0 < t < 1

Solve the model for the equilibrium income. Derive the model’s multipliers.

4 Market Equilibrium Model

37

20. The following is a numerical version of the model in question (19): C = 10 + 0.85Y D I = 20 + 0.20Y T = 30 + 0.15Y (a) If G 0 = 48, solve the model for the equilibrium values of all the endogenous variables. (b) If government gives a tax cut by reducing T0 from 30 to 20, what would be the impact on the equilibrium income, consumption, and investment. (c) What would be the impact on the equilibrium income, consumption, and investment if government reduces the average tax rate t from 15 % (0.15) to 12 %? 21. Consider the following aggregate consumption and investment function for a closed economy: C = 15 + 0.70Y I = 30 + 0.1Y − 7.5 r Here, investment is expressed as a function of income Y and interest rate r (expressed as percentage). If G 0 = 50, write an equation relating Y and r and graph it. This graph is called the IS curve. 22. In problem (21) assume that the rate of interest r is 5.5 % and that the government expenditure is 50. Determine the equilibrium income, consumption and investment. 23. In problem (21) assume that the Federal Reserve in order to stimulate the economy reduces the interest rate to 4.5 %. Determine the new equilibrium income, consumption and investment, and compare them with the results in problem (22). Did the Fed achieve its objective? What is the rate of growth of the economy? 24. Below is a numerical version of the model presented in Example 4: C = 60 + 0.75Y D T = 20 + 0.20Y I = 30 + 0.1Y − 7r Assume that the rate of interest r is 5.0 % and that the government expenditure is 80. Determine the equilibrium income, consumption and investment. What are the multipliers? What is the amount of budget deficit or surplus?

38

4 Market Equilibrium Model

25. In problem (24) assume that the government expenditure is increased to 90 while the Federal Reserve reduces the interest rate to 3.5 %. Determine the new equilibrium income, consumption and investment. Answers to Chapter 4, 4.4 Exercises #1 (a) The market demand is the aggregate of 5000 households demands Qd =

5000 

Q id =

5000 

(30 − 2P + 0.001PCDI i − 0.028Pc )

i=1

i=1

Since households are identical all PCDIs are the same and we have Q d = 150000 − 10000P + 5PCDI − 140Pc

(4.1SM)

Similarly we arrive at the market supply function by aggregating the supply functions of 100 producers Qs =

100  j=1

Q sj =

100 

(−50 + 10P − 0.5PL − 0.1PE )

j=1

Q s = −5000 + 1000P − 50PL − 10PE

(4.2SM)

(b) After substituting 8000, 20, 100, and 80 for PCDI, Pc , PL , and PE in the market demand and supply Eqs. (4.1SM) and (4.2SM), we have Q d = 150000 − 10000P + 5(8000) − 140(20) = 187200 − 10000P Q s = −5000 + 1000P − 50(100) − 10(80) = −10800 + 1000P Invoking the equilibrium condition Q d = Q s , we have 187200 − 10000P = −10800 + 1000P leading to 11000P = 198000 Thus, the equilibrium price is P∗ =

198000 = 18 11000

and the equilibrium quantity is Q ∗ = 7200 units #2 By substituting 8500 for PCDI in Eq. (4.1SM), the new market demand equation is Q d = 150000 − 10000P + 5(8500) − 140(20) = 189700 − 10000P

4 Market Equilibrium Model

39

The new equilibrium price is 189700 − 10000P = −10800 + 1000P −→ 11000P = 200500 P∗ =

200500 = 18.22727 and Q ∗ = 7427.3 units 11000

(We must calculate the equilibrium price to 5 or more decimal places. Otherwise we may get different answers for the equilibrium quantity from the demand and the supply functions.) Comparing this equilibrium price and quantity with that of part (b) in problem 1 shows that increase in income, ceteris paribus, leads to higher market quantity and price (this is the comparative static analysis). Note that graphically increase in PCDI leads to a rightward shift in the demand curve, resulting in a higher market quantity and price. #3 Doubling the number of households and number of producers doubles the size of the market, reflected in doubling the equilibrium quantity, without change in the equilibrium price. Doubling the household and producers changes Eqs. (4.1SM) and (4.2SM) in problem 1 to Q d = 300000 − 20000P + 10PCDI − 280Pc Q s = −10000 + 2000P − 100PL − 20PE And after plugging 8000, 20, 100, and 80 for PCDI, Pc , PL , and PE , we have Q d = 374400 − 20000P Q s = −21600 + 2000P 374400 − 20000P = −21600 + 2000P −→ 22000P = 396000 P∗ =

396000 = 18 but Q ∗ = 374400 − 20000(18) = 14400 22000

#4 An increase in the prices of labor and energy increases the cost of production. With their current “working capital” producers cannot buy the same amount of resources, leading to a cutback in the output of the firm. This reduces the market supply and increases the equilibrium price. Increase in price of labor and energy changes the supply Eq. (4.2SM) in problem 1 to Q s = −5000 + 1000P − 50PL − 10PE = −5000 + 1000P − 50(130) − 10(112) Q s = −12620 + 1000P

40

4 Market Equilibrium Model

The new equilibrium is 187200 − 10000P = −12620 + 1000P −→ 11000P = 199820 P ∗ = 18.16545 and Q ∗ = 5545.45 units Graphically, increase in prices of resources shift the supply curve to the left, leading to a higher price and lower quantity. #5 (a) Given that the weekly market equilibrium quantity is 7,200 units and there are 7200 = 5000 identical households then the weekly consumption of a household is 5000 1.44 units. (b) The weekly budget allocation of a household for this good is 18 ∗ 1.44 = 25.92. #6 The producers collectively sell 7200 units to the market, generating 18 ∗ 7200 = 129600 of revenue. The cost of producing this 7200 units is Labor Cost + Energy Cost

→

100 ∗ 1000 + 80 ∗ 500 = $14,0000

The difference between the total revenue and costs is the producer’s collective loss of $10,400. Each individual producer loss is $104. This is, of course, a shortrun situation. Based on the narrative of perfect competition, in the long-run some of the firms which cannot produce more efficiently leave the market. This creates excess demand in the market that leads to higher prices and elimination of losses. More likely alternative, and again based on the narrative of perfect competition, is that the producers strive for a more efficient use of labor and energy by raising their productivity. Assume, for example, that the firms could utilize more efficient production technology and raise labor and energy productivity by 10 and 20 %, respectively. In that case the production cost is 100 ∗ (1000 − 100) + 80 ∗ (500 − 100) = 122000 and firms generate a profit. #7 Using the equilibrium condition, we have 20000 − 400P = −4000 + 800P −→ 1200P = 24000 P∗ =

24000 = 20 and Q ∗ = 12000 1200

#8 (a) With the average income of $45,000 the demand equation reduces to Q d = 200 − 5P + 0.002 ∗ 45000 = 290 − 5P

4 Market Equilibrium Model

41

290 − 5P = −100 + 8P −→ 13P = 390 −→ P ∗ = 30 and Q ∗ = 140 (b) Figure 4.1 is the graph of supply and demand function. Figure 4.2 is the graph of the same supply and demand functions but in a more familiar format; with price on the vertical axis. (c) The new demand equation is Q d = 200 − 5P = 0.002 ∗ 50000 = 300 − 5P Figure 4.3 is the graph of the supply function and the new demand function.

Fig. 4.1 Graph of supply and demand functions

Fig. 4.2 Graph of supply and demand functions

42

4 Market Equilibrium Model

Fig. 4.3 Shift in the demand function

(d) The new equilibrium is 300−5P = −100+8P −→ 13P = 400 −→ P ∗ = 30.77 and Q ∗ = 146.15 The impact of rise in income is the higher equilibrium price and quantity. Figure 4.3 captures the shift in the demand curve as the result of increase in income. #9 The market equilibrium price and quantity in part (a) is P ∗ = 2 and Q ∗ = 120. For part (b) P ∗ = 20 and Q ∗ = 200. #10 Figures 4.4 and 4.5 depict the supply and demand curves for parts (a) and (b). From Fig. 4.4 we easily read the equilibrium price and quantity as P ∗ = 5 and Q ∗ = 10. We get the same answer for P by solving 20 − 2P = −10 + 4P. Figure 4.5 shows that the demand and supply curves intersect in the third quadrant, leading to a negative market quantity. This model is certainly inconsistent and must be discarded. We get the same result by invoking the equilibrium condition and solving for 30 − 2P = −50 + 3P, which yield the answer P = 16. But substituting for P = 16 in either supply or demand equation leads to Q = −2. #11 (a) Using the equilibrium condition, we have 30 − P 2 = −10 + 4P leading to the quadratic equation P 2 + 4P − 40 = 0 with the acceptable solution P ∗ = 4. The market equilibrium quantity is Q ∗ = 14.

4 Market Equilibrium Model

43

Fig. 4.4 Supply and demand curves for part (a)

Fig. 4.5 Supply and demand curves for part (b)

(b) Using the equilibrium condition, we have 95−3P 2 = −10 + 6P leading to the quadratic equation 3P 2 + 6P − 105 = 0 with the acceptable solution P ∗ = 5. The market equilibrium quantity is Q ∗ = 20.

44

4 Market Equilibrium Model

#12 (a) Good 1 and good 2 are substitute goods. (b) Using the equilibrium condition, we have 30 − 8P1 + 4P2 = −60 + 6P1 −→ 14P1 − 4P2 = 90

(4.3SM)

200 + 4P1 − 4P2 = −40 + 6P2 −→ −4P1 + 10P2 = 240

(4.4SM)

Next we must solve Eqs. (4.3SM) and (4.4SM) simultaneously. If we multiply both sides of Eq. (4.3SM) by 2.5 and add it to the Eq. (4.4SM), we eliminate P2 and reduce the system to one equation with one unknown. 31P1 = 465 −→ P1∗ = 15 By substituting this value of P1∗ in Eq. (4.4SM), we calculate P2 as −4 ∗ 15 + 10P2 = 240 −→ 10P2 = 300 and P2∗ = 30 By plugging values of P1∗ and P2∗ in the supply equations, we have Q ∗1 = 30 and Q ∗2 = 140 #13 This problem is similar to #12. After using the equilibrium condition we end up with the following system of two equations with two unknowns:  9P1 − 2P2 = 50 −4P1 + 6P2 = 80 We eliminate P2 by multiplying both sides of the first equation by 3 and adding it to the second equation. This leads to 23P1 = 230 −→ P1∗ = 10. Subsequently, P2∗ = 20, Q ∗1 = 20, and Q ∗2 = 30. #14 (a) Endogenous variables: Y, C, T ; Exogenous variables: C0 , I0 , G 0 ; Parameters: b, t (b) By substituting Y − T for Y D in the consumption function, we can write Y = C + I0 + G 0 as Y = C0 + b(Y − tY ) + I0 + G 0 = b(1 − t)Y + C0 + I0 + G 0 [1 − b(1 − t)] Y = C0 + I0 + G 0 −→ Y ∗ = The autonomous expenditure multiplier is

1 (C0 + I0 + G 0 ) 1 − b(1 − t)

4 Market Equilibrium Model

m=

45

1 1 = 1 − b(1 − t) 1 − b + bt

(c) To avoid the model generating absurd results the multiplier must be positive. This means that 1 − b + bt > 0 leading to b < 1. We must also have m > 1 therefore 1 − b + bt < 1 −→ bt − b < 0 −→ t < 1 In order to have a meaningful model, we need the restrictions b < 1 and t < 1 on the model’s parameters. These restrictions, of course, make perfect economic sense because consumption and taxes cannot be greater than national output or income. #15 (a) From Exercise 14, we have Y∗ =

1 (30 + 70 + 60) = 840.336 1 − 0.92(1 − 0.12)

T ∗ = 0.12(840.336) = 100.840 and C ∗ = 710.336 billion (b) Given that the multiplier is m=

1 = 5.252 1 − 0.92(1 − 0.12)

then Y = m I0 −→ 5.252 ∗ 5 = 26.26 (c) Y = m(C0 + I0 + G 0 ) = m(C0 + I0 + G 0 ) = m(0 − 10 + 10) = 0 Since a decline in the autonomous investment is offset by an equal increase in government expenditure, then there is no change in the equilibrium values of the endogenous variables. #16 (a) C = 5 + 0.75Y D = 5 + 0.75(Y − T ) = 5 + 0.75(Y − 0.2Y ) = 5 + 0.6Y Y = C + I0 + G 0 + X 0 − M0 = 5 + 0.6Y + 100 + 110 + 90 − 105 = 0.6Y + 200 (1 − 0.6)Y = 200 −→ Y ∗ =

200 = 500 billion 0.4

Note that consumption is C = 5 + 0.6Y = $305 billion and the multiplier is 1 = 2.5 m = 0.4

46

4 Market Equilibrium Model

(b) T = 0.2Y −→ T = 0.2(500) = 100 Since $100 billion tax revenue is less than the government expenditure of $110 billion, the budget in this economy is not balanced. The government has 110 − 100 = 10 budget deficit. (c) Given that S = Y − T − C = 500 − 100 − 305 = 95 and investment is 100, then S = I . The $5 billion difference between saving and investment is partially offset by $15 billion difference between import and export. (d) Y = C + I0 + G 0 + X 0 − M0 = 5 + 0.6Y + 100 + 110 + 90 − 85 = 0.6Y + 220 220 = 550 Y∗ = 0.4 Change in Y is $50 billion. Alternatively, Y = m(−M0 ) = −mM0 Y = −2.5(−20) = 50. This means purchasing more domestic goods stimulates the economy. #17 (a) C = C0 + b(Y − T ) = C0 + b(Y − tY ) = C0 + b(1 − t)Y Y = C0 + b(1 − t)Y + I0 + iY + G 0 = [b(1 − t) + i]Y + (C0 + I0 + G 0 ) {1 − [b(1 − t) + i]}Y = C0 + I0 + G 0 Thus, the equilibrium income is Y∗ =

1 (C0 + I0 + G 0 ) 1 − b(1 − t) − i

and the expenditure multiplier is m=

1 1 − b(1 − t) − i

(b) In order to have a meaningful multiplier we must have m > 1. m > 1 implies that 1 − b(1 − t) − i < 1 −→ −b + bt − i < 0 In Exercise 14 we established that we need b < 1. Therefore, with the given restrictions that 0 < i < 1 and 0 < t < 1, the condition −b + bt − i < 0 always holds. But we need the additional restriction i < 1 − b + bt to make sure that the multiplier is a positive number.

4 Market Equilibrium Model

47

#18 From the model we have C0 = 10, b = 0.60, I0 = 20, i = 0.2, and t = 0.1. With G 0 given as 48, we have Y∗ =

1 1 (C0 + I0 + G 0 ) = (10 + 20 + 48) 1 − b(1 − t) − i 1 − 0.6(1 − 0.1) − 0.2

Y∗ =

1 ∗ 78 = 300, subsequently 0.26

T ∗ = 0.1Y = 30, I ∗ = 20 + 0.2 ∗ 300 = 80 and C ∗ = 10 + 0.6(300 − 30) = 172

To check our answers, we form C ∗ + I ∗ + G 0 = 172 + 80 + 48 = 300 and verify that it is equal to the equilibrium value of Y ∗ . #19 We derive the model C = C0 + b(Y − T ) = C0 + b(Y − T0 − tY ) = C0 + b(1 − t)Y − bT0 Y = C0 +b(1−t)Y −bT0 + I0 +iY +G 0 = [b(1−t)+i]Y +(C0 + I0 +G 0 −bT0 ) {1 − [b(1 − t) + i]}Y = C0 + I0 + G 0 − bT0 Thus, the equilibrium income is Y∗ =

b 1 (C0 + I0 + G 0 ) − T0 1 − b(1 − t) − i 1 − b(1 − t) − i

The expenditure multiplier is the same as m=

1 1 − b(1 − t) − i

The autonomous tax multiplier, m ′ , is m′ = −

b 1 − b(1 − t) − i

#20 The numerical version of the model is C = 10 + 0.85Y D I = 20 + 0.2Y T = 30 + 0.15Y

48

4 Market Equilibrium Model

(a) Y = C + I + G0 and C = 10 + 0.85(Y − T ) = 10 + 0.85(Y − 30 − 0.15Y )

(4.5SM)

C = 10 + 0.7225Y − 25.5 = −15.5 + 0.7225Y With G 0 = 48, we have Y = −15.5 + 0.7225Y + 20 + 0.2Y + 48 = 0.9225Y + 52.5 Y − 0.9225Y = 52.5 −→ Y ∗ =

52.5 = 677.419 0.0775

I ∗ = 20 + 0.2 ∗ 677.419 = 155.484 T ∗ = 30 + 0.15 ∗ 677.419 = 131.613 C ∗ = 10 + 0.85(677.419 − 131.613) = 473.935 (b) Here we have T = 20 + 0.15Y. By substituting for T in Eq. (4.5SM), we have C = 10 + 0.85(Y − 20 − 0.15Y ) = −7 + 0.7225Y Subsequently, Y = −7 + 0.7225Y + 20 + 0.2Y + 48 = 61 + 0.9225Y −→ Y ∗ = 787.097 I ∗ = 177.419,

T ∗ = 138.065, and C ∗ = 561.678

Note that a reduction in autonomous taxes leads to higher income, consumption, investment, and interestingly even higher taxes. The rise in income generates more tax revenue for the government, overcompensating a cut in the autonomous taxes. In this model the autonomous expenditure and the tax multiplier, m and m ′ , as we derived them in exercise number 19, are m=

1 1 1 = = = 12.903 1 − b + bt − i 1 − 0.85 + 0.85 ∗ 0.15 − 0.2 0.0775

m′ = −

b = −bm = −10.968 1 − b + bt − i

4 Market Equilibrium Model

49

Since we have Y = m ′ T0 then Y = −10.968 ∗ (−10) = 109.68 which is exactly the change in Y from part (a) to part (b) 787.097 − 677.419 = 109.678. (c) Now we have T = 30+0.12Y. Again by substituting for T in Eq. (4.5SM), we get C = 10 + 0.85(Y − 30 − 0.12Y ) = −15.5 + 0.748Y Y = C + I + G 0 = −15.5 + 0.748Y + 20 + 0.2Y + 48 = 52.5 + 0.948Y 0.052Y = 52.5 −→ Y ∗ = 1009.615 I ∗ = 20 + 0.2 ∗ 1009.615 = 221.923 T ∗ = 30 + 0.12 ∗ 1009.615 = 151.154 C ∗ = 10 + 0.85(1009.615 − 151.154) = 739.692 It is clear that a tax cut in the form of reducing the average marginal tax rate has a positive impact on the economy; an impact larger than reducing the autonomous taxes. Note, again, that increase in the GDP leads to higher tax revenue for the government, over-compensating the reduction in the tax rate. #21 After substituting for C, I , and G 0 in the national income identity, we have Y = 15 + 0.70Y + 30 + 0.10Y − 7.5r + 50 Y = 95 + 0.80Y − 7.5r −→ 0.2Y = 95 − 7.5r Y = 475 − 37.5r The graph of this function, given in Fig. 4.6, is the IS curve. This curve shows the inverse relationship between national income and the interest rate. A low interest rate stimulates investment in plants, equipment, construction, and part of the consumption which is mostly financed (home buying and purchase of household big ticket items like furniture and home appliances). This leads to higher GDP.

50

4 Market Equilibrium Model

Fig. 4.6 The IS Curve

#22 Using the equation of IS curve from problem #21, we have Y ∗ = 475 − 37.5 ∗ 5.5 = 268.75 Subsequently, I ∗ = 30 + 0.1 ∗ 268.75 − 7.5 ∗ 5.5 = 15.625 and C ∗ = 15 + 0.7 ∗ 268.75 = 203.125 #23 From Exercise #21 we have Y = 475 − 37.5r

→

Y ∗ = 475 − 37.5 ∗ 4.5 = 306.25

I ∗ = 30 + 0.1 ∗ 306.25 − 7.5 ∗ 4.5 = 26.875 and C ∗ = 15 + 0.7 ∗ 306.25 = 229.375 The Federal Reserve success in reducing the interest rate from 5.5 to 4.5 % indeed stimulated the economy and the GDP grew from 268.75 to 306.25, a growth rate of 306.25 − 268.75 ∗ 100 = 13.95 % 268.75 [There is no time frame specified for this problem and one should be hesitant to interpret this rate as annual rate. It is more realistic to assume that the economy grew from 268.75 to 306.25 over several years, say four years. In that case the average

4 Market Equilibrium Model

51

annual rate of growth would be 13.95 ÷ 4 = 3.49 %, a much more realistic number for the US or other Western economies.] #24 In Example 4 of the text we determined the equilibrium value of Y as 1 b k (C0 + I0 + G 0 ) − T0 − r0 1 − b + bt − i 1 − b + bt − i 1 − b + bt − i

Y∗ =

In this problem C0 = 60, b = 0.75, T0 = 20, 0.1, k = 7, r = 5, and G 0 = 80. Therefore Y∗ =

I0 = 30,

i =

1 (60 + 30 + 80) 1 − 0.75 + 0.75 ∗ 0.2 − 0.1 −

Y∗ =

t = 0.2,

7 0.75 (20) − (5) 1 − 0.75 + 0.75 ∗ 0.2 − 0.1 1 − 0.75 + 0.75 ∗ 0.2 − 0.1

1 0.75 7 (170) − (20) − (5) = 400.00 0.3 0.3 0.3

Subsequently, T ∗ = 20 + 0.2 ∗ 400 = 100; I ∗ = 30 + 0.1 ∗ 400 − 7 ∗ 5 = 35; and C ∗ = 60 + 0.75(400 − 100) = 285 Multipliers are m=

1 = 3.333; 0.3

m′ = −

0.75 = −2.5; 0.3

m ′′ = −

7 = −23.333 0.3

#25 C = 60 + 0.75Y D I = 30 + 0.10Y − 7r T = 20 + 0.20Y Given r = 3.5 % and G 0 = 90, we have Y = 60 + 0.75(Y − 20 − 0.2Y ) + 30 + 0.10Y − 7(3.5) + 90 = 140.50 + 0.70Y 0.3Y = 140.5 −→ Y ∗ = 468.333 I ∗ = 52.333, T ∗ = 113.667, C ∗ = 326.0

52

4 Market Equilibrium Model

Chapter 4 Supplementary Exercises 1. Find the equilibrium price and quantity for the following market models: (a) Q d = 350 − 10P

(b) Q d = 14000 − 60P

Q s = −50 + 30P

Q s = −400 + 300P

2. Graph the following demand and supply functions. Determine the equilibrium solution from the graph. Solve the equation to verify your answers. (a) Q d = 200 − 2P

(b) Q d = 3000 − 20P

Q s = −40 + 4P

Q s = −500 + 30P

3. Find the equilibrium solution for the following models: (a) Q d = 300 − 2P 2

(b) Q d = 750 − 5P 2

Q s = −20 + 10P

Q s = −100 + 60P

4. A market consists of 6500 identical households and 200 identical producers. The demand equation for a typical household over a week is given by Q id = 60 − 2P

i = 1, 2, 3, . . . , 6500

And the supply equation for a typical firm over a week is given by Q sj = −270 + 15P

j = 1, 2, 3, . . . , 200

(a) Write the market demand and supply equations. (b) Determine the market equilibrium price and quantity. 5. In problem (4) (a) What is the weekly consumption of a household? (b) What is a household weekly budget allocated to this good? 6. Consider the following model suggested for a commodity Q d = 2000 − 300P Q s = −4000 + 500P Is this model consistent?

4 Market Equilibrium Model

53

7. Assume the daily market demand and supply functions for a good are Q d = 3000 − 6P + 0.002 INC 0 Q s = −1000 + 4P where INC 0 is the exogenously determined average household income. (a) Find the equilibrium price and quantity if the average income is 50,000 dollars. (b) Assume average income rises to $55,000, write and graph the new demand function. (c) Find the new equilibrium solution. What is the impact of rise in income? 8. If market in problem (7) consists of 400 households and 25 producers (a) In part (a), what is the daily consumption of a typical household? (b) In part (a), what is a household daily expenditure on this good? (c) In part (b), what is a typical producer’s share of the market? 9. A market consists of 4000 identical households and 200 identical producers. The demand equation for a typical household over a month is given by Q id = 40 − 2P − 0.028Pc

i = 1, 2, 3, . . . , 4000

And the supply equation for a typical firm over a month is given by Q sj = −50 + 10P − 0.6Pr

j = 1, 2, 3, . . . , 200

where Pr is the average price of all resources used in the production of this good. (a) Write the market demand and supply equations. (b) Assume that Pc and Pr are $20 and $100. Determine the market equilibrium price and quantity. 10. In problem (9) (a) What is the monthly consumption of a household? (b) What is the monthly budget allocation of a household to this good? (c) What is the monthly production of a firm? 11. In problem (9) assume that a typical firm uses 10 units of resources each month (a) What is the monthly cost of production of a firm? (b) What is the monthly profit of a firm?

54

4 Market Equilibrium Model

12. A market consists of 7000 identical households and 300 identical producers. The demand equation for a typical household over a week is given by Q id = 44.293 − 2.5P + 0.001PCDI i + 0.3PS

i = 1, 2, 3, . . . , 7000

And the supply equation for a typical firm over a week is given by Q sj = −40 + 12P − 0.4PL − 0.15PE

j = 1, 2, 3, . . . , 300

where PCDI is the average per capita disposable income, Ps is the price of a substitute good, PL is the price a unit of labor, and PE is the price of a unit of energy. (a) Write the market demand and supply equations. (b) Assume the average household income is $50,000 and the average size of households is 4. Further assume that Ps , PL , and PE are $30, $80, and $70, respectively. Determine the market equilibrium price and quantity. 13. What would be the equilibrium price and quantity in problem (12) if the households’ income increased to $55,000, ceteris paribus? 14. How would the equilibrium solution in problem (12) change if you doubled the number of households and producers? 15. In problem (12) assume that producers collectively employ 12000 units of labor and use 3000 units of energy per week. Assuming no other production costs, what is the weekly profit of a producer? 16. The demand and supply functions of two related goods are given by Q d1 = 40 − 5P1 + 6P2 Q s1 = −50 + 4P1 Q d2 = 120 + 3P1 − 4P2 Q s2 = −30 + 3P2 (a) What is the relationship between good one and good two? (b) Find the equilibrium prices and quantities. 17. The demand and supply functions of two goods are Q d1 = 82 − 3P1 + P2

Q d2 = 92 + 2P1 − 4P2

Q s1 = −5 + 15P1

Q s2 = −6 + 32P2

4 Market Equilibrium Model

55

(a) How these goods are related to each other? (b) Find the equilibrium prices and quantities. 18. The demand and supply functions of two related goods are given by Q d1 = 300 − 50P1 + 20P2 Q s1 = −200 + 40P1 Q d2 = 700 + 40P1 − 40P2 Q s2 = −100 + 20P2 Find the equilibrium prices and quantities. 19. Consider the demand and supply functions of two goods given in the previous problem. Show that if the quantity demanded and supplied increased by a factor n, the equilibrium prices remain unchanged but the market equilibrium quantities will increase by the same factor n. 20. Consider a close economy with the following consumption and tax function C = C0 + bY D T = T0 + tY (a) Identify the endogenous and exogenous variables and the parameters (b) Assume I0 = 60, G 0 = 50, C0 = 55, T0 = 10, b = 0.95, and t = 0.15 (I) Find the equilibrium national income, consumption, and tax (II) What is the MPC? What is the value of multiplier? (III) Is government budget in this economy balanced? If not, what is the amount of deficit or surplus? 21. Assume a closed economy with the following national income model: Y = C + I + G0 C = C0 + bY D T = T0 + tY I = I0 + iY where I0 = 20, G 0 = 120, C0 = 55, T0 = 30, b = 0.90, t = 0.20, and i = 0.08.

56

4 Market Equilibrium Model

(a) (b) (c) (d) (e) (f)

Identify the endogenous and exogenous variables and the parameters. Find the equilibrium national income, consumption, and tax. What is the numerical value of the multiplier? What is the MPS? Is government budget in this economy balanced? What would be the new equilibrium income if the government expenditure changes by 20?

22. Consider a national income model with the following consumption, investment, and tax functions C = 55 + 0.85Y D T = 20 + 0.05Y I = 10 + 0.1Y − 8r where r is given as 6 % and G 0 = 100. (a) Find the equilibrium national income, consumption, investment, and tax. (b) What is the numerical values of the multipliers? (c) Is government budget in this economy balanced? 23. Now assume that the budget hawks in the country depicted in problem (22) pressure the government to implement an austerity plan by cutting the government expenditure by 20 % and raising the marginal tax rate t to 10 %. (a) Find the new equilibrium national income, consumption, investment, and tax. (b) Is government budget in this economy balanced now? 24. The result of the austerity policy is an almost balanced budget, unfortunately accompanied by a big decline in the country’s GDP and rise in unemployment. Under pressure from opponents of austerity, the government decides to follow a more balanced approach by the following combination of monetary and fiscal policies: Cut the interest rate to 4 % and set the marginal tax rate at 8 % and the government expenditure at 90. (a) Find the new equilibrium national income, consumption, investment, and tax. (b) Determine whether the budget is balance or not. 25. What would be the equilibrium national income and the budget in Exercise 24 if while maintaining the tax rate at 8 % and the government expenditure at 90, the monetary authority cut the interest rate to 2 %?

Chapter 5

Rates of Change and the Derivative

Chapter 5, 5.4 Exercises 1. What are the y-intercept(s) and x-intercept(s) of graphs of the following functions? (a) y = 3x − 10 (c) (e)

1 f (x) = 12 + x 3 f (x) = x2 + 5x + 6

(b) y = −5x + 10 1 (d) g(x) = 12 − x 3 2 (f) y = x + x − 6

(g) y = x2 − 8x + 16

(h) y − 3x + 8 = 0

(i) y = −0.005x + 0.01

(j) y = 10

(k) x = −4

(l) y = x

(m) Q = 20 − 0.5P

(n) Q = −10 + 2P

2. Write the equation of the line joining points P(5, 10) and Q(20, 30). 3. A line joins point A(2, 12) to point B(4, 22) and point B to point C(8, 38). Is this line a straight line? 4. Find the point of intersection of the lines with equations y = 3x + 4 y = 5x − 10 5. Write the equation of the line that passes through the point A(4, 6) and is parallel to a line with equation y = 2x − 5 © Atlantis Press and the authors 2015 S. Vali, Principles of Mathematical Economics II, Mathematics Textbooks for Science and Engineering 4, DOI 10.2991/978-94-6239-088-1_5

57

58

5 Rates of Change and the Derivative

6. Use the rules of differentiation to find the derivative of the following functions (a) y = 3x − 10

(b) y = −5x + 10

(c)

f (x) = 12x2 + 1/3x

(d) g(x) = 12 − 1/3x2

(e)

f (x) = 1/3x3 + 5x2 − 6x

(f) y = x2 + x − 6 3x − 5 x2 + 10 3w2 − 2 (j) ϕ(w) = w−1

(g) y = (x2 − 8x + 16)2

(h) y =

(i) y = 10

7. Find the rate of change of the following functions at the given value (a) y = 3x − 10

at x = 5

(b) y = −5x + 10

at x = −10

(c)

f (x) = 12x2 + 1/3x

at x = 3

(d)

f (x) = 1/3x3 + 5x2 − 6x

at x = −2

(e) y = (w2 − 8w + 16)2 3t − 5 t 2 + 10 √ (g) g(x) = 12 x − 1/3x2

(f)

f (t) =

at w = 4 at t = 2 at x = 9

8. Differentiate 3t −4

(a)

f (t) =

(c)

f (x) = ax3 + bx2 + cx + d

t2

(e) y = (3x2 + 2)4 (g) z =

αx − βx2 x−1

(b) y = ax2 + bx + c z2 √ 3+ z  4 (f) y = 2x x2 − 3 (d) g(z) =

1

(h) w = 5u 3 + 2u

9. The total cost function of a steel company is TC(Q) = 0.0003Q 3 − 0.02Q 2 + 0.5Q + 300 What is the firm’s marginal cost at the level of output 50? At the level of output 100?

5 Rates of Change and the Derivative

59

10. A firm has the following total cost function TC(Q) = Q 3 − 24Q 2 + 200Q + 500 (a) What is this firm’s fixed cost? (b) At what level(s) of output does the firm’s marginal cost equal $50? 11. A monopolist faces the following demand function in the market  P = 15500 − 100 Q Find its marginal revenue if it produces 10000th units.

12. A firm’s profit, π , is the difference between its total revenue TR and its total costs TC. If TR = f (Q) and TC = g(Q) write the profit function and find the firm’s marginal profit. What is the economic interpretation of marginal profit? 13. Market demand for a good produced by a monopolist is P=

1500 Q 0.8

What is the monopolist’s marginal revenue from selling the 100th unit? 14. A firm has the following total cost TC function TC = Q 3 − 30Q 2 + 400Q + 500 At what level of output the firm’s marginal cost MC is $100? 15. A monopolistically competitive firm is facing the following demand for its product Q=

1000 P2

What is the firm’s marginal revenue from selling the 100th unit? 16. A firm has the following quadratic cost function TC = 0.025Q 2 + 12Q + 500 Find the coordinates of the point where the marginal cost curve intersects the average total cost curve. Answers to Chapter 5, 5.4 Exercises #1 (a) x = 0

(b) x = 0 (c) x = 0

y = −10;

y = 10; f (x) = 12;

y=0

y=0 f (x) = 0

10 3 x=2 x = −36 x=

60

5 Rates of Change and the Derivative

(d) x = 0 g(x) = 12; g(x) = 0 x = 36 (e) x = 0 f (x) = 6; f (x) = 0 x2 + 5x + 6 = 0. This quadratic equation has two zeros x = −3 and x = −2 (f) x = 0 y = −6; y = 0 x = −3 and x = 2. (g) x = 0 y = 16; y=0 x=4 8 (h) x = 0 y = −8; y=0 x= 3 (i) x = 0 y = 0.01; y=0 x=2 (j) x = 0 y = 10. Graph of this function is a line parallel to the x axis and has no x-intercept. This function has one element in its range, 10. R A = {y ∈ R | y = 10} (k) Graph of this function is line parallel to the y axis. The domain of this function has only one element, −4. D = {x ∈ R | x = −4}. (l) x = 0 y = 0. Graph of this function is a line through the origin, a ray. (m) P = 0 Q = 20; Q = 0 P = 40 (n) P = 0 Q = −10; Q=0 P=5 #2 Denoting the coordinate of P as (x1 , y1 ) and Q as (x2 , y2 ), we have y − y1 =

y2 − y1 (x − x1 ) x2 − x1

y − 10 =

4 4 10 30 − 10 (x − 5) = (x − 5) −→ y = x + 20 − 5 3 3 3

#3 We write equation of the line joining points A and B and the line joining points B and C. If these lines have the same slope, then the line joining all three points is a straight line. Alternatively, we can write equation of the line joining points A and B, then check to see if the coordinate of point C satisfy the equation. Let us use this alternative. The equation of the line joining A(2, 12) and B(4, 22) is y − 12 =

22 − 12 (x − 2) 4−2

→

y − 12 = 5(x − 2)

→

y = 5x + 2

We now test whether point C(8, 38) is a point on this line y = 5 ∗ 8 + 2 = 42 = 38 Therefore point C is not on this line and the line joining these three points is not a straight line. #4 We set 3x + 4 equal to 5x − 10 and solve for x, which is x = 7. From both equation we get y = 25.

#5 The slope of the line parallel to the line y = 2x − 6 is 2. The equation of a line with slope 2 passing through point A(4, 6) is y − 6 = 2(x − 4) −→ y = 2x − 2

5 Rates of Change and the Derivative

61

#6 (a) (c) (e) (g) (h) (i)

dy dy =3 (b) = −5 dx dx 1 d 2 d f (x) = 24x + (d) g(x) = − x dx 3 dx 3 d dy f (x) = x2 + 10x − 6 = 2x + 1 (f) dx dx dy = 2(2x − 8)(x2 − 8x + 16) = 4x3 − 48x2 + 192x − 256 dx 3(x2 + 10) − 2x(3x − 5) dy −3x2 + 10x + 30 = = dx (x2 + 10)2 (x2 + 10)2 d 6w(w − 1) − (3w2 − 1) 3w2 − 6w + 1 dy =0 (j) ϕ(w) = = dx dx (w − 1)2 (w − 1)2

#7 (a) (b) (c) (d) (e)

dy = 3 for all values of x dx dy = −5 for all values of x dx d 1  1 ¯ f (x) = 24x +  = 72 + = 72.33 dx 3 x=3 3  d  = 4 − 20 − 6 = −22 f (x) = x2 + 10x − 6 x=−2 dx  dy  = 2(2w − 8)(w2 − 8w + 16) =0 w=4 dw

42 − 4 19 3(t 2 + 10) − 2t (3t − 5) −→ f ′ (2) = = 2 2 2 (t + 10) 14 98 6 2 (g) g ′ (x) = √ − x −→ g ′ (9) = 2 − 6 = −4 x 3 (f) f ′ (t) =

#8 (a) f ′ (t) =

3(t 2 − 4) − 2t (3t) 3t 2 + 12 = − (t 2 − 4)2 (t 2 − 4)2

(b) y ′ = 2ax + b (c) f ′ (x) = 3ax2 + 2bx + c (d) g ′ (z) =

z2 √ z) − √ 12z z + 3z 2 2 z = √ √ √ (3 + z)2 2 z(3 + z)2

2z(3 +

√

62

5 Rates of Change and the Derivative

(e) y ′ = 4(6x)(3x2 + 2)3 = 24x(3x2 + 2)3  4 (f) y ′ = 2 x2 − 3 +  4

x2

3x2 − 6 =  4 (x2 − 3)3 (x2 − 3)3

(α − 2βx)(x − 1) − (αx − βx2 ) −βx2 + 2βx − α = (x − 1)2 (x − 1)2 5 2 (h) w′ = u − 3 + 2 3

(g) z ′ =

#9 The marginal cost function of the firm is the derivative of its total cost function MC(Q) =

dTC = 0.0009Q 2 − 0.04Q + 0.5 dQ

The marginal costs at the levels of outputs 50 and 100 are MC(50) = 0.0009(502 ) − 0.04 ∗ 50 + 0.5 = 0.75 MC(100) = 0.0009(1002 ) − 0.04 ∗ 100 + 0.5 = 5.5 #10 (a) The firm’s fixed cost is 500. (b) To determine the level(s) of output when the marginal cost of the firm is 50, we must take the derivative of the total cost function and set it equal to 50. MC(Q) =

dTC = 3Q 2 − 48Q + 200 dQ

We must now solve the following quadratic equation 3Q 2 − 48Q + 200 = 50 −→ 3Q 2 − 48Q + 150 = 0 This equation has two real solutions Q = 4.2583 and Q = 11.742. At these two levels of output the firm’s marginal cost is 50 (See Fig. 5.1). #11 The marginal revenue function of the firm is the derivative of its total revenue function. The total revenue function is TR(Q) = PQ. Substituting for P from the demand function, we have   3 TR(Q)=PQ = (15500−100 Q)Q = 15500Q−100Q Q = 15500Q−100Q 2

The marginal revenue function MR(Q) is MR(Q) =

 1 3 dTR = 15500 − (100Q 2 ) = 15500 − 150 Q dQ 2

5 Rates of Change and the Derivative

63

Fig. 5.1 MC curve and Levels of Output where MC = 50

The marginal revenue of the 10000th unit is √ MR(10000) = 15500 − 150 10000 = 15500 − 150 ∗ 100 = 500 #12 A firm’s profit is π = TR − TC. If TR = f (Q) and TC = g(Q), then π(Q) = f (Q) − g(Q) The marginal profit is d d d π(Q) = f (Q) − g(Q) dQ dQ dQ π ′ (Q) = f ′ (Q) − g ′ (Q) = TR′ − TC ′ = MR − MC Then the marginal profit at any output level is the difference between the marginal revenue and the marginal cost at that level. #13 We write the total revenue function as   1500 Q = 1500Q 0.2 TR(Q) = PQ = Q 0.8   300 MR(Q) = TR′ (Q) = 0.2 1500Q 0.2−1 = 0.8 Q The marginal revenue of the 100th unit is MR(100) =

300 = 7.536 1000.8

64

5 Rates of Change and the Derivative

#14 MC = TC ′ = 3Q 2 − 60Q + 400 = 100 −→ 3Q 2 − 60Q + 300 = 0 This quadratic equation has one solution, Q = 10. #15 We must first solve Q = 1000 −→ P = P2 = Q

TR(Q) = PQ =



1000 for P in terms of Q P2



1000 Q



Q=

1000 Q



1000Q

1000 500 MR(Q) = TR′ (Q) = √ =√ 2 1000Q 1000Q The marginal revenue from selling 100th unit is MR(100) = √

500 1000 ∗ 100

= 1.581

#16 We have, MC = TC ′ = 0.05Q + 12 ATC =

0.025Q 2 + 12Q + 500 TC = Q Q

At the point of intersection of the marginal cost and the average total cost MC = ATC −→ 0.05Q + 12 =

0.025Q 2 + 12Q + 500 Q

This leads to 0.05Q 2 + 12Q = 0.025Q 2 + 12Q + 500 −→ 0.025Q 2 = 500 Then Q = 141.421 and MC = ATC = 19.0711 are the coordinates of the point of intersection of the two curves. Figure 5.2 depicts the two curves. As we will see later in the chapter, the marginal cost curve cuts the average cost curve at its lowest (minimum) point.

5 Rates of Change and the Derivative

65

Fig. 5.2 Marginal Cost and Average Total Cost curves

Chapter 5, 5.6 Exercises 1. Find the relative maximum and/or minimum values of the following functions (a) y = 2x2 − 4x + 10 (b) y = −2x2 − 4x + 10 (c) y = x3 − 3x + 10 (d) y = x3 − 9x2 − 48x + 50 2. A baseball player hits a home run. The height, in feet, of the ball t second after it is hit is given by h = −15t 2 + 96t + 4 (a) How high does the ball goes before returning to the ground? (b) How many seconds is the ball in the air? 3. A travel agency offers a one week Caribbean tour for $1,000 per person. If more than 50 passengers join the tour, the price is reduced by $5 per additional passenger. The tour can accommodate 150 passengers. What number of passengers maximizes the travel agency’s total revenue? What price does each passenger pays? 4. A cruise line offers a one week tour of Alaska for $2,400 per person. If more than 200 passengers join the tour, the price is reduced by $3 per additional passenger. The tour can accommodate 1000 passengers. What number of passengers maximizes the cruise line’s total revenue? What price does each passenger pays? 5. Find the relative maximum or minimum of the function y = x(1 − x)4

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5 Rates of Change and the Derivative

Answers to Chapter 5, 5.6 Exercises #1 (a) The first and second order conditions are FOC : SOC :

dy = 4x − 4 = 0 dx d2 y =4>0 dx 2

−→

x=1

Therefore the function y = 2x2 − 4x + 10 has a local minimum at x = 1. At this point the value of the function is y = 8. (b) The first and second order conditions are FOC : SOC :

dy = −4x − 4 = 0 dx d2 y = −4 < 0 dx 2

−→

x = −1

Therefore the function y = −2x2 − 4x + 10 has a local maximum at x = −1. At this point the value of the function is and y = 12. (c) The first and second order conditions are FOC : SOC:

dy = 3x2 − 3 = 0 −→ x2 = 1 −→ x = ±1 dx   d2 y d2 y   = 6x = 6x = 6 > 0 and = −6 < 0   x=1 x=−1 dx 2 dx 2

Therefore the function y = x3 − 3x + 10 has a local minimum at x = 1 and a local maximum at x = −1. The minimum and maximum values of the function are 8 and 12, respectively. (d) The FOC and SOC are FOC : SOC :

y ′ = 3x2 − 18x − 48 = 0 −→ x = −2 x = 8     = −30 < 0 and y ′′  = 30 > 0 y ′′ = 6x − 18  x=−2

x=8

Thus, the function y = x3 − 9x2 − 48x + 50 has a local maximum at x = −2 and a local minimum at x = 8. Figure 5.3 is the graph of this function. #2 (a) We must find maximum of h = −15t 2 + 96t + 4. The FOC is dh = −30t + 96 = 0 −→ t = 3.2 dt

5 Rates of Change and the Derivative

67

Fig. 5.3 Graph of function y = x3 − 9x2 − 48x + 50

The SOC for a maximum is satisfied because d2h dx 2

= −30 < 0

After 3.2 seconds the ball reaches its highest point of h = −15(3.2)2 + 96 ∗ 3.2 + 4 = 157.6 feet. (b) This is the “hang time”. After ball reaches h = 157.6 feet high it starts its descend. Then h becomes smaller and smaller and finally ball hits the ground, i.e. h becomes 0. Then h = −15t 2 + 96t + 4 = 0 −→ t = 6.44 s The ball is 6.44 seconds in the air. Figure 5.4 is the graph of h.

Fig. 5.4 Graph of h = −15t 2 + 96t + 4

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5 Rates of Change and the Derivative

#3 Let denote the number of passengers by N . Then the price charged to passengers can be expressed as  P = 1000 for N ≤ 50 P = 1000 − 5(N − 50) = 1250 − 5N for 50 < N ≤ 150 The total revenue is then  TR = 1000N for N ≤ 50 TR = PN = (1250 − 5N )N = 1250N − 2N 2 for 50 < N ≤ 150 The FOC for a maximum is dTR = 1250 − 10N = 0 −→ N = 125 dN The SOC for a maximum is satisfied since d 2 TR dN 2

= −10 < 0

Therefore 125 passengers each paying P = 1250 − 5 ∗ 125 = $625 generate a maximum of 625 ∗ 125 = $78,125 revenue for the travel agency. Figure 5.5 is the graph of the total revenue function. Note that the segment of the graph for N ≤ 50 is a straight line, depicting the total revenue for 1000N . When N is more than 50, the discount kicks in and the total revenue becomes 1250N − 2N 2 . The curvy part of the graph is the depiction of this part of the total revenue function. Since the travel agency cannot accommodate more 150 passengers, the graph does not go beyond N = 150.

Fig. 5.5 Graph of TR = 1250N − 5N 2

5 Rates of Change and the Derivative

69

#4 This problem is similar to problem number 3. By denoting the number of passengers by N , the price charged to passengers can be expressed as 

P = 2400 for N ≤ 200 P = 2400 − 3(N − 200) = 1800 − 3N for 200 < N ≤ 1000

The total revenue is then  TR = 2400N for N ≤ 200 TR = PN = (1800 − 3N )N = 1800N − 3N 2 for 200 < N ≤ 1000 The FOC for a maximum is dTR = 1800 − 6N = 0 −→ N = 300 dN The SOC for a maximum is satisfied since d 2 TR = −6 < 0 dN2 Therefore 300 passengers each paying P = 1800 − 3 ∗ 300 = $900 generate a maximum of 900 ∗ 300 = $270,000 revenue for the travel agency. #5 The first derivative of the function is y ′ = (1 − x)4 − 4x(1 − x)3 = (1 − x)3 (1 − 5x) For the FOC y ′ = 0 we have either (1 − x)3 = 0, which leads to a solution x = 1, or (1 − 5x) = 0, which leads to the second solution x = 0.2. The SOC is y ′′ = −3(1 − x)2 (1 − 5x) − 5(1 − x)3 = −(1 − x)2 (8 − 20x)   y ′′ 

x=1

  = 0 and y ′′ 

x=0.2

= −2.56 < 0

It is clear from the second order condition that this function has a maximum at x = 0.2. But status of the extreme point at x = 1 is unclear. Recall that y ′ evaluated at any x is the slope of the tangent to the curve of the function at that point. One approach to determine whether the function at the point x = 1 has a minimum, maximum, or an inflection point is to check the sign of y ′ as we move from values of x less than 1 to 1 and then to values more than 1. Values of y ′ for selected values of x in the vicinity of 1 are presented in Table 5.1. As the table shows, values of y ′ change from negative to 0 and then to positive, indicating that the function has a minimum at x = 1. Figure 5.6 is the graph of the function.

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5 Rates of Change and the Derivative

Table 5.1 Values of x and y ′

x 0.7 0.8 0.9 1.0 1.1 1.2 y ′ −0.0675 −0.0240 −0.0035 0.0000 0.0045 0.0400

Fig. 5.6 Graph of y = x (1 − x)4

Chapter 5, 5.8 Exercises 1. If f (x, y) = 3x2 − 2xy3 + 10 find (a) f (1, 2) (b) f (−1, 3) (c) f (x + 1, 2y) 2w + z find g(2, −2) 2. If g(w, z) = 5 − wz 3. Find the partial derivatives of the following functions (a) f (x, y) = 3x3 − 2x2 y 3 + y 4 − 6xy2 + 10 (b) g(w, z) = −2w3 + 5z 2 − 5wz + 10 x2 − 2y + 3x2 − 2y (c) z = 3x + y 2  (d) f (x, y) = 2 x2 − 5y + 5 x + 2y 2 (e) z = (w2 + x2 )3 4. The graph of a bivariate function f (x, y) is a surface in three-dimensional space. The level curves or the contour plot for f (x, y) are curves in the xy plane defined by f (x, y) = c, where c is a constant. Graph the level curves for f (x, y) = x2 + y 2 − 2y for c = 2 and c = 10. 5. Find the extreme values of the following functions (a) f (x, y) = x2 + y 2 − 2y (b) z =  3x3 + 3x2 y − y 2 + 5 (c) z = x2 + y 2

5 Rates of Change and the Derivative

71

Answers to Chapter 5, 5.8 Exercises #1 (a) f (1, 2) = 3 − 16 + 10 = −3

(b) f (−1, 3) = 3 + 2 ∗ 27 + 10 = 67

(c) f (x+1, 2y) = 3(x+1)2 − 2(x+1)(2y)3 +10 = 3(x+1)2 −16(x+1)y 3 +10 #2 g(2, −2) =

2 2 ∗ 2 + (−2) = 5 − 2(−2) 9

#3 (a) ∂f = f x = 9x2 − 4xy3 − 6y 2 ∂x

∂f = f y = −6x2 y 2 + 4y 3 − 12xy ∂y

∂2 f ∂2 f = = f xy = f yx = −12xy2 − 12y ∂x∂ y ∂ y∂x (b) ∂g = f w = −6w2 − 5z ∂w

∂g = f z = 10z − 5w ∂z

f wz = −5

(c) ∂z 2x(3x + y 2 ) − 3(x2 − 2y) 3x2 + 2xy2 + 6y = + 6x = + 6x 2 2 ∂x (3x + y ) (3x + y 2 )2 −2(3x + y 2 ) − 2y(x2 − 2y) ∂z −6x − 2x2 y + 2y 2 = − 2 = −2 ∂y (3x + y 2 )2 (3x + y 2 )2 (d) 2x 5 fx =  +  2 x − 5y 2 x + 2y 2

(e)

f xy = f yx = 

5

(x2 − 5)3

−

∂z = 6w(w2 + x2 )2 ∂w ∂2z = 24wx(w2 + x2 ) ∂w∂x

5y

fy = 

(x + 2y 2 )3

∂z = 6x(w2 + x2 )2 ∂x

10y + − 5y x + 2y 2

−5

x2

72

5 Rates of Change and the Derivative

Fig. 5.7 Graph of x2 + y 2 − 2y = 2

#4 We write x2 + y 2 − 2y = 2 as y 2 − 2y + x2 − 2 = 0, which is a quadratic equation in terms of y, and solve it for y. y=

2±

  4 − 4(x2 − 2) = 1 ± 3 − x2 2

Each branch of y is a semi-circle and √ both together form a circle shown in Fig. 5.7. For c = 10 the equation is y = 1 ± 11 − x2 . See part (a) of problem #5 for the three dimensional graph of the function x2 + y 2 − 2y and its contour lines. Note that in that graph the level curves are all circles. #5 (a) The first order condition requires that we determine a combination of x and y that simultaneously satisfy f x = 0 and f y = 0. f x = 2x = 0 leading to x = 0 and f y = 2y − 2 = 0 leading to y = 1 This function has an extreme point at (0,1), which is f(0, 1) = −1. For the second order condition we need to find value of , where  is

2  = f xx (0, 1) f yy (0, 1) − f xy (0, 1)

Given that

f xx = 2

f yy = 2

f xy = 0

5 Rates of Change and the Derivative

73

Fig. 5.8 Graph of f (x, y) = x2 + y 2 − 2y and its Contour Plot

we have =2∗2−0=4>0 Since f xx > 0 and  > 0 this function has a minimum at point (0,1). Figure 5.8 is a graph of this function. This graph, and the next 2 for parts (b) and (c), are produced using the Internet based freely accessible mathematical software Wolfram Alpha. Readers are encouraged to read the Appendix to Chap. 6 where this software is introduced and some of its simple commands for generating plots are given.

74

5 Rates of Change and the Derivative

(b) To calculate the components of the first order conditions, we have ∂z ∂z = 9x2 + 6xy = 0 and = 3x2 − 2y = 0 ∂x ∂y We must now solve the following two equations with two unknowns 

9x2 + 6xy = 0 3x2 − 2y = 0

In both equations we solve for y in terms of x 

y = −1.5x y = 1.5x2

We now have −1.5x = 1.5x2 −→ 1.5x(x + 1) = 0 leading to x = 0 or x = −1 It is clear from the system above that if x = 0 then y = 0, otherwise, if x = −1 then y = 1.5. We have, therefore, two sets of values (0, 0) and (−1, 1.5) for (x, y) that satisfy the first order condition. We now calculate the required components for checking the second order condition  ∂2z  =0 = 18x + 6y  (0,0) ∂x2 ∂2z = −2 ∂ y2

and

 ∂2z  = −9 = 18x + 6y  (−1,1.5) ∂x2

  ∂2z ∂2z   = 6x  = 6x  = 0 and = −6 (0,0) (−1,1.5) ∂xy ∂xy

For (0,0)

 = 0 ∗ (−2) − (0)2 = 0

For (−1, 1.5)

 = (−9) ∗ (−2) − (−6)2 = −18 < 0

As it was mentioned in the text when  = 0 the second order condition gives no information. The function at (0, 0) could have a local maximum, minimum, or a saddle point. But the case for (x, y) = (−1, 1.5) is very clear. Since  < 0 then the function at (−1, 1.5) is a saddle point. Figure 5.9 is the graph of the function. (c) In  this part we are asked to determine the extreme value(s) of function z = x2 + y 2 . It should be immediately clear that the lowest value of this function is 0 when both x and y are zero. For any other values of x = 0 and y = 0 we

5 Rates of Change and the Derivative

75

Fig. 5.9 Graph of z = 3x3 + 3x2 y − 2 y + 5

Fig.  5.10 Graph of z = x2 + y 2

have z > 0 Therefore, this function has a minimum (which is actually a global minimum) at (0,0). Figure 5.10 is the graph of this function. Chapter 5 Supplementary Exercises 1. Write the equation of the line joining points A(−3, 5) and B(3, 3). 2. Write the equation of a line that passes through the point A(2, 4) and is perpendicular to the line with equation y = 5 + 4x. 3. Find the derivative of the following functions (a) f (x) = 3x2 − 10x + 5 (b) y = (x3 − 2x2 + 10)2 (c) g(z) = 3z − 10

(d) g(x) =

x2 − 1 x+1

 (e) z = 2 y 2 − 1

76

5 Rates of Change and the Derivative

√ 3 (i) φ(x) = 3 x2 − 2x + 1 3 y − 5y (j) g(y) = 2 y −2

w2 − 3w + 2 (f) f (w) = w−1 (g) y = (x3 − 3x2 + 10x − 5)3 2x (h) f (x) = √ x−1

4. Find the rate of change of the following functions at the given value. (a) y = 3x2 − 5x + 1 (b) y = −3x + 10 2

at x = 2

2

(c) f (w) = (w − 6w + 10)  (d) g(x) = 6 x2 − 4

2t 2 − 3t + 1 t −2 300 (f) Q = 2 P

(e) f (t) =

at x = −2 at w = 5 at x = 4 at t = 2 at P = 20

5. Show that the function y = 2x3 −6x2 +6x+1 has an inflection point at x = 1. 6. Find the local maximum or minimum of the following functions. If the second derivative test fails, use the first derivative test. (a) (b) (c) (d)

y y y y

= x4 − 18x2 + 40 = x3 − 6x2 − 10 = x4 − 8x2 + 20 = x6 + 4x3 − 10

7. A monopolistically competitive firm faces the following demand function for its product in the market P=

2160 Q 0.75

What is the firm’s marginal revenue from selling the 81st unit? 8. The demand function for product of a noncompetitive firm is given by Q=

3125 P 2.5

What is the firm’s marginal revenue from selling the 100th unit? 9. A noncompetitive firm faces the following demand function for its product  P = 2400 − 16 Q + 12

Find the combination of price and quantity that maximizes the firm’s revenue.

5 Rates of Change and the Derivative

77

10. A firm has the following total cost function TC = 0.8Q 3 − 20Q 2 + 220Q + 500 (a) (b) (c) (d) (e)

What is the average total cost if output is 15 units? At what level of output the average cost is minimum? At what level(s) of output(s) the marginal cost is 100? At what level of output the marginal cost is the same as the average fixed cost? At what level of output the difference between the average total cost and the average variable cost is 50? (f) At what level of output the difference between the average total cost and the average fixed cost is 100?

11. Find the partial derivatives of the following functions (a) (b) (c) (d)

f (x, y) = 3x − 4y + 5 + 3y 3 − 4xy + 10 f (x, y) = −2x2 √ 2 g(u, v) = 5u − v + 1 + 2uv z = x3 y − xy3 + x2 − y 2 x2 − 2 (e) f (x, y) = 2 y +1 (f) f (x, y, z) = 2xy2 − xz + z 2 y − 3xyz z−y x−z + (g) g(x, y, z) = 2y + 1 2x − 1

12. Evaluate the partial derivatives of the following functions at the given values (a) φ(x, y) =



x2 − y 2x + y

(b) g(x, y) = 3x2 y 3 − 5x3 + 4y 2  (c) z = x2 − 2xy + 10 3

3

(d) f (x, y) = x + y − xy

at x = 2

y=1

at x = −1 y = 2 at x = 1

y=2

at x = −1 y = −1

13. The market demand and supply functions for two related goods are given by Q d1 = 40 − 10P1 + 7P2

Q s1 = 70 + P1 − P2

Q d2 = 30 + 7P1 − 5P2

Q s2 = −60 − P1 + 2P2

Determine the equilibrium prices and quantities. 14. Show that the function  f (x, y) = 49 − x2 − y 2

has a maximum of 7 at point (0,0) (see Fig. 5.11).

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5 Rates of Change and the Derivative

Fig.  5.11 Graph of 49 − x2 − y 2

15. Find the extreme values of the following function z = x3 + y 3 − 6xy 16. An apple grower charges $22 for a 40-pound crate of apple. If customers buy more than 5 crates, he reduces the price by $2 per additional crate. What number of crate per customer maximizes the grower’s revenue per customer? What is the revenue maximizing price? 17. An appliance manufacturer produces two models of refrigerators: small dorm-room size 4 cu ft (cubic feet) and small family size 8 cu ft. It costs $200 to produce a unit of 4 cu ft. Cost of producing a unit of 8 cu ft is $300. It is estimated that the demand functions for each model is Q 1 = 7500 − 40P1 + 25P2

Q 2 = 8000 + 20P1 − 30P2

where Q 1 and Q 2 are the quantity demanded and P1 and P2 are the prices of first and second model. Determine the quantity/price combination of models that maximizes the producer’s profit. What size has the higher mark up? 18. A firm produces two types of ball bearing: 1 in. diameter and 2.5 inch diameter. It sells the smaller ball bearing for $2 a unit and the larger one for $3 a unit. If the total cost function of the firm is given by TC(Q 1 , Q 2 ) = Q 21 + 2Q 22 − 2Q 1 Q 2 + 6Q 1 − 9Q 2 + 15 where Q 1 and Q 2 are the daily output of the ball bearings in 1000, determine the profit maximizing level of output for each ball bearings. 19. A consumer staples company produces two types of soap: regular and moisturizing. It costs $3 to produce a half-dozen pack of regular soap. The cost of

5 Rates of Change and the Derivative

79

producing a half-dozen pack of moisturizing soap is $4. Assume the demand function for half-dozen pack of soaps are given by Q 1 = 58 − 15P1 + 10P2 Q 2 = 72 + 8P1 − 12P2 Determine the price/quantity combination for each soap to maximize profit. 20. A company produces two types of product. Assume the firm’s total cost function is given by TC(Q 1 , Q 2 ) = 4Q 21 + 3Q 22 + 5Q 1 Q 2 where Q 1 and Q 2 are units of outputs. Due to facilities and logistic constraints, the company is limited to production of a total of 50 units of both products a day, that is, Q 1 + Q 2 = 50. Determine the combination of outputs that minimizes the firm’s cost.

Chapter 6

Optimal Level of Output and Long Run Price

Chapter 6, 6.1 Exercises (pp. 138–140) 1. Find the stationary points of the following quadratic functions. (a) (b) (c) (d)

y = 3x 2 + 65 y = 3x 2 y = 3x 2 − 10x + 65 y = −2.5x 2 − 30x + 80

2. Determine whether the following quadratic functions could be employed as legitimate total cost functions. Specify any restrictions needed for these functions for legitimate profit optimization. (a) (b) (c) (d) (e) (f) (g)

TC TC TC TC TC TC TC

= 2Q2 + 15Q + 50 = −2Q2 + 15Q + 250 = −Q2 − 15Q + 50 = 5Q2 + 15Q = 10Q2 = 10Q2 − 40Q + 150 = 8Q2 + 25Q + 60

3. A competitive firm has the following quadratic total cost function TC = aQ2 + 10Q + 50 determine a if the market price is $50 per unit and the firm’s profit maximizing level of output is 10 units. 4. A competitive firm has the following quadratic total cost function: TC = 3Q2 + bQ + 50 Determine b if the market price is $70 per unit and the firm’s profit maximizing level of output is 10 units. © Atlantis Press and the authors 2015 S. Vali, Principles of Mathematical Economics II, Mathematics Textbooks for Science and Engineering 4, DOI 10.2991/978-94-6239-088-1_6

81

82

6 Optimal Level of Output and Long Run Price

5. A competitive firm has the following quadratic total cost function: TC = 5Q2 − 25Q + 300 (a) Determine the firm’s break-even points if the market price is P = $80. (b) Show that the profit maximizing level of output is the average of the two break-even points. 6. To generalize from the last problem, show that for a firm with a quadratic cost function the optimal level of output is the average of its break-even points, as long as the market price is greater than b. 7. Assume a firm has the following total cost function TC = 10Q2 − 25Q + 150 (a) Specify the bounds of output and price for the function. (b) Find the break-even points of output if the market price is $55. (c) Find the profit maximizing level of output and show that it is the average of the two break-even points in part (b). (d) Find the price and output when the total revenue line is tangent to the TC curve. (e) Show that the market equilibrium price and the firm’s output in the long run are the same as in part (d). 8. Assume all firms in an industry have the following total cost function TC = aQ2 + 25Q + 60 Determine a if the market equilibrium price and a single firm’s output in the long run are 49 and 5, respectively. Answers to Chapter 6, 6.1 Exercises #1 d2y dy = 6x = 0, leading to x = 0. The SOC is 2 = 6 > 0. Then the function dx dx has a local minimum at x = 0, which is 65. d2y dy = 6x = 0, leading to x = 0. The SOC is 2 = 6 > 0. Then the function (b) dx dx has a minimum at x = 0, which is 0. 5 d2y dy = 6x − 10 = 0, leading to x = . The SOC is 2 = 6 > 0. Then the (c) dx 3 dx 5 function has a minimum at x = , which is 56.667. 3 d2y dy = −5x − 30 = 0, leading to x = −6. The SOC is 2 = −5 < 0. Then the (d) dx dx function has a maximum at x = −6, which is 170. (a)

6 Optimal Level of Output and Long Run Price

83

#2 For a function (including quadratic) to be a legitimate total coat function it must be an increasing function of output Q, that is TC(Q) > 0 ∀ Q ∈ R and TC(Q2 ) > TC(Q1 ) if Q2 > Q1 Also, the average total cost, the average variable cost, and the marginal cost functions must be all positive at all levels of output. If a quadratic cost function has a linear term, in order to have a legitimate profit optimization, the market price must be greater than the coefficient of the linear term. Using these criteria, the quadratic functions in parts (b) and (c) are not legitimate, the rest are. Functions in part (d) and (e) do not have a fix cost, so they must be treated as a long-run total cost function. In case of (a), (d), and (g) the market price must be greater than 15, 15, and 25, respectively. Function (f) requires a restriction on the values of Q, namely Q > 4. Note that the average variable cost is 10Q − 40. We need the restriction Q > 4 to make sure that AVC is positive, which in turn leads to the restriction of the market price greater than 40. #3 We want to determine a in TC = aQ2 + 10Q + 50, given that if P = 50 the firm’s profit maximizing level of output is 10. The profit function is π = TR − TC = PQ − aQ2 − 10Q − 50 = 50Q − aQ2 − 10Q − 50 = −aQ2 + 40Q − 50

The FOC for a local maximum is dπ = −2aQ + 40 = 0 dQ Since profit it maximized at Q = 10, then −2a ∗ 10 + 40 = 0 #4

leading to

a=2

Using profit maximizing condition MC = P in a competitive market, we have MC = 6Q + b = P

−→

6 ∗ 10 + b = 70

−→

b = 10

#5 (a) The break-even level(s) of output is (are) when the total cost is equal to the total revenue. Thus, TR = TC

−→

80Q = 5Q2 −25Q+300

−→

5Q2 −105Q+300 = 0

This quadratic equation has two solutions Q1 = 17.59 and Q2 = 3.41. The average of these two levels of output is (17.59 + 3.41)/2 = 10.5.1 1

Readers should be alerted that Q = 3.41 violates the restriction of Q > 5 on this function.

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6 Optimal Level of Output and Long Run Price

(b) Next, we determine the profit maximizing level of output when price is 80. Utilizing MC = P, we have 10Q − 25 = 80

−→

Q=

105 = 10.5 10

#6 Assume TC = aQ2 + bQ + c is the total cost function and P the market price. For the break-even level(s) of output, we have TR = TC

PQ = aQ2 + bQ + c

→

−→

aQ2 − (P − b)Q + c = 0

This quadratic equation has two solutions P−b+

Q1 =



(P − b)2 − 4ac P−b− and Q2 = 2a

The average of these solutions is of output is MC = P

→



(P − b)2 − 4ac 2a

P−b Q1 + Q2 = . The profit maximizing level 2 2

2aQ + b = P leading to Q =

P−b 2

#7 (a) To ensure that the average variable cost 10Q − 40 is positive we must specify bound of output as Q > 4. To satisfy this restriction for profit maximization, the market price must be greater than 40. (b) TR = TC (c)

→

55Q = 10Q2 −40Q+150

−→

10Q2 −95Q+150 = 0

This quadratic equation has two solutions Q = 2 and Q = 7.5. 20Q − 40 = 55

−→

Q = 4.75 which is the average of 2 and 7.5

(d) When the total revenue line is tangent to the total cost curve then we have the following two conditions simultaneously satisfied, 

TC = TR MC = P

→ 10Q2 − 40Q + 150 = PQ → 20Q − 40 = P

We eliminate P by multiplying both sides of the second equation by −Q and adding it to the first equation,

6 Optimal Level of Output and Long Run Price

85

√ −10Q2 + 150 = 0 −→ Q = 15 = 3.873. We then calculate P from the second equation 20 ∗ 3.873 − 40 = P, which leads to P = 37.46. (e) As it was shown in the text, the point of tangency between TR line and TC curve provides the firm’s long-run equilibrium output and price. The second approach is to determine the minimum of the average total cost function. ATC(Q) =

TC 10Q2 − 40Q + 150 150 = = 10Q − 40 + Q Q Q

ATC ′ (Q) = 10 −

FOC :

150 =0 Q2

leading to the same solution as Q = calculated as

√ 15 = 3.873. The long-run price is then

P = ATC(3.873) = 10(3.873) − 40 +

150 = 37.46, same as in part(d). 3.873

#8 For the long-run output and price we find the minimum of the average total cost ATC = FOC :

60 aQ2 + 25Q + 60 = aQ + 25 + Q Q ATC ′ = a −

60 =0 Q2

−→

a=

60 Q2

Given that Q = 5, a = 60 ÷ 5 = 2.4. As a check, we calculate the long-run price   ATC 

Q=5

= 2.4 ∗ 5 + 25 +

60 = 49 5

Chapter 6, 6.8 Exercises 1. For the following total cost functions, determine the total cost TC average total cost ATC, average variable cost AVC, average fixed cost AFC, and marginal cost MC for the specified level of output (a) (b) (c) (d) (e) (f)

TC = 4Q2 + 100 1 TC = Q2 + 150Q + 20000 2 TC = 0.01Q2 + 140Q + 50000 3

2

TC = 0.02Q − 6Q + 6800Q + 20000 3

2

TC = 0.03Q − 0.6Q + 5Q + 3700

TC = 0.08Q3 − 1.8Q2 + 50Q + 678

Q = 20 Q = 500 Q = 1000 Q = 100 Q = 10

Q = 20

86

6 Optimal Level of Output and Long Run Price

2. For the cubic total cost functions given in problem (1), find (a) the minimum average total cost. (b) the minimum average variable cost. (c) the minimum marginal cost. 3. A firm operating in a competitive market has the following quadratic total cost function TC = 10Q2 + 5Q + 400 If the market price of the good is $165 per unit. (a) What is the firm’s profit maximizing level of output? What is its profit? (b) What is the long run price established in this market? (c) What is the output and profit of the firm at the long run price? 4. A competitive firm has the following quadratic TC function TC = 5Q2 + 200 Assume the market price P is $175 per unit. (a) (b) (c) (d) (e) (f)

How much it costs the firm to produce the 11th unit of output? What is the marginal profit of the 11th unit of output? What is the firm’s profit maximizing level of output? What is its profit? What is marginal profit of the last unit of output? What is the long-run price established in this market? What is the output and profit for the firm in the long-run?

Revenue Maximization In some industries a firm’s profit maximization objective is basically a revenue maximization problem. A good example is the operation of sport clubs, especially when the franchises do not own the arena or stadium they play in. With a stadium/arena lease and players’ contracts in place, the cost of operation of a team for a season is very much known. Therefore, profit is largely determined by revenue from ticket sales and merchandising. Managing rental apartment complexes and hotels are other variations on this theme. 5. The New York-area MLS (Major League Soccer) team, Red Bulls (formerly MetroStar), plays in a newly built soccer-specific stadium located in Harrison, New Jersey. The new stadium has a 25,000 seating capacity. If the average ticket price is $30, the average attendance per game is 15000. When the ticket price is $25, the average attendance is 20000. (a) Find the demand function for the team’s ticket, assuming it is linear. (b) What price should the team charge for a ticket in order to maximize revenue?

6 Optimal Level of Output and Long Run Price

87

(c) What would the attendance be at the revenue maximizing ticket price? (d) Show that the price elasticity of demand at the revenue maximizing price and quantity is 1. 6. Easy Living Corporation manages a 1000-unit rental apartment complex. If rent is set at $2,000 a month, the complex would have a 90 % occupancy rate. A 100 % occupancy rate is achieved by reducing the rent to $1,800 a month. What are the revenue maximizing rent and the occupancy rate, assuming a linear demand function? 7. Grand Paradise resort and casino in Las Vegas has 2500 rooms. The hotel management knows from past experience that a room rate of $100 a night results in full occupancy. It is also known that a 10 % increase in rate leads to a 1 % decline in the occupancy rate. If demand is linear, what rate should the hotel charge to maximize revenue? What would the occupancy rate be? 8. All firms operating in a perfectly competitive market have the variable cost function VC = 0.2Q2 + 2Q and the firms’ fixed cost is 120. The market demand and supply functions for this good, denoted by Qmd and Qms are given by Qmd = 107 − 4P

Qms = −37 + 2P

where Qmd and Qms are in units of 1000. (a) (b) (c) (d)

What is a typical firm’s profit maximizing level of output? What is a firm’s maximum profit? How many firms are in the market? If similar firms enter the market, what would the long-run price be in this market? (e) What is a firm’s output in the long-run?

9. The variable cost function of a competitive firm is given by VC = Q3 − 40Q2 + 1600Q Assume the firm’s fixed cost is $1,500. (a) If the market price for a unit of Q is $1,400, determine the firm’s profit maximizing level of out and its profit. (b) What is the market long run price? (c) What is the firm’s shutdown price?

88

6 Optimal Level of Output and Long Run Price

10. Using the rules developed in the text, evaluate the following costs functions (a)

TC =

1 3 Q + 25Q2 + 1600Q 5

(b)

TC =

1 3 Q − 25Q2 + 1600Q 5

(c)

TC = 20Q2 − 20Q − 100

(d)

TC = 2Q2 − 10Q + 100

11. A competitive firm has a quadratic cost function TC = aQ2 + bQ + c. Is it possible to determine this firm’s shutdown price? 12. A monopolistically competitive firm has the following total cost function TC = 0.03Q2 + 12Q + 500 If the demand function for the firm’s product is P = 33 − 0.005Q (a) Find the firm’s profit maximizing level of output and its profit. (b) What is the firm’s long run price-quantity combination? (c) What is the equation of the dislocated demand curve associated with the long run situation? 13. A monopolistically competitive firm has the following total cost function TC = 4Q2 + 30Q + 400 If the demand function for the firm’s product is P = 600 − 30Q

where

Q < 20

(a) Find the firm’s profit maximizing level of output and its profit. (b) What is the firm’s long-run price-quantity combination? (c) What is the equation of the dislocated demand curve associated with the long run situation? 14. A monopolistically competitive firm has the following total cost function TC = Q3 − 10Q2 + 45Q + 100

6 Optimal Level of Output and Long Run Price

89

If the demand function for the firm’s product is P = 100 − 10Q (a) Find the firm’s profit maximizing level of output, price, and profit. (b) What is the firm’s long run price-quantity combination? (c) What is the equation of the dislocated demand curve? 15. A monopolist has the following cubic cost function TC = 0.08Q3 − 8.2Q2 + 2000Q + 32000 If the market demand function for the firm’s product is P = 3400 − 14Q What is the monopolist’s profit maximizing combination of price and quantity? What is the firm’s profit? 16. A monopolist has the following cubic cost function TC = 0.2Q3 − 7Q2 + 136Q + 100 If the market demand function for the firm’s product is P = 1000 − 7Q What is the monopolist’s profit maximizing combination of price and quantity? What is the firm’s profit? 17 Due to adverse economic conditions, demand for goods produced by the monopolist in problem (16) declines, to the extent that the firm’s above normal profit is eliminated. What is the monopolist’s new optimal combination of price and quantity? What is the equation of the dislocated market demand curve? 18. A monopolist firm with the following cubic cost function TC = 3Q3 − 40Q2 + 250Q + 900 faces the following market demand function for its product P = 2000 − 10Q The management of the firm decides to follow a three-stage market strategy 1- find the level of output that minimizes its average total cost 2- find the level of output that maximizes its total revenue 3- choose the average of the two levels for actual production

90

6 Optimal Level of Output and Long Run Price

(a) What is the level of output in this strategy compared to the profit maximizing output? (b) What is the price level in this case compared to the profit maximizing case?

Average Cost Pricing Instead of pursuing profit maximization through marginal cost pricing, the regulated natural monopolists are required to follow the average cost pricing. In the average cost pricing, a natural monopolist firm sets its price at the level where the demand curve intersects the average cost ATC curve. This policy, compared to marginal cost pricing, leads to production of a larger volume of output at a relatively lower price. 19. What is the level of output and price if a natural monopolist with the cost function TC = 0.3Q3 − 8Q2 + 120Q and market demand function P = 100 − Q follows the average cost pricing? (From 2 possible solutions choose the larger one.) Compare this solution to the profit maximizing solution. 20. A firm is competitive in both product and resource market. The firm’s short run production function is given by Q = 0.5L 1/2 where L is the labor input. Assume the wage rate is $10 and the firm’s fixed cost is 100. If the price of the good in the market is $1,200 per unit (a) What is the firm’s profit maximizing level of output and employment? (b) What is the firm’s total wage bill? (c) What is its profit? 21. Continued from Problem (20): Assume new firms enter this market and due to additional supply price declines to $1,000 per unit. Redo parts (a), (b), and (c). 22. Continued from Problem (21): Assume that due to strong demand for labor the wage rate in the labor market is increased by 10 %. Redo parts (a), (b), and (c). 23. A monopolistically competitive firm has a short run production function given by Q = 0.5L 1/2

6 Optimal Level of Output and Long Run Price

91

Assume the wage rate is $5 and the firm’s fixed cost is 828. If demand for the firm’s output is P = −3Q + 460 (a) What is the firm’s profit maximizing level of output and employment? (b) What is the firm’s total wage bill? (c) What is its profit? 24. Continued from Problem (23): If new firms enter the mark and as the result the firm’s demand curve is dislocated (a) What is the firm’s long-run level of output and employment? (b) What is the equation of the firm’s dislocated demand curve? 25. A monopolistically competitive firm has a short run production function given by Q = 2L 0.5 Assume the wage rate is $10, the firm’s fixed cost is 500, and demand for the firm’s output is P = 600 − 3Q Suppose the firm’s goal is to maximize its sales revenue. (a) What is the firm’s revenue maximizing output, price, employment, and profit? (b) What is the firm’s profit maximizing output, price, employment, and profit? 26. Write the average (A) and marginal (M) functions of the following functions. Show that the value of the average function A at its extrema (minimum or maximum) is the same as of the M function, that is, M intersects A at its extreme point. (1)

y = x 2 + 2x + 2

(2)

z = −2w3 + 5w2 + 10w

(3)

TC = 5Q2 − 25Q + 240

27. A firm operating in a competitive market has the following long run total cost function TC = Q3 − 24Q2 + 200Q Find the long run equilibrium price for this good.

92

6 Optimal Level of Output and Long Run Price

28. A firm operating in a competitive market has the following production function Q = 30L 2 − L 3 If this firm uses 18 units of labor and the price of the good in the market is $20, what wage must this firm pay its workers? If this firm’s fixed cost is 30,000, what is its profit? 29. A firm in a non-competitive market has the total cost function TC =

1 3 Q − 6Q2 + 50Q + 45 3

The demand function for this firm’s product is P = 35 − 2Q (a) Write the firm’s marginal revenue function. (b) Find the firm’s profit maximizing level of output and price. 30. A firm operating in a competitive market has the following production function Q = 10L − L 2 If the firm’s output sells for $10 per unit and it pays $40 per unit of labor, what is the firm’s equilibrium level of labor utilization? 31. Suppose the market demand function is Qd = 200 − 5P (a) Determine the price elasticity of demand at prices $10, $15, and $30. (b) Determine the price at which the demand function is unit elastic. (c) Determine the range of prices over which the demand function is (a) elastic and (b) inelastic. 32. Suppose the market supply function is Qs = −500 + 50P

where P > 10

(a) Determine the price elasticity of supply η at prices $15, $20, and $25. (b) Is there a price at which this supply function is unit elastic? 33. Suppose the market supply function is given by Qs = −c + dP

where c, d > 0

and P >

Show that this supply function is elastic at all prices above

c . d

c d

6 Optimal Level of Output and Long Run Price

93

34. Suppose the market demand function is Qd = aPλ

where a > 0

and

λ0  dQ2 Q3 Q=167.765

The minimum of average total cost is 6475.526.

6 Optimal Level of Output and Long Run Price

95

(b) The first order condition for a local minimum of the AVC function is dAVC = 0.04Q − 6 = 0 dQ

−→

Q = 150

The second order condition for a local minimum is satisfied d 2 AVC = 0.04 > 0 dQ2 The minimum of average variable cost is 6350. (c) FOC for minimum of MC function is dMC = 0.12Q − 12 = 0 dQ

→

Q = 100 and SOC is

d 2 MC = 0.12 > 0 dQ2

The minimum of the MC function is 6200. The minimum of ATC, AVC, and MC for function in parts (e) and (f) are similarly determined. #3 (a) The profit function of the firm is π = TR − TC = 165Q − 10Q2 − 5Q − 400 = −10Q2 + 160Q − 400 FOC :

dπ = −20Q + 160 = 0 dQ

→

Q = 8,

SOC :

d2π = −20 < 0 max dQ2

The firm’s maximum profit is π = 240. (b) The long-run price will be the minimum of the ATC function ATC = 10Q + 5 +

400 Q

→

FOC :

dATC 400 = 10 − 2 = 0 dQ Q

Solution to the quadratic equation 10Q2 − 400 = 0 is Q = 6.325 and the long-run price, PLR , is PLR = ATC(6.325) = 131.491 (c) Output of the firm at the long-run price is Q = 6.325 units and its profit is zero. TR − TC = 131.491 ∗ 6.325 − 10(6.3252 ) − 5 ∗ 6.325 − 400 ≈ 0.

96

6 Optimal Level of Output and Long Run Price

#4 (a) Given TC = 5Q2 + 200 then MC = 10Q and at Q = 11, MC = $110. (b) To determine the marginal profit of the 11th unit we write the total profit as π = TR−TC = PQ−5Q2 −200 = 175Q−5Q2 −200, then the marginal profit dπ  dπ = 175 − 10Q. Evaluating the marginal profit at 11, we have is =  dQ dQ Q=11 ′ ′ ′ 175 − 10 ∗ 11 = $65. Alternatively, π = TR − TC = MR − MC = P − MC = 175 − 110 = $65 dπ , which is the marginal profit, equal to zero, we have (c) By setting dQ π ′ = 175 − 10Q = 0

−→

Q = 17.5 units

π = 175 ∗ 17.5 − 5(17.5)2 − 200 = $1331.25 (d) Given that π ′ = 175 − 10Q, then the marginal profit of the last unit of output is 175 − 10 ∗ 17.5 = 0 (e) The long run price is the minimum of the average total cost (ATC) function, then ATC =

5Q2 + 200 200 TC = = 5Q + Q Q Q

The FOC for a minimum is √ dATC 200 = 5− 2 = 0 or 5Q2 −200 = 0 leading to Q = 40 = 6.325 dQ Q Then the long-run price PLR is PLR = ATC(6.325) = 5(6.325) +

200 = 63.25 6.325

(f) In the long-run firm’s level of output is 6.325 units and its profit is zero. #5 (a) Given that the demand function assumed to be linear, we can derive it using the two points on the line. Denoting the number of people who attend the games by N, the coordinate of these points are (P1 , N1 ) = (30, 15000) and (P2 , N2 ) = (25, 20000). Equation of the line joining these two points is our demand function. N −N1 =

N2 − N1 (P−P1 ) P2 − P1

−→

N −15000 =

20000 − 15000 (P−30) 25 − 30

N = 45000 − 1000P, or after solving for P, P = 45 − 0.001N

6 Optimal Level of Output and Long Run Price

97

(b) The total revenue function is TR = PN = 45N − 0.001N 2 The FOC for a maximum is dTR = 45 − 0.002N = 0 dN

N=

−→

45 = 22500 0.002

And the revenue maximizing ticket price is P = 45 − 0.001 ∗ 22500 = $22.5. d 2 TR The SOC for a maximum is satisfied: = −0.002 < 0. dN 2 (c) At price $22.5 the attendance would be N = 22500. (d) To determine the price elasticity of demand, we express the demand equation in terms of N, N = 45000 − 1000P. Now we have,   1000 ∗ 22.5 P dN P =− = −1000 = −1 ǫ= dP N 45000 − 1000P 45000 − 1000 ∗ 22.5 #6 This problem and problem # 7 are similar to problem # 5. We have two points on the linear demand curve so we can write the demand function. Denoting the number units rented by Q, the coordinates of these two points are (P1 , Q1 ) = (2000, 900) and (P2 , Q2 ) = (1800, 1000). The linear demand function is Q − Q1 =

Q2 − Q1 (P − P1 ) P2 − P1

Q − 900 =

→

1000 − 900 (P − 2000) 1800 − 2000

And the demand function is Q = 1900 − 0.5P. The total revenue function is TR = PQ = P(1900 − 0.5P) = 1900P − 0.5P2 . Maximizing this function leads to FOC :

dTR = 1900 − P = 0 dP

−→

P = 1900 and Q = 950 units

The profit maximizing rent is $1,900 with the occupancy rate of 95 %. #7 Based on the given information, the two points on the demand curve are (P1 , Q1 ) = (100, 2500) and (P2 , Q2 ) = (110, 2475). Subsequently, the demand function is Q−2500 = FOC :

−25 (P −100) 10

→

TR′ = 2750 − 5P = 0

Q = 2750−2.5P and TR = 2750P −2.5P2 →

P = 550 and Q = 1375

#8 (a) We first determine the market price and then find the profit maximizing level of output for the firm. Using the equilibrium condition Qmd = Qms we have,

98

6 Optimal Level of Output and Long Run Price

107 − 4P = −37 + 2P

−→

6P = 144

−→

P = 24

and the market equilibrium quantity Qm = −37 + 2 ∗ 24 = 11 or 11000 units. The firm’s profit function is π = TR − TC → π = 24Q − 0.2Q2 − 2Q − 120 → π = −0.2Q2 + 22Q − 120

The FOC for a max is dπ = −0.4Q + 22 = 0 dQ

−→

Q = 55 units

dπ 2 = −0.4 < 0, is satisfied. dQ2 (b) The firm’s total profit is π = −0.2(55)2 + 22 ∗ 55 − 120 = 485. (c) Each firm produces 55 units and the total market quantity is 11000 units, then there must be 11000 ÷ 55 = 200 firms in the market. (d) The long-run price in a perfectly competitive market is established at the minimum of the average total cost function, therefore Note that the SOC for a maximum,

ATC =

0.2Q2 + 2Q + 120 120 TC = = 0.2Q + 2 + Q Q Q

The FOC for a minimum is dATC 120 = 0.2 − 2 = 0 dQ Q

−→

Q2 = 600 and Q = 24.495

PLR = ATC(24.495) = 0.2(24.495) + 2 +

120 = 11.8 24.495

(e) The firm’s output in the long run is 24.495 units and it earns a normal profit. #9 (a) We determine the profit maximizing level of output by using MC = P, MC = 3Q2 −80Q+1600 → 3Q2 −80Q+1600 = 1400 → 3Q2 −80Q+200 = 0

The solution to this quadratic equation, which satisfies the second order condition for a maximum, is Q = 23.874 units. Note that   MC ′ = 6Q − 80 and MC ′  = 6(23.874) − 80 > 0 Q=23.874

which satisfies the SOC for profit maximization. The profit of the firm is 2916.5. (b) The market long-run price is the minimum of the average total cost ATC,

6 Optimal Level of Output and Long Run Price

99

1500 Q

ATC = Q2 − 40Q + 1600 +

1500 dATC = 2Q − 40 − 2 = 0 dQ Q

→

2Q3 − 40Q2 − 1500 = 0

A real root of this cubic function is Q = 21.61. The long-run price is PLR = ATC(21.61) = 1272. (c) The firm’s shutdown price is the minimum of the average variable cost AVC, AVC = Q2 − 40Q + 1600

→

AVC ′ = 2Q − 40 = 0

→ Q = 20

PSD = AVC(20) = 1200 #10 1 3 Q + 25Q2 + 1600Q is not a legitimate total cost function; it violates 5 R2 that requires b, coefficient of the quadratic term, to be negative. 1 (b) In function TC = Q3 − 25Q2 + 1600Q, R1, R2, and R3 are satisfied, so 5 the function is a legitimate total cost function. Note that c = 0, therefore this function should be treated as a long-run TC function. (c) This function is not a legitimate total cost function. The constant term, which represents the fix cost, is negative. (d) This function is not a legitimate total cost function, it violates the requirement of TC(Q2 ) > TC(Q1 ), if Q2 > Q1 . Note that at output level Q = 1 the total cost is 92 while at out level 2 it is 88. (a) TC =

#11 The shutdown price is the minimum of the average variable cost. The average variable cost of a quadratic cost function is a linear function AVC = aQ + b, and a linear function’s maximum or minimum occurs at the end point of an interval. It is clear that the AVC here is the lowest when Q = 0, which implies that the firm is not producing anything, already. #12 (a) The profit function of the firm is π = TR−TC = 33Q−0.005Q2 −0.03Q2 −12Q−500 = −.035Q2 +21Q−500 The first order condition for a local minimum or maximum is FOC :

dπ = −0.07Q + 21 = 0 −→ dQ

Q = 300

We check the second order condition for a local maximum

100

6 Optimal Level of Output and Long Run Price

SOC :

d2π = −0.7 < 0 and it is satisfied. dQ2

From the demand function, we determine the price as P = 33 − 0.005 ∗ 300 = 31.5 With this price/quantity combination, the profit is π = 31.5 ∗ 300 − 0.03(3002 ) − 12 ∗ 300 − 500 = 2650 (b) The long-run or normal-profit narrative of a monopolistically competitive market is that due to adverse economic condition and/or competition demand for the firm’s product declines, and the normal-profit state is reached when the dislocated demand curve becomes tangent to the average total cost curve. At the point of tangency of the dislocated demand curve and the ATC curve both have the same rate of change or slope. Then we have, ATC =

500 TC = 0.03Q + 12 + Q Q

500 dATC dP = 0.03 − 2 = = −0.005 dQ Q dQ where −0.005 is the slope of the demand curve. The normal-profit level of output, Qnp , is 500 = 0.035 Q2

−→

Qnp = 119.523

And the normal-profit price, Pnp , is determined by Pnp = ATC(119.523) = 19.77 (c) Let express the equation of the dislocated demand curve as P = a − 0.005Q We have to determine the new intercept a. For the demand curve associated with the long-run or normal-profit, we have 19.77 = a − 0.005(119.523) P = 20.368 − 0.005Q

−→

a = 20.368

6 Optimal Level of Output and Long Run Price

101

#13 (a) The profit function is π = TR − TC = 600Q − 30Q2 − 4Q2 − 30Q − 400 = −34Q2 + 570Q − 400 FOC : π ′ = −68Q + 570 = 0 → Q = 8.382,

SOC : π ′′ = −68 < 0

P = 600 − 30 ∗ 8.382 = 348.54 (b) The long-run quantity is determined by dP dATC = dQ dQ 4−

400 = −30 Q2

−→

Qnp = 3.43

Pnp = ATC(3.43) = 160.34 (c) The demand equation associated with the normal profit situation is 160.34 = a − 30 ∗ 3.43

−→

a = 263.24

P = 263.24 − 30Q #14 (a) π = 100Q − 10Q2 − Q3 + 10Q2 − 45Q − 100 = −Q3 + 55Q − 100 π ′ = −3Q2 + 55 = 0

−→

Q = 4.2817

P = 100 − 10 ∗ 4.2817 = 57.183 π = 56.997 (b) ATC = Q2 − 10Q + 45 + ATC ′ = 2Q − 10 −

100 Q

100 = −10 Q2

Pnp = ATC(3.684) = 48.876

→

2Q3 − 100 = 0

→

Qnp = 3.684

102

6 Optimal Level of Output and Long Run Price

(c) P = a − 10Q

−→

48.876 = a − 10 ∗ 3.684

−→

a = 85.716

P = 85.716 − 10Q #15 π = 3400Q − 14Q2 − 0.08Q3 + 8.2Q2 − 2000Q − 32000 π = −0.08Q3 − 5.8Q2 + 1400Q − 32000 π ′ = −0.24Q2 − 11.6Q + 1400 = 0

Q = 55.942

−→

P = 3400 − 14 ∗ 55.942 = 2616.81 and profit is π = 14162 #16

We form the profit function

π = TR − TC = (1000 − 7Q)Q − 0.2Q3 + 7Q2 − 136Q − 100 π = −0.2Q3 + 864Q − 100 FOC : SOC :

π ′ = −0.6Q2 + 864 = 0 −→   π ′′ = −1.2Q 0 then the SOC for a local minimum is satisfied. This cubic equation has a real solution Q = 8.665. Since ATC ′′ = 6 + Stage 2- Determining the level of output that maximizes total revenue: TR = PQ = (2000 − 10Q)Q = 2000Q − 10Q2 TR′ = 2000 − 20Q = 0

→

Q = 100

SOC : TR′′ = −20 < 0

8.665 + 100 = 54.333 2 (a) The profit maximizing level of output is Stage 3- Average of the two outputs =

π = 2000Q−10Q2 −3Q3 +40Q2 −250Q−900 = −3Q3 +30Q2 +1750Q−900 π ′ = −9Q2 + 60Q + 1750 = 0

→

Q = 17.671

(b) Price associated with the three stage strategy is P = 2000 − 10(54.333) = 1456.67 while the price associated with profit maximization is P = 2000 − 10(17.671) = 1823.29

104

6 Optimal Level of Output and Long Run Price

#19 For average cost pricing we set the average total cost function equal to the demand function and solve for Q ATC = P

−→

0.3Q2 − 8Q + 120 = 100 − Q

−→

0.3Q2 − 7Q + 20 = 0

The larger solution to this quadratic function is Q = 20, leading to P = 80. For the profit maximizing quantity and price we follow the familiar routine MC = MR

→

0.9Q2 − 16Q + 120 = 100 − 2Q

and solve for 0.9Q2 − 14Q + 20 = 0. The larger solution to this quadratic equation Q = 13.964 is the right answer, which also satisfies the second order condition MC ′′ (13.964) = 1.8(13.964) − 16 > 0. The price is P = 100 − 13.964 = 86.04. The optimal price/quantity combination of the average cost pricing compared to that of profit maximizing, shows that if a natural monopolist follows the average cost pricing its level of output is more than the profit maximizing level of output, while the price is less than the profit maximizing price. The average cost pricing solution is a break-even solution. Note that if we multiply both sides of ATC = P by Q, we arrive at TC = TR. #20 We can solve this problem by directly expressing the firm’s profit as a function of L. Since labor is considered the only input in the production process, the total cost function of the firm can be expressed as TC = wL + FC = 10L + 100. The profit function is then √ π = TR − TC = PQ − 10L − 100 = 1200(0.5 L) − 10L − 100 √ = 600 L − 10L − 100 To maximize profit, we have FOC : SOC :

600 dπ = √ − 10 = 0 dL 2 L

→

300 √ − 10 = 0 L

→

L = 900

d2π

150 =− √ 0 dL L L 2

√ (a) The firm’s profit maximizing level of output is Q = 0.5 900 = 15 units and it employs 900 units of labor. (b) The firm’s wage bill is wL = 10 ∗ 900 = $9,000. (c) The firm’s profit is TR − TC = 1200 ∗ 15 − 9000 − 100 = $8,900. Alternative, we can solve this problem by expressing profit as a function of output. To do this, we first solve the production function Q = 0.5L 0.5 for L in terms of Q as L = 4Q2 and then write the profit function as π = TR − TC = 1200Q − 10L − 100 = 1200Q − 10(4Q2 ) − 100

6 Optimal Level of Output and Long Run Price

105

π = −40Q2 + 1200Q − 10 For maximizing π, we have π ′ = −80Q + 1200 = 0 leading to the same solution for Q = 15 and L = 900

The third alternative utilizes the fact that when the product and resource markets are competitive, resources receive the value of their marginal product, the marginal revenue product.2 In our case the resource used in production is labor. Therefore w = MRPL . Given the production function Q = 0.5L 0.5 , the marginal physical product of labor, MPPL is MPPL =

dQ 1 = 0.25L −0.5 = √ dL 4 L

And the value of the marginal product of labor, MRPL is 1 300 MRPL = P ∗ MPPL = 1200 √ = √ 4 L L Given that the wage rate w must be the marginal revenue product of labor MRPL , we have MRPL = w

→

300 √ = 10 L

→

√

L = 30 and L = 900

#21 After the price drops to $1,000, the profit function of Exercise 20 changes to √ π = TR − TC = PQ − 10L − 100 = 1000(0.5 L) − 10L − 100 √ = 500 L − 10L − 100 (a) The profit maximizing level of labor utilization is now 250 π ′ = √ − 10 = 0 L

−→

L = 625 units

√ with output Q = 0.5 625 = 12.5 units. (b) The wage bill is 10 ∗ 625 = 6250. (c) The firm’s profit is 1000 ∗ 12.5 − 6250 − 100 = 6150. Note that a 16.7 % decline in the market price leads to an almost 31 % drop in the firm’s profit.

2

See Chap. 10 for a more detail discussion of this topic.

106

6 Optimal Level of Output and Long Run Price

#22 An increase in the number of firms in the market leads to more demand for labor and an increase in the wage rate. With the wage rate at $11, the profit function in Exercise 21 changes to √ π = TR − TC = PQ − 11L − 100 = 1000(0.5 L) − 11L − 100 √ = 500 L − 11L − 100 (a) The profit maximizing level of labor utilization is now 250 π ′ = √ − 11 = 0 L

L = 516.53 units

−→

√ with output Q = 0.5 516.53 = 11.36 units. (b) The wage bill is 11 ∗ 516.53 = 5681.83. (c) The firm’s profit is 1000 ∗ 11.36 − 5681.83 − 100 = 5578.17. Note that a 10 % increase in the wage rate leads to about 9 % drop in the level of output and profit. #23 We use the production function and write L = 4Q2 , and then express the firm’s profit function as π = TR − TC = PQ − wL − 828 = (−3Q + 460)Q − 5(4Q2 ) − 828 = −23Q2 + 460Q − 828

(a) The profit maximizing level of output and employment are π ′ = −46Q + 460 = 0

−→

Q = 10 and L = 4(10)2 = 400

From the demand function we have P = −30 + 460 = 430. (b) The firm’s total wage bill is 5 ∗ 400 = 2000. (c) The firm’s profit is 430 ∗ 10 − 2000 − 828 = 1472. #24 The long-run occurs at the point of tangency between the demand and the average total cost curves. The total cost function is TC = wL + FC = 5L + 828 = 5(4Q2 ) + 828 = 20Q2 + 828 And the average total cost function is ATC = 20Q +

828 Q

(a) At the point of tangency the rate of change of ATC is the same as the slope of the linear demand function, therefore ATC ′ = 20 −

828 = −3 Q2

→

Q2 =

828 = 36 and Q = 6 units 23

6 Optimal Level of Output and Long Run Price

107

L = 4Q2 = 144 units, and PLR = ATC(6) = 20 ∗ 6 +

828 = 258 6

(b) We write the dislocated demand function as P = a − 3Q and determine a by plugging in the long-run values of output and price a = P + 3Q = 258 + 3 ∗ 6 = 276

−→

P = −3Q + 276

#25 (a) The revenue function is TR = PQ = 600Q − 3Q2 FOC :

dTR = 600 − 6Q = 0 dQ

−→

Q = 100

Subsequently, P = 600 − 3 ∗ 100 = 300. By using the production function, we have 100 = 2L 0.5 , leading to the revenue maximizing employment of L = 2500 units. The firm’s profit is π = TR − TC = 300 ∗ 100 − 10 ∗ 2500 − 500 = 30000 − 25500 = 4500 (b) To find the profit maximizing output, we solve for L in the production function, L = 0.25Q2 , and write the profit function as π = TR − TC = 600Q − 3Q2 − wL − FC = 600Q − 3Q2 − 2.5Q2 − 500 π = −5.5Q2 + 600Q − 500 FOC :

π ′ = −11Q + 600 = 0 leading to Q = 54.55

SOC :

π ′′ = −11 < 0 Max

L = 0.25Q2 = 0.25(54.552 ) = 743.93; and π = 15863.64 #26 (1) The average and marginal functions are A=x+2+

2 x

M = 2x + 2

P = 600 − 3 ∗ 54.55 = 436.35

108

6 Optimal Level of Output and Long Run Price

Fig. 6.1 Graph of 2 A = x + 2 + and x M = 2x + 2

The extreme point(s) of A is (are) FOC :

A′ = 1 −

SOC :

A′′ =

2 =0 x2

4 x3

−→

x=

√ 2 = ±1.4142

A′′ (1.4142) > 0 and A′′ (−1.4142) < 0

Thus the average function A has a minimum at x = 1.4142, which is 4.8284, and a maximum at x = −1.4142, which is −0.8284. We check and see that both of these points are on the marginal function M M = 2 ∗ (1.4142) + 2 = 4.8284

M = 2 ∗ (−1.4142) + 2 = −0.8284

This means that the graph of M function intersect the graph of A function at its extreme points. Figure 6.1 is the graph of both functions. (2) The average and marginal functions are A = −2w2 + 5w + 10

M = −6w2 + 10w + 10

The average function has a maximum at w = 1.25, which is 13.125. This point is on the graph of the marginal function, indicating that it intersect the average function at this point. M = −6(1.252 ) + 10 ∗ 1.25 + 10 = 13.125 Figure 6.2 is the graph of both functions. (3) In this case, the average and marginal functions are the average total cost ATC and the marginal cost MC functions.

6 Optimal Level of Output and Long Run Price

109

Fig. 6.2 Graph of A = −2w2 + 5w + 10 and M = −6w2 + 10w + 10

ATC =

240 TC = 5Q − 25 + and MC = 10Q − 25 Q Q

ATC minimum occurs at FOC :

dATC 240 =5− 2 =0 dQ Q

SOC :

d 2 ATC 480 = 3 >0 ∀ Q>0 dQ2 Q

Q=

−→

ATC(6.9282) = 5 ∗ 6.9282 − 25 +

√ 48 = 6.9282

240 = 44.282 6.9282

We now check to see whether the ATC’s minimum point is a point on MC curve MC = 10 ∗ 6.9282 − 25 = 44.282 It is, proving that these two curve intersect at the minimum point of the ATC curve. Figure 6.3 is the graph of both functions. #27

We must determine the minimum of the firm’s average total coat function ATC =

TC = Q2 − 24Q + 200 Q

FOC :

ATC ′ = 2Q − 24 = 0

PLR = ATC(12) = 56

−→

QLR = 12

110

6 Optimal Level of Output and Long Run Price

Fig. 6.3 Graph of ATC = 5Q − 25 + MC = 10Q − 25

240 Q

and

#28 The firm’s profit function in terms of L is π = TR − TC = PQ − wl − FC = 20(30L 2 − L 3 ) − wL − FC = −20L 3 + 600L 2 − wL − FC Profit is maximized when π ′ = −60L 2 + 1200L − w = 0 We know that the firm is using 18 units of labor, therefore, −60(182 ) + 1200 ∗ 18 − w = 0

−→

w = 2160

If the firm fixed cost is 30000, its profit is   π = 20 30 ∗ (182 ) − 183 −2160∗18−30000 = 77760−38880−30000 = 8880

(An alternative solution to this problem will be offered in Chap. 10, at the end of answer to problems in Exercise 10.7) #29 (a) We first write the firm’s revenue function as TR = PQ = (35 − 2Q)Q = 35Q − 2Q2 The marginal revenue function is then MR = TR′ = 35 − 4Q

6 Optimal Level of Output and Long Run Price

111

(b) The firm’s profit function is 1 1 π = TR−TC = 35Q−2Q2 − Q3 +6Q2 −50Q−45 = − Q3 +4Q2 −15Q−45 3 3 FOC : π ′ = −Q2 + 8Q − 15 = 0

−→

Q = 3 and Q = 5

Of these two solutions Q = 5 satisfies the second order condition for a maximum. The profit maximizing price is P = 35 − 2 ∗ 5 = 25. #30 The profit function of the firm is π = TR − TC = PQ − wL − FC = 10(10L − L 2 ) − 40L − FC = −10L 2 + 60L − FC

FOC : π ′ = −20L + 60 = 0

−→

L = 3 and Q = 21

Another approach to solving this problem is to set the marginal revenue product of labor equal to wage, that is MRPL = PMPPL = w From the production function, we have MPPL =

dQ = 10 − 2L dL

MRPL = 10 ∗ (10 − 2L) = 40

−→

leading to the same level of labor utilization L = 3. #31 We write the demand elasticity function and use it to determine elasticity of demand at different prices ǫ(p) =

dQ P = −5 dP Q



P 200 − 5P



=−

5P 200 − 5P

(a) Using ǫ(P), we have ǫ(10) = −

50 1 =− 150 3

ǫ(15) = −

75 3 =− 125 5

ǫ(30) = −

150 = −3 50

(b) We need to determine the price such that ǫ(P) = −1 −

5P = −1 200 − 5P

−→

200 − 5P = 5P

−→

P = 20

(c) The unit elastic price is the demarcation point between elastic and inelastic segments of a demand curve. Here |ǫ(20)| = 1, therefore

112

6 Optimal Level of Output and Long Run Price

    (a) For P > 20 ǫ(P) > 1 and demand is elastic #32

    (b) For P < 20 ǫ(P) < 1 and demand is inelastic

The price elasticity of supply is

η(P) =

dQ P = 50 dP Q





P −500 + 50P

=

50P −500 + 50P

(a) The supply elasticity for the 3 given prices are η(15) =

750 =3 250

1000 =2 500

η(20) =

η(25) =

1250 = 1.67 750

(b) We need to find P such that η(P) = 1 η(P) =

50P =1 −500 + 50P

−→

50P = −500 + 50P

It is obvious that there is no solution to this equation. This leads to the conclusion that the linear supply functions can never be unit elastic. #33 dQ P =d η(P) = dP Q



P −c + dP



dP −c + dP

=

Note that in the above fraction the numerator dP is always greater than the denominator dP − c, therefore the price elasticity of a linear supply function is always c c due to the fact that is a number greater than 1. We need the restriction P > d d the P−intercept of the supply curve, where Q is zero. Recall from Chap. 4 that for c P ≤ , Q = 0. d #34 dQ P ǫ(P) = = aλPλ−1 dP Q



P aPλ



=

aλPλ−1 P =λ aPλ

#35 ǫ(p) =

dQ P = −4P dP Q

ǫ(5) = −

100 = −0.1818 550



P 600 − 2P2



ǫ(6) = −

=−

4P2 600 − 2P2

144 = −0.2727 528

ǫ(8) = −

256 = −0.5424 472

6 Optimal Level of Output and Long Run Price

Demand is unit elastic when

−

113

4P2 = −1 600 − 2P2

4P2 = 600 − 2P2

−→

leading to P = 10. #36 (a) These two goods are substitute goods. (b) We calculate the partial income elasticity as ǫ(I) =

∂Q I 5I = ∂I Q 5I + 10P1 − 0.4P2

Given that income is expressed in $1,000, we must enter 50 for $50,000 in the equation 5 ∗ 50 250 = 1.316 = 5 ∗ 50 + 10 ∗ 10 − 0.4 ∗ 202 190

ǫ(50) =

Similarly, the partial cross price, and own price elasticities are ǫ(P1 ) = ǫ(P) =

10P1 ∂Q P1 = ∂P1 Q 5I + 10P1 − 0.4P2

∂Q P 0.8P2 =− ∂P Q 5I + 10P1 − 0.4P2

→ ǫ(10) =

100 = 0.526 190

→ ǫ(20) = −

320 = −1.684 190

(c) We have −

0.8P2 = −1 5I + 10P1 − 0.4P2

→

0.8P2 = 5I + 10P1 − 0.4P2

1.2P2 = 5I + 10P1 = 5 ∗ 50 + 10 ∗ 10 = 350

→

P = 17.08

#37 η(P) =

2.4P2 ∂Q P 960 = = 21.33 = 2 ∂P Q 1.2P − 0.5PE − 10PL 480 − 35 − 400

η(PE ) =

0.5PE ∂Q PE 35 =− = −0.78 =− 2 ∂PE Q 480 − 35 − 400 1.2P − 0.5PE − 10PL

η(PL ) =

∂Q PL 400 10PL =− =− = −8.89 ∂PL Q 480 − 35 − 400 1.2P2 − 0.5PE − 10PL

114

6 Optimal Level of Output and Long Run Price

#38 (a) dQ P 1 P P = −√ = −1 √ =− √ dP Q P 100 − 2 P 100 P − 2P

ǫ(P) =

√ P = 100 P − 2P

→

√ 3P = 100 P

→

9P = 10000

→

P = 111.11

(b) −

√

P

100 P − 2P

= −1.5

→

√ P = 150 P − 3P

→

P = 1406.25

#39 (a) These two goods are complementary goods. (b) ǫ(I) =

∂Q I 8I = ∂I Q 8I − 9P1 − 0.5P2

Given that income is expressed in $1,000, we must enter 40 for $40,000 in this equation ǫ(40) =

320 8 ∗ 40 = = 2.723 2 8 ∗ 40 − 9 ∗ 10 − 0.5 ∗ 15 117.5

ǫ(P1 ) =

9P1 90 ∂Q P1 =− = −0.766 =− ∂P1 Q 8I − 9P1 − 0.5P2 117.5

ǫ(P) =

∂Q P P2 225 =− = −1.915 =− 2 ∂P Q 8I − 9P1 − 0.5P 117.5

(c) ǫ(P) = −

P2 = −1 8I − 9P1 − 0.5P2

→

P2 = 230 − 0.5P2

1.5P2 = 230 and P = 12.38 #40 ǫ(P) =

∂Q P P = (−2.4P − 0.4) = −1 ∂P Q 6I − 1.2P2 − 0.4P

6 Optimal Level of Output and Long Run Price

2.4P2 + 0.4P = 6 ∗ 57 − 1.2P2 − 0.4P

115

→

3.6P2 + 0.8P − 342 = 0

P = 9.64 Chapter 6 Supplementary Exercises 1. A competitive firm has the following quadratic cost function: TC = aQ2 + bQ + 100 Determine a and b if the following two conditions are satisfied: I. When the market price is 60, the profit maximizing level of output is 15. II. When the market price is 40, the break-even level of output is 10. 2. A firm has a quadratic total cost function given by TC = 6Q2 − 12Q + 80 (a) Specify the bounds of output and price for the function. (b) Find the break-even levels of output if the market price is $38. (c) Show that the profit maximizing level of output when price is $38 is the average of the break-even levels. 3. Assume all firms operating in a competitive market have the following total cost function: TC = 2Q2 − 8Q + 60 (a) What is the long-run price established in the market? (b) Show that at the long-run price the total revenue line is tangent to the total cost curve. 4. Consider a firm with the total cost function TC = 0.00025Q3 − 0.10Q2 + 20Q + 4000 (a) What is the firm’s average cost of producing 500 units? (b) What is the firm’s marginal cost of producing 500th unit? 5. In problem #4, show that the marginal cost curve intersects the average total and the average variable cost curves at their minimum points. 6. There has been an explosion of use of antibiotic for both medical and veterinary (especially in meat and poultry industry) purpose. The following cubic equation is an estimate of billion dozes of antibiotic used each year over 15 year time periods (t = 1 for year 1995),

116

6 Optimal Level of Output and Long Run Price

A(t) = 0.192 t 3 − 2.305 t 2 + 10.95 t + 12.06 (a) Graph A(t) and its average and marginal functions. (b) Calculate A(10) and A′ (10). What is the interpretation of A′ (10)? 7. The maintenance cost of capital equipment—machineries, tools, and buildings— increases as they get older. Assume the maintenance cost of capital equipment, M(t), for a manufacturer is given by M(t) = 30 t 3 + 5000 t + 400 where t is years of equipment in service. (a) Write the average and marginal functions of the maintenance cost. (b) Calculate M(5) and M ′ (5). What is the interpretation of M ′ (5)? 8. A firm operating in a competitive market has the following total cost function: TC = 2.5Q3 − 35.5Q2 + 210.5Q + 320 Find the profit maximizing level of output of this firm if the market price of this good is $165.5. What is the firm’s shutdown price? 9. Assuming all firm in the industry have the same total cost function given in the previous exercise, find the long-run price established in the market. 10. The new Yankees Stadium opened in 2009 has 50,000 seating capacity. If the average ticket price is $60, the average attendance is 25000. When the ticket price is $45, the average attendance increases to 35000. (a) (b) (c) (d)

Find the demand function for Yankees ticket, assuming it is linear. What price should the team charge for a ticket in order to maximize revenue? What would the attendance be at the revenue maximizing ticket price? Show that the price elasticity of demand at the revenue maximizing price and quantity is 1.

11. A hotel in a major metropolitan area has 1500 rooms. The hotel management knows from the past experience that a room rate of $120 a night results in 92 % occupancy. It is also known that a 5 % increase in rate leads to a 3 % decline in the occupancy rate. If demand is linear, what rate should the hotel charge to maximize revenue? What would be the occupancy rate? 12. All firms in a competitive market have the following total cost function TC = 0.5Q2 + 4Q + 120 The market demand and supply functions are given by Qmd = 1700 − 20P where Qmd and Qms are in 100 units.

Qms = −100 + 10P

6 Optimal Level of Output and Long Run Price

117

(a) What is a typical firm’s profit maximizing level of output? (b) What is the firm’s profit? (c) How many firms operate in this market? 13. Assume no change in the cost structure of the firms and in demand in the previous problem, (a) (b) (c) (d)

Determine the long-run price established in this market. What is a typical firm’s output in the long-run? In the long-run how many firms operate in this market? What is the long-run equation of the supply function?

14. A noncompetitive firm with the cost function TC =

1 3 Q − 12Q2 + 185Q + 400 3

is facing the following demand function for its product Q=



6400 − 2P2

(a) Determine the firm’s profit maximizing combination of price and quantity. (b) Determine the firm’s revenue maximizing combination of price and quantity. (c) Show that at the total revenue maximizing price the demand elasticity is unitary. (d) What is the demand elasticity at the profit maximizing price? 15. Show that the price elasticity of demand functions of the form Q=



a − bP2

at the revenue maximizing price is unitary. 16. To generalize from the previous exercise, show that the price elasticity of demand function given as Q = f (P)

dQ 15

find the price elasticity of supply at prices $20 and $25. 26 The demand function for a good is given by Qd = 100 + 0.01I − 5P + 3P1 where I is the average household income and P1 is the price of a related good. (a) What is the relationship between these two goods? (b) calculate the income, cross price, and own price elasticity of demand if I = $35,000, P1 = $20, and P = $35. 27. Consider the demand function for a good given as Qd = 150 + 4I − 0.2P2 − 5P1 where income I is in $1,000 and P1 is the price of a related good. (a) What is the relationship between these two goods? (b) calculate the income, cross price, and own price elasticity of demand if I = $ 45,000, P1 = $30, and P = $25.

120

6 Optimal Level of Output and Long Run Price

(c) At what price demand is own price unit elastic if income is $40,000 and P1 is $35? 28. The demand function for a good is given by Q = 400 − 25P − 0.4P2 At what price the demand elasticity is −0.8? 29. Find the elasticity function for the following demand function and graph it. Determine at what price the function is unit elastic. Qd = 100 −

√ 3P

30. Business cost analysts call the elasticity of profit with respect to output ψ(Q) = dπ Q the degree of operating leverage. Derive the degree of operating leverage dQ π function for problem #24. Show that ψ(Q) is zero at the profit maximizing level of output, while it is extremely large near break-even points. Is there an explanation for this observation? 31. A monopolistically competitive firm has short-run production function given by Q = 5L 0.8 Assume the wage rate is $12 and the firm’s fixed cost is $500. If demand for the firm’s output is P = 350 − 4Q (a) Determine the firm’s profit maximizing level of output, price, and employment. (b) Determine the firm’s profit. 32. In the previous problem, assume the demand for the firm’s product declines to the extent that the firm only makes normal profit. (a) Find the firm’s normal profit combination of output, price, and employment. (b) Find the equation of dislocated demand curve. 33. A firm’s Minimum Efficient Scale (MES) of operation in the short-run is the level of output at which the average cost is minimized. For well-behaved cost functions, MES is reached at the level of output where the output elasticity of dTC Q is equal to 1. Use the cost function given in exercise #24 cost φ(Q) = dQ TC and determine the firm’s MES by minimizing ATC function. Show that at this level of output φ(Q) = 1.

6 Optimal Level of Output and Long Run Price

121

34. Assume the total cost of a firm is given by TC(L, K, R) = 2L 2 +3K 2 +4R2 +3LK +5LR+6KR−35L−39K −53R+54 where L, K, and R are the units of labor, capital and the raw materials (all in 1000 units) used in production. Find the combination of labor, capital and raw materials that minimizes the firm’s total cost.

Chapter 7

Nonlinear Models

Chapter 7, 7.4 Exercises 1. Assume the cubic function

f (x) = x 3 − 15x 2 − 600

(a) Create a table for the values of x from 6 to 20 and their corresponding values for f (x). (b) Graph the f (x) function. (c) Use the information from parts (a) and (b) and find the first and second degree approximations to the cubic function using the Taylor expansion series. 2. Write the linear expansion of the following demand and supply functions about P = 6 separately and then solve the linear system for the market equilibrium price and quantity. Compare your results with the results from Example 2. Qd =

100P + 1500 P 2 + 10

Q s = 10P 0.7

3. Assume all firms in a perfectly competitive market have the following total cost function T C(Q) =

1 3 Q − 2.5Q 2 + 30Q + 100 12

Follow steps discussed in Example 3 and find the linear approximation of the market long run equilibrium price. 4. Find linear and quadratic approximations for the market long run price if all firms in a competitive market have the following T C function T C = Q 3 − 15Q 2 + 120Q + 600 5. The market demand function for product of a monopolist with the total cost function © Atlantis Press and the authors 2015 S. Vali, Principles of Mathematical Economics II, Mathematics Textbooks for Science and Engineering 4, DOI 10.2991/978-94-6239-088-1_7

123

124

7 Nonlinear Models

T C(Q) = 2Q 3 − 10Q 2 + 30Q + 100 is given by  P = 65000 − 8 Q

(a) Find the firm’s profit-maximizing level of output and price. (b) Assume the economy experiences a severe recession and the demand for the monopolist’s product declines to the level that the firm can only earn a normal profit. Determine the normal profit price and quantity and the equation of the dislocated demand function. 6. Assume the following market demand and supply functions Q d = 100 − 5P 0.7 Q s = (2P − 10)1.1 (a) Graph the functions. (b) Graph the excess demand function. (c) Use the Taylor expansion and find the first degree or linear approximation of market equilibrium price and quantity. 7. Consider the following market demand and supply functions Qd =

700 − 10 2P + 3

Q s = (2P − 15)0.7 (a) Graph the excess demand function. (b) Use the Taylor expansion of the excess demand function and find the first degree approximation for the market equilibrium price and quantity. 8. Assume the following market demand and supply functions Q d = 1000



P + 25 P2 + 5



Q s = 100P 0.4 (a) Graph the excess supply function. (b) Use the Taylor expansion of the excess supply function and find the first degree approximation for the market equilibrium price and quantity.

7 Nonlinear Models

125

(c) Write the linear expansion of the demand and supply functions separately and then solve the linear system for the market equilibrium price and quantity. Compare your results with those from part (b). 9. A firm with a certain amount of resources produces two goods. If the firm produces Q 1 units of good one, then it could produce Q 2 = 85 −

700 900 − Q 21

Q 1 < 30

units of good two. Assume price of good one P1 is $60 per unit and that of good two P2 is $90 per unit. (a) Use the first degree Taylor approximation and determine the combination of good one and two that maximizes the firm’s revenue. (b) Use the quadratic Taylor approximation and determine the combination of good one and two that maximizes the firm’s revenue. 10. Consider the following market supply and demand functions Q s = (2P − 10)0.6 Qd =

700 − 10 2P + 3

(a) Graph the excess supply function. (b) Find the linear approximation of excess supply function and determine the equilibrium price and quantity. 11. A monopolist operates two plants. Total cost functions for the two plants, denoted as T C(Q 1 ) and T C(Q 2 ), are T C(Q 1 ) = 3Q 31 − 5Q 21 + 50Q 1 + 400 T C(Q 2 ) = Q 32 − 5Q 22 + 20Q 2 + 100 Assume that the market demand for the monopolist’s product is P = 400 − 5Q Determine the monopolist’s profit maximizing level of output and price, and find the optimal division of output between its two plants. 12. A monopolist operates two plants. Assume that the first and second plant total cost functions, denoted as T C(Q 1 ) and T C(Q 2 ), are T C(Q 1 ) = 0.6Q 31 − 3Q 21 + 6Q 1 + 50

126

7 Nonlinear Models

T C(Q 2 ) = Q 32 − 10Q 22 + 45Q 2 + 100 Assume that the market demand for the monopolist’s product is Q = 250 − 2P Determine the monopolist’s profit maximizing level of output and price, and find the optimal division of output between its two plants. 13. A monopolist operates two plants. Assume that the first and second plant total cost functions, denoted as T C(Q 1 ) and T C(Q 2 ), are T C(Q 1 ) = 0.2Q 31 − 7Q 21 + 136Q 1 + 100 T C(Q 2 ) = 3Q 32 − 4Q 22 + 25Q 2 + 900 Assume that the market demand for the monopolist’s product is P = 1000 − 5Q Determine the monopolist’s profit maximizing level of output and price, and find the optimal division of output between its two plants. 14. In Example 6, linear approximate the M R function over the interval 125 < Q < 130, and then find the profit-maximizing number of housing units along with their price. 15. In Example 6, linear approximate the demand function over the interval 125 < Q < 140 and then find the profit-maximizing number of housing units along with their price. 16. Assume the following market supply and demand functions Q s = 10 e0.1P − 20 Q d = 150 e−0.2P Determine the equilibrium price and quantity. 17. A firm operating in a noncompetitive market has the following cost function T C = Q 3 − 20Q 2 + 150Q + 1000 The market demand function for the firm’s product is Q = 3000 − 200P 0.8 Find the firm’s profit-maximizing combination of price and quantity.

7 Nonlinear Models

127

18. Consider a market for a commodity with the following demand and supply functions: Qd =

100P + 1500 P 2 + 10

Q s = P(3)0.02P − 10 Use the Taylor expansion and find the first degree or linear approximation of the market equilibrium price and quantity. 19. Find the equilibrium of the following market Q d = 100 − 10P 0.3 Q s = −10 + 5P 1.1 20. Assume a Keynesian model for a closed economy with C = 506 + 0.9177Y − 0.0001355Y 2 I = 35 + 0.15Y If G 0 = 545, determine the equilibrium values of the endogenous variables. 21. Assume a Keynesian model for a closed economy with C = 506 + 0.9177Y − 0.0001355Y 2 √ I = 30 + 0.15Y + 4.3 Y

Assume G 0 = 545 and write the reduced form of the model. Linearly approximate Y by expanding the reduced form around 3390. 22. Using time series data for the United States from 1990 to 2006,1 the following Consumption and Investment functions are estimated C = −238.64 + 0.683Y − 0.0001355Y 2 I = −219.0 + 0.18Y Assume a government expenditure of $2216.8 and a net export of −$615.4, (a) determine the equilibrium national income, consumption and investment. (b) What is the model multiplier? 1

Economic Report of the President, 2008. Statistical Tables.

128

7 Nonlinear Models

23. By considering consumption C a function of disposable income Yd and using the same data set as in problem (22), the following consumption function is estimated C = −1016.48 + 0.966Yd − 0.0000087Yd2 An investment and tax function is also estimated I = −219.0 + 0.18Y T = 0.1043Y Assume a government expenditure of $2216.8 and a net export of −$615.4, (a) determine the equilibrium national income, consumption and investment. (b) find the multiplier. Answers to Chapter 7, 7.4 Exercises #1 The following table shows values of f (x) for selected values of x. x 6 8 10 12 14 16 18 20 f (x) −924 −1048 −1100 1032 −796 −344 372 1400

Figure 7.1 is the graph of the function for values in the table. It is clear from the table and the graph that this function has a zero at a value of x between 16 and 18. Let us expand the function around 17,

Fig. 7.1 Graph of f (x) = x 3 − 15x 2 − 600

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129

f (17) = −22 f ′ (x) = 3x 2 − 30x f ′′ (x) = 6x − 30

→ →

f ′ (17) = 357 f ′′ (17) = 72

Using these pieces, the linear and quadratic approximations to the function are fl (x) = −22 + 357(x − 17) f q (x) = −22 + 357(x − 17) +

72 (x − 17)2 = 36(x − 17)2 + 357x − 6091 2

By setting the linear function fl (x) equal to zero, we obtain x = 17.062 as a solution to x 3 − 15x 2 − 600 = 0. Similarly, by setting f q (x) equal to zero, we find the quadratic approximation x = 17.061. Microsoft Mathematics, using cubic formula, provides a solution x = 17.061. Clearly the linear solution is a very good approximation, and the quadratic is right on the money. #2 We first calculate the necessary components of linear expansions Q d (6) = Q ′d =

600 + 1500 = 45.6522 46

100(P 2 + 10) − 2P(100P + 1500) (P 2 + 10)2

Q ′d (6) =

4600 − 25200 = −9.7353 2116

The linear approximation to the demand function about 6, Q ld , is Q ld = 45.6522 − 9.7353(P − 6) = 104.064 − 9.7353P Q s (6) = 10(6)0.7 = 35.0514 Q ′s = 7(P)−0.3 =

7 P 0.3

Q ′s (6) =

7 = 4.0893 (6)0.3

The linear approximation to the supply function about 6, Q ls , is Q ls = 35.0514 + 4.0893(P − 6) = 10.5156 + 4.0893P By using the market equilibrium condition, we have 104.064 − 9.7353P = 10.5156 + 4.0893P

→

13.8246P = 93.5458

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And the market price is P = 6.7668, with the corresponding market equilibrium quantity Q = 38.1871. The results of linear approximation of function in Eq. (7.2) in the text are P = 6.96 and Q = 38.23. #3 To find the long-run price we must determine the minimum of the average total cost. AT C = F OC :

100 TC 1 2 = Q − 2.5Q + 30 + Q 12 Q AT C ′ =

100 1 Q − 2.5 − 2 = 0 6 Q

1 3 Q − 2.5Q 2 − 100 = 0 6

(7.1SM)

which is equivalent of finding the zero(s) of the function f (Q) =

1 3 Q − 2.5Q 2 − 100 6

The following table shows value of the function f (Q) for selected values of Q, Q 5 10 15 16 17 18 19 f (Q) −141.67 −183.33 −100.00 −57.33 −3.67 62.00 140.67

Figure 7.2 is the graph of the function f (Q). The above table and Fig. 7.2 indicate that the zero of the function is close to 17. We then expand the function about 17 for a linear approximation.

Fig. 7.2 Graph of 1 3 2 6 Q − 2.5Q − 100

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f ′ (Q) =

131

1 2 Q − 5Q 2 f ′ (17) = 59.5

f (17) = −3.6667

fl (Q) = −3.6667 + 59.5(Q − 17) = −1015.1667 + 59.5Q The linear approximation to f (Q) is −1015.1667 + 59.5 = 0

Q = 17.0616

−→

Microsoft Mathematics solution to the equation (7.1SM), using cubic formula, is 17.0612. #4 Similar to problem #3 we must find the minimum of the average total cost function, AT C = Q 2 − 15Q + 120 +

600 Q

600 =0 Q2

AT C ′ = 2Q − 15 −

−→

2Q 3 − 15Q 2 − 600 = 0

As usual, solution to this cubic equation is zero(s) of the following cubic function f (Q) = 2Q 3 − 15Q 2 − 600 A table of values for Q indicates that zero of this function is between 10 and 11. We expand the function around 10 for linear and quadratic approximation. f (10) = −100 f ′ (Q) = 6Q 2 − 30Q f ′′ (Q) = 12Q − 30

−→ −→

f ′ (10) = 300 f ′′ (10) = 90

fl (Q) = −100 + 300(Q − 10) = −3100 + 300Q fl (Q) = 0

−→

Q = 10.333

f q (Q) = −100 + 300(Q − 10) + f q (Q) = 0

−→

90 (Q − 10)2 = 45Q 2 − 600Q + 1400 2

45Q 2 − 600Q + 1400 = 0 and Q = 10.318

The quadratic approximation is quite accurate.

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#5 (a) The profit function is  π = T R − T C = 65000Q − 8Q Q − 2Q 3 + 10Q 2 − 30Q − 100 π = −2Q 3 + 10Q 2 − 8Q 1.5 + 64970Q − 100  π ′ = −6Q 2 + 20Q − 12 Q + 64970 = 0

Solution to this√equation is Q = 105.64 and the corresponding price is P = 65000 − 8 105.64 = 64917.77. (b) At the normal profit situation, we have dP d AT C = dQ dQ 4Q − 10 −

4 100 = −√ 2 Q Q

4Q 3 − 10Q 2 + 4Q 1.5 − 100 = 0 Solution to this equation is Q = 3.756 and P = AT C(3.756) = 47.279. The equation of dislocated demand is √ 47.279 = a − 8 3.756  P = 62.783 − 8 Q

−→

a = 62.783

which shows a dramatic decline in demand for the product of this monopolist. This example highlights the possibility of a monopolist losing its market. This often happens when a new and technologically more advance product comes to market as substitute for the monopolist’s product. An interesting example is a total plunge in demand for “slide ruler” when hand held calculator was introduced to the market. #6 (a) Figure 7.3 is the graph of supply and demand functions. In case you wish to generate supply and demand curves in standard format, P on the vertical axis, you must first solve each equation for P in terms of Q, as follows Q d = 100 − 5P 0.7

→

P = (20 − 0.2Q)1.4286

5P 0.7 = 100 − Q

→

P 0.7 = 20 − 0.2Q (7.2SM)

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133

Fig. 7.3 Graph of supply and demand

Q s = (2P − 10)1.1

→

2P − 10 = Q 0.909

P = 0.5Q 0.909 + 5

(7.3SM)

Now, if you graph (7.2SM) and (7.3SM) together you get the more familiar demand and supply curves. (b) The excess demand function is E D(P) = Q d − Q s = 100 − 5P 0.7 − (2P − 10)1.1 Figure 7.4 is the graph of the excess demand function. (c) It can be deduced from the graph of the excess demand function that zero of the function is about 23. We expand the function around this number for deriving the linear approximation. E D(23) = 3.592 E D ′ (Q) = −3.5P −0.3 − 2.2(2P − 10)0.1

Fig. 7.4 Graph of excess demand

−→

E D ′ (23) = −4.514

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Fig. 7.5 Graph of E D(P) = 700 − 10 − (2P − 15)0.7 2P + 3

E Dl (Q) = 3.592 − 4.514(P − 23) = 107.414 − 4.514P E Dl (Q) = 0

−→

P = 23.796 and Q = 54.02

#7 The excess demand function is E D(P) = Q d − Q s =

700 − 10 − (2P − 15)0.7 2P + 3

(a) Figure 7.5 is the graph of the excess demand function. (b) As the graph indicates, the excess demand function intersect the P axis half way between 15 and 20. So we expand the function about 17.5. E D(17.5) = 0.27924 E D ′ (P) = −

1400 −1.4(2P −15)−0.3 (2P + 3)2

→

E D ′ (17.5) = −1.5395

The first degree or the linear approximation of the excess demand function is E Dl (P) = E D(17.5)+E D ′ (17.5)(P−17.5) = 0.27924−1.5395(P−17.5) E Dl (Q) = 27.22 − 1.5395P The zero of this function is the approximate market equilibrium price 27.22 − 1.5395P = 0

→

P = 17.68

And the market equilibrium quantity is Q = 8.24

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135

Fig. 7.6 Graph of  E S(P)  = 100P 0.4 − 1000 P+25 P 2 +5

#8 The excess supply function is E S(P) = Q s − Q d = 100P

0.4

− 1000



P + 25 P2 + 5



(a) Figure 7.6 is the graph of the excess supply function E S(P). (b) The graph indicates that the zero of the excess supply function is in the vicinity of P = 11. We expand the function around this value, E S(11) = −24.764 − 1000



P 2 + 5 − 2P(P + 25) (P 2 + 5)2

E S ′ (P) = 40P −0.6 − 1000



−P 2 − 50P + 5 (P 2 + 5)2

′

E S (P) = 40P

−0.6





E S ′ (11) = 36.196 Thus the linear approximation to the function around 11.5 is E Dl (P) = −24.764 + 36.196(P − 11) = −422.92 + 36.196P Zero of this linear function is the approximate market equilibrium price E Dl (P) = 0

→

P = 11.68

And the market equilibrium quantity is approximately Q = 267.32. (c) In this case we must expand both functions about P = 11.

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Q d (11) = 285.714 Q ′d

−P 2 − 50P + 5 = 1000 (P 2 + 5)2 



Q ′d (11) = −26.707

→

Q dl = 285.714 − 26.707(P − 11) = 579.49 − 26.707P Q s (11) = 260.95 Q ′s = 40P −0.6

Q ′s (11) = 9.489

→

Q sl = 260.95 + 9.489(P − 11) = 156.571 + 9.489P Using the equilibrium condition Q dl = Q sl , we have 579.49 − 26.707P = 156.571 + 9.489P

→

36.196P = 422.919

which leads to exactly the same answer we obtained from linearizing the excess supply function about 11, P = 11.68. #9 We formulate this problem as Maximize

T R(Q 1 , Q 2 ) = P1 Q 1 + P2 Q 2 = 60Q 1 + 90Q 2

Subject to: Q 2 = 85 −

700 900 − Q 21

Q 1 < 30

After substituting for Q 2 in the total revenue function, we have 

700 T R(Q 1 ) = 60Q 1 + 90 85 − 900 − Q 21



= 60Q 1 −

63000 + 7650 900 − Q 21

To maximize revenue, T R ′ (Q 1 ) = 60 −

126000Q 1 =0 (900 − Q 21 )2

After simplifying the above equation, we arrive at Q 41 − 1800Q 21 − 2100Q 1 + 810000 = 0 Solution to this forth degree equation is (are) zero(s) of the function f (Q 1 ) = Q 41 − 1800Q 21 − 2100Q 1 + 810000

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137

Fig. 7.7 Graph of f (Q 1 ) = Q 41 − 1800Q 21 − 2100Q 1 + 810000

The graph of this function, Fig. 7.7, indicates that the function has two solutions, one around 25 and the other one around 35. Since the second solution violates the constraint Q 1 < 30, we concentrate on expanding the function around 25. f (25) = 23125 f ′ (Q 1 ) = 4Q 31 − 3600Q 1 − 2100 f ′′ (Q 1 ) = 12Q 21 − 3600

−→

−→

f ′ (25) = −29600

f ′′ (25) = 3900

The linear approximation is fl (Q 1 ) = 23125 − 29600(Q 1 − 25) and fl (Q 1 ) = 0

−→

Q 1 = 25.781

The quadratic approximation is f q (Q 1 ) = 23125 − 29600(Q 1 − 25) + 1950(Q 1 − 25)2 = 1950Q 21 − 127100Q 1 + 1981875

This function has two zeros, 25.826 and 39.353. The second solution must be discarded, it violates Q 1 < 30 restriction. The quadratic approximation, therefore, is 25.826. This approximation is very good, given that Wolfram Alpha’s solution to the original fourth degree equation is 25.828. #10

The excess supply function is

E S(P) = Q s − Q d = (2P − 10)0.6 −

700 + 10 2P + 3

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Fig. 7.8 Graph of 700 (2P − 10)0.6 − + 10 2P + 3

(a) Figure 7.8 is the graph of the excess supply function E S(P). (b) As the graph of the excess supply function indicates this function has a zero near P = 19. We expand the function about this value, E S(19) = 0.311 E S ′ (P) = 1.2(2P − 10)−0.4 +

1400 (2P + 3)2

→

E S ′ (19) = 1.1493

E Sl (P) = 0.311 + 1.1493(P − 19) = −21.5257 + 1.1493P Zero of this function is −21.5257 + 1.1493P = 0

→

P = 18.73

The market equilibrium quantity is Q = 7.298 units. #11 Let denote the sum of the two plants output by Q, that is Q 1 + Q 2 = Q. At optimal level of output it must be true that MC(Q 1 ) = MC(Q 2 ) = M R(Q) MC(Q 1 ) = 9Q 21 − 10Q 1 + 50 MC(Q 2 ) = 3Q 22 − 10Q 2 + 20

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139

Expressing the marginal revenue function as T R(Q) = P Q = 400Q − 5Q 2 −→ M R(Q) = 400 − 10Q = 400 − 10(Q 1 + Q 2 ) we then have to solve the following set of two equations with two unknowns:

9Q 21 − 10Q 1 + 50 = 400 − 10(Q 1 + Q 2 ) 3Q 22 − 10Q 2 + 20 = 400 − 10(Q 1 + Q 2 )

This system is simplified to

9Q 21 + 10Q 2 − 350 = 0 3Q 22 + 10Q 1 − 380 = 0

We solve the first equation for Q 2 in terms of Q 1 and substitute in the second equation 10Q 2 = 350 − 9Q 21

−→

Q 2 = 35 − 0.9Q 21

(7.4SM)

3(35 − 0.9Q 21 )2 + 10Q 1 − 380 = 3675 + 2.43Q 41 − 189Q 21 + 10Q 1 − 380 = 0 2.43Q 41 − 189Q 21 + 10Q 1 + 3295 = 0 Solution to this forth degree equation is the zero(s) of the function f (Q 1 ) = 2.43Q 41 − 189Q 21 + 10Q 1 + 3295 The graph of this function, Fig. 7.9, indicates that this function has two positive zeros about 5 and 7. By using Wolfram Alpha we found the positive solutions as 5.2225 and 7.0830. By plugging 5.2225 for Q 1 in equation (7.4SM) we obtain 10.5834 for Q 2 . The second solution for Q 1 leads to a negative value

Fig. 7.9 Graph of f (Q 1 ) = 2.43Q 41 − 189Q 21 + 10Q 1 + 3295

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for Q 2 , therefore must be discarded. The combination (Q 1 , Q 2 ) = (5.2225, 10.5834) is the optimal division of output between two plants. The total output is Q = Q 1 + Q 2 = 15.8059. It could be easily verified that at this combination M R(Q) = MC(Q 1 ) = MC(Q 2 ). At the optimal level of output Q = 15.8059 the price is P = 400 − 5 ∗ 15.8059 = 320.97. #12 Exercise #12 is similar to #11. To determine the marginal revenue, we must first solve Q = 250 − 2P for P in terms of Q. P = 125 − 0.5Q −→ T R = 125Q − 0.5Q 2 −→ M R = 125 − Q After substituting Q 1 + Q for Q in the marginal revenue function, we have M R = 125 − Q 1 − Q 2 Next, we set the marginal costs of plant 1 and 2 equal to the marginal revenue M R(Q 1 ) = 1.8Q 21 − 6Q 1 + 6 = 125 − Q 1 − Q 2 M R(Q 2 ) = 3Q 22 − 20Q 2 + 45 = 125 − Q 1 − Q 2 After simplification, we obtain the following 2 equations with 2 unknowns:

1.8Q 21 − 5Q 1 + Q 2 − 119 = 0 3Q 22 − 19Q 2 + Q 1 − 80 = 0

If in the first equation we solve for Q 2 and substitute that in the second equation, we arrive at 3(−1.8Q 21 + 5Q 1 + 119)2 − 19(−1.8Q 21 + 5Q 1 + 119) + Q 1 − 80 = 0 After some manipulation, we obtain the following fourth degree polynomial: 9.72Q 41 − 54Q 31 − 1176Q 21 + 3476Q 1 + 40142 = 0 This equation has two positive solutions, Q 1 = 9.33 and Q 1 = 9.7252. We must discard the second solution because it leads to a negative value for Q 2 . Using the first solution for Q 1 , we obtain Q 2 = 8.962. Thus, the total output is Q = 18.292 and the price is P = 115.854. #13 This problem is similar to the last two problems. After equating the marginal costs of Q 1 and Q 2 with the marginal revenue and manipulating the resultant equations, we arrive at a fourth degree polynomial. 0.0216Q 41 − 0.2883Q 31 − 60.072Q 21 + 416.88Q 1 + 42140.8 = 0

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141

This equation has two positive solutions Q 1 = 38.788 and Q 1 = 43.152. We must discard the second solution because it generates a negative value for Q 2 . Thus, the profit maximizing division of output between to plants is Q 1 = 38.788 and Q 2 = 11.684 with total output Q = 50.47 and price P = 747.64. #14

The answer to this exercise is given in the text on page 211.

#15 The nonlinear demand function in Example 6 is given as Q d = 1000 − 10P 0.4 To linearize this function over the interval 125 < Q < 140, we must first solve the function for P in terms of Q. 10P 0.4 = 1000 − Q

−→

P = (100 − 0.1Q)2.5

For Q = 125 and Q = 140 we have the corresponding values of P = 71617.66 and P = 68587.68. Therefore, the equation of the line joining the points (125, 71617.66) and (140, 68587.68) is the linear approximation of the demand function over the interval. The equation of this line is P−71617.66 =

68587.68 − 71617.66 (Q−126) 140 − 125

→

P = 96867.66 − 202Q

with the associated marginal revenue function M R = 96867.66 − 404Q. The firm’s total cost function in Example 6 is T C = Q 3 − 20Q 2 + 950Q + 650000. Thus, the profit maximizing number of units is obtained from M R = MC 96867.66 − 404Q = 3Q 2 − 40Q + 950

→

3Q 2 + 364Q − 95917.66

The acceptable solution to this quadratic equation is Q = 129 units. #16 By invoking the equilibrium condition, we obtain 10 e0.1P − 20 = 150 e−0.2P 10 e0.1P − 20 =

150 e0.2P

10 e0.3P − 20 e0.2P − 150 = 0 Solution to this natural exponential equation is P = 12.07. Subsequently, the equilibrium quantity is Q = 13.4257. #17 To write the profit function we must first express P as function of Q Q = 3000 − 200P 0.8

−→

200P 0.8 = 3000 − Q

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P = (15 − 0.005Q)1.25 Now we can express the profit function as π = T R − T C = Q(15 − 0.005Q)1.25 − Q 3 + 20Q 2 − 150Q − 1000 The first order condition for a local maximum is π ′ = (15 − 0.005Q)1.25 + 1.25Q(−0.005)(15 − 0.005Q)0.25 − 3Q 2 + 40Q − 150 = 0 (15 − 0.005Q)1.25 − 0.00625Q(15 − 0.005Q)0.25 − 3Q 2 + 40Q − 150 = 0

This equation has two solutions, with the largest value Q = 8.7192 satisfying the second order condition. The equilibrium price is P = (15 − 0.005 ∗ 8.7192)1.25 = 29.41 #18 To solve this problem using linear approximation requires derivative of an exponential function which is discussed in Chap. 9. Example 3 of Chap. 9 (pages 298– 300) is the solution to this problem using both the linear and quadratic approximation. We solve this problem here by using Wolfram Alpha and refer you to Example 3 of Chap. 9 for the linear approximation. Using the equilibrium condition, we have 100P + 1500 = P(3)0.02P − 10 P 2 + 10   100P + 1500 = (P 2 + 10) P(3)0.02P − 10 = P 3 (3)0.02P −10P 2 + 10P(3)0.02P − 100

P 3 (3)0.02P − 10P 2 + 10P(3)0.02P − 100P − 1600 = 0 Solution to this equation is equivalent to zero(s) of the function f (P) = P 3 (3)0.02P − 10P 2 + 10P(3)0.02P − 100P − 1600 The graph of this function, Fig. 7.10, indicates that this function has a zero about 16. Using Wolfram Alpha, we obtain P = 15.63. The corresponding equilibrium quantity is Q = 12.04. #19 Invoking the equilibrium condition, we have 100 − 10P 0.3 = −10 + 5P 1.1 5P 1.1 + 10P 0.3 − 110 = 0

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143

Fig. 7.10 Graph of f (P) = P 3 (3)0.02P − 10P 2 + 10P(3)0.02P − 100P − 1600

Fig. 7.11 Graph of f (P) = 5P 1.1 + 10P 0.3 − 110

Solution to the above equation, using Microsoft Mathematics, is P = 13.58. The market equilibrium quantity is Q = 78.13. Figure 7.11 is the graph of f (P) = 5P 1.1 + 10P 0.3 − 110 #20 Y = C + I + G 0 = 506 + 0.9177Y − 0.0001355Y 2 + 35 + 0.15Y + 545 Y = 1086 + 1.0677Y − 0.0001355Y 2 −0.0001355Y 2 + 0.0677Y + 1086 = 0 The acceptable solution to this quadratic equation is Y = 3091.851. The equilibrium values of the other endogenous variables are C = 2048.074 and I = 498.778.

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#21 √ Y = C + I + G 0 = 506 + 0.9177Y − 0.0001355Y 2 + 30 + 0.15Y + 4.3 Y + 545 √ Y = 1081 + 1.0677Y − 0.0001355Y 2 + 4.3 Y

√ −0.0001355Y 2 + 0.0677Y + 4.3 Y + 1081 = 0 The equilibrium nation income level is Y = 3394.52 and C = 2059.82 and I = 789.71. To linearly approximate by expanding around 3390, we write √ f (Y ) = −0.0001355Y 2 + 0.0677Y + 4.3 Y + 1081 f (3390) = 3.6854 4.3 f ′ (Y ) = −0.0002710Y + 0.0677 − √ 2 Y f ′ (3390) = −0.8879 fl (Y ) = 3.6854 − 0.8879(Y − 3390) = 3013.6664 − 0.8879Y fl (Y ) = 0

Y =

−→

3013.6664 = 3394.15 0.8979

#22 (a) Denoting the net export by N X , we write the national income equation as Y = C + I + G0 + N X0 Y = −238.64 + 0.683Y − 0.0001355Y 2 − 219.0 + 0.18Y + 2216.8 − 615.4 0.0001355Y 2 + 0.137Y − 1143.76 = 0 Solution to this quadratic equation is Y = 2443.5. C and I are, 220.83 and 621.24, respectively. Note that a large part of consumption and investment in this country is financed by other countries. (b) In this problem f (Y ) = 0.683Y − 0.0001355Y 2 and g(Y ) = 0.18Y . The multiplier, m, is calculated as m=

1 1−

f ′ (Y ) − g ′ (Y )

f ′ (Y ) = 0.683 − 0.0002710Y and g ′ (Y ) = 0.18

7 Nonlinear Models

m=

145

1 1 = 1 − 0.683 + 0.0002710Y − 0.18 0.137 + 0.0002710Y

Evaluating the multiplier at equilibrium Y , we have m=

1 = 1.2513 0.79918

#23 (a) Given that Yd = Y − T, we write the consumption function as C = −1016.48 + 0.966(Y − 0.1043Y ) − 0.0000087(Y − 0.1043Y )2 C = −1016.48 + 0.86525Y − 0.00000698Y 2 We now have Y = −1016.48 + 0.86525Y −0.00000698Y 2 − 219 + 0.18Y + 2216.8− 615.4 leading to the following quadratic equation −0.00000698Y 2 + 0.04525Y + 365.92 = 0 This equation has a positive solution Y = 11174.30. Values of the other endogenous variables are C = 7780.53, I = 1792.37, and T = 1165.48. (b) We have f (Y ) = 0.86525Y − 0.00000698Y 2 and g(Y ) = 0.18Y f ′ (Y ) = 0.86525 − 0.00001396Y and g ′ (Y ) = 0.18 Thus, we have the multiplier m=

1 1 = 1 − f ′ (Y ) − g ′ (Y ) 1 − 0.86525 + 0.00001396Y − 0.18

m=

1 −0.04525 + 0.00001396Y

Evaluating the multiplier at equilibrium Y , we have m=

1 = 9.02 −0.04525 + 0.156

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7 Nonlinear Models

Chapter 7 Supplementary Exercises 1. Find the marker equilibrium price and quantity if the market demand and supply functions are given as Q d = 2500 − 24P

Q s = −2000 + 2P 2

2. Assume the market demand and supply functions are given by Qd =

3000 2P 2 + 5

Q s = −50 + 5P 1.2

Determine the market equilibrium price and quantity. 3. In the previous problem, expand the demand and supply function about P = 9 and write the linear approximation for both functions. Determine the market equilibrium price and quantity using the linear functions. Compare your results to the results of the previous exercise. 4. Consider a noncompetitive firm with the total cost function T C = 0.5Q 3 − 15Q 2 + 160Q + 200 facing the following demand function for its product  P = 2500 − 3 Q 2 + 100

Figure 7.12, the graph of the derivative of profit function, indicates that there is real solution near Q = 48. Expand the demand function around Q = 48 and write the linear approximation to the function. Use the approximated linear demand function and determine the approximate equilibrium price and quantity. 5. In problem #4, use Wolfram Alpha or Excel and find more accurate values for the equilibrium price and quantity. How good is your approximation in problem #4? Fig. 7.12 Graph of the derivative of profit function

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147

6. Assume due to bad economic environment, the firm in problem #4 can only earn a normal profit. Determine its normal profit level of output and price. Write the equation of the dislocated demand curve. 7. The offer functions for quantity (in million units) of goods x and y traded between two countries are given by y=

x(60 − x)2 1000

x = 7y − 0.35y 2

What is the volume of each good traded at equilibrium? What is the terms of trade? 8. Determine the equilibrium price and quantity for the following market Q d = 350 − 15P 0.5 Q s = −77.583 + 10P 1.3 9. In the previous problem, linear approximate both supply and demand equation about the solution you determined for the market price. Use the linearized equations and solve for the market equilibrium price and quantity. You should get exactly the same answers as in problem #8. 10. In problem #8 linear approximate both supply and demand around a value 0.5 point less than the actual solution you found for the market price. Used the linearized supply and demand equation and find the market equilibrium price and quantity. What is the error rate of your approximation? 11. A manufacturer produces an item in two different plants. The cost function for each plant is given by T C(Q 1 ) = Q 31 − 15Q 21 + 120Q 1 + 200 T C(Q 2 ) = 2Q 32 − 25Q 22 + 160Q 2 + 400 If the market demand for the firm’s product is given by P = 500 − 6Q find the profit maximizing level of output at each plant and price. What is the firm’s profit?2

2

For this, and similar problems that involves solving a system of nonlinear equations, you can use Microsoft Mathematics. Click on “Equation Solver”, then choose “Solve a System of 2 Equations”. Type the equations in the boxes for Eq. 1 and Eq. 2 and then click “Solve”.

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7 Nonlinear Models

12. Similar to the previous problem, assume a multi-plant firm is operating 2 plants with the following cost functions: T C(Q 1 ) =

1 3 Q − 10Q 21 + 110Q 1 + 500 3 1

T C(Q 2 ) =

1 3 Q − 20Q 22 + 280Q 2 + 600 2 2

If the market demand for the firm’s product is given by Q = 300 − 0.5P find the profit maximizing level of output at each plant and price. What is the firm’s profit? 13. A firm produces two goods. If it produces Q 1 unit of good 1, then it can produce Q 2 = 120 −

900 1100 − 1.5Q 21

Q 1 < 27.08

units of good 2. If prices of goods 1 and 2 are $80 and $100 per unit, determine the combination of two goods that maximizes the firm’s revenue. 14. The joint product equation for a firm producing two goods is Q 1 = 250 −

150 300 − 2Q 1.5 2

Q 2 < 28.23

If the price of good 1 and good 2 are $150 and $200 a unit, find the revenue maximizing combination of two goods. 15. Figure 7.13 is the graph of the derivative of the total revenue function in problem # 14. The graph indicates that the derivative has a zero about Q 2 = 25 Fig. 7.13 Graph of the Derivative of Revenue function

7 Nonlinear Models

149

(the other solution violates the restriction on Q 2 ). Find the linear approximation to the derivative around 25 and estimate the revenue maximizing combination of the two goods. 16. Assume the market demand for a good is Q = 2500 − 20P 0.8 This good is produced by a monopolist with the following cost function TC =

1 3 Q − 12Q 2 + 210Q + 5000 3

Find the profit maximizing level of output and price. 17. Assume the market demand for the monopolist in problem #16 declines to extent that it can only earn a normal profit. Determine the firm’s normal profit price and output combination. Write the equation of dislocated demand curve. 18. In problem #16, write the market demand equation associated with the shutdown price.

Chapter 8

Additional Topics in Perfect and Imperfect Competition

Chapter 8, 8.6 Exercises 1. Assume a competitive industry consists of 300 firms each with the following cost function: TC = Q3 − 10Q2 + 45Q + 100 If the market demand function is Qd = 4000 − 20P what is the market short run equilibrium price and quantity? 2. Determine the long-run market equilibrium price and quantity in Problem 1. How many new firms enter or leave the market? 3. Assume a noncompetitive firm with a total cost function TC = Q3 − 20Q2 + 150Q + 1000 facing the following linear demand function for its product in the market P = 250 − 2Q Derive the trace function of the firm. 4. All 150 firms in a competitive industry have the same cost structure captured by the following total cost function: TC = 2Q3 − 60Q2 + 950Q + 600

© Atlantis Press and the authors 2015 S. Vali, Principles of Mathematical Economics II, Mathematics Textbooks for Science and Engineering 4, DOI 10.2991/978-94-6239-088-1_8

151

152

8 Additional Topics in Perfect and Imperfect Competition

If the market demand function is Qd = 4568.20 − 3.5P determine the market equilibrium price and quantity. 5. If one of the firms in Problem 4 acquires the other 149 firms in the industry and forms a giant monopoly, what would the market equilibrium price and quantity be? Average Cost Pricing Instead of pursuing profit maximization through marginal cost pricing, the regulated natural monopolists are required to follow the average cost pricing. In the average cost pricing, a natural monopolist firm sets its price at the level where the demand curve intersects the average cost ATC curve. This policy, compared to marginal cost pricing, leads to the production of a larger volume of output at a relatively lower price. 6. If the newly formed monopolist in Problem 5 is regulated and required to follow the average cost pricing, what would the market equilibrium price and quantity be? 7. A monopolistically competitive firm has the following total cost function: TC =

1 3 Q − 20Q2 + 350Q + 100 3

If the demand function for the firm’s product is P = 350 − 10Q (a) Derive the firm’s trace function. (b) What is the equation of the dislocated demand curve associated with normal profits? (c) What is the equation of the dislocated demand curve associated with a price of $206? 8. A monopolist has the following cubic cost function: TC = 0.8Q3 − 32Q2 + 1000Q + 32000 If the market demand function for the firm’s product is P = 3400 − 14Q (a) What is the monopolist’s profit-maximizing combination of price and quantity?

8 Additional Topics in Perfect and Imperfect Competition

153

(b) What is the firm’s profit? (c) Derive the firm’s trace function. (d) What is the equation of the dislocated demand curve associated with normal profits? (e) What is the equation of the dislocated demand curve associated with a price of $3,200? 9. A monopolist has the following cubic cost function1 : TC = 2Q3 − 20Q2 + 136Q + 100 If the market demand function for the firm’s product is P = 1000 − 7Q (a) What is the monopolist’s profit-maximizing combination of price and quantity? (b) What is the firm’s profit? (c) Derive the firm’s trace function. (d) What is the equation of the dislocated demand curve associated with normal profits? (e) What is the equation of the dislocated demand curve associated with a price of $3,200? 10. A monopolist cubic cost function is TC = 0.08Q3 − 8.2Q2 + 2000Q + 32000 If the market demand function for the firm’s product is P = 3400 − 4Q (a) Derive the firm’s trace function. (b) What is the equation of the dislocated demand curve associated with normal profits? (c) What is the equation of the dislocated demand curve associated with a price of $3,200? 11. A monopolist has the following cubic cost function: TC = 0.4Q3 − 7Q2 + 70Q + 100

1

There are couple of typo errors in the text. Coefficient of Q3 and Q2 in TC are given as 0.2 and 10. They must be 2 and 20. These errors are corrected here.

154

8 Additional Topics in Perfect and Imperfect Competition

If the market demand function for the firm’s product is P = 2000 − 10Q What is the monopolist’s profit maximizing combination of price and quantity? What is the firm’s profit? 12. Due to adverse economic conditions the demand for the good produced by the monopolist in Problem 11 declines to the extent that the firm’s above normal profit is eliminated. What is the monopolist’s new optimal combination of price and quantity? What is the equation of dislocated market demand curve? 13. A monopolist firm with the following cubic cost function: TC = 3Q3 − 40Q2 + 250Q + 900 faces the following market demand function for its product P = 2000 − 10Q The management of the firm decides to follow a three-stage market strategy, 1- find the level of output that minimizes its average total cost 2- find the level of output that maximizes its total revenue 3- choose the average of the two levels for actual production (a) What is the level of output in this strategy compared to the profit-maximizing output? (b) What is the price level in this case compared to the profit-maximizing case? 14. What is the level of output and price if a natural monopolist with the cost function TC = 0.3Q3 − 8Q2 + 120Q and market demand function P = 100 − Q follows the average cost pricing? (From two possible solutions choose the larger one). Compare this solution to the profit-maximizing solution. 15. If one of the firms in Problem 1 acquires the remaining 299 firms and creates a big monopoly, what would happen to the market equilibrium quantity and price? 16. A monopolistically competitive firm has the following total cost function: TC = 4Q2 + 30Q + 400 If the demand function for the firm’s product is

8 Additional Topics in Perfect and Imperfect Competition

P = 600 − 30Q

155

Q < 20

(a) Find the firm’s profit maximizing level of output, and its profit. (b) What is the firm’s long run (price, quantity) combination? (c) What is the equation of the dislocated demand curve associated with the long run situation? 17. A monopolistically competitive firm has the following total cost function: TC = Q3 − 10Q2 + 45Q + 100 If the demand function for the firm’s product is P = 100 − 10Q (a) Find the firm’s profit maximizing level of output, price, and profit. (b) What is the firm’s long run (price, quantity) combination? (c) What is the equation of dislocated demand curve associated with the long run situation? 18. A competitive industry consists of 500 identical firms with the total cost function given by TC = 0.6Q3 − 3Q2 + 6Q + 50 Assume the market demand function is Qmd = 6000 − 10P (a) (b) (c) (d) (e)

What is the market supply function? Determine the short run market equilibrium price and quantity. What is the profit-maximizing level of output for each firm? What is the long run market equilibrium price and quantity? In the long run how many firms enter into (or exit from) the market?

19. Assume all n firms in a competitive industry have the following quadratic total cost function: TCi = aQi2 + bQi + c

i = 1, 2, . . . , n

Assume the market demand function is Q = α − βP (a) Find the market equilibrium price PC and quantity QC .

156

8 Additional Topics in Perfect and Imperfect Competition

(b) Now assume that all n firms consolidate into a giant monopoly. Find the market equilibrium price PM and quantity QM . (c) Show that QM < QC and PM > PC . 20. Assume the market demand and supply functions are P = D(Q)

dP 0 dQ

where Q > 0. We form the total revenue function TR TR = PQ = D(Q)Q and the marginal revenue function MR MR(Q) = TR′ =

dT = D′ (Q)Q + D(Q) dQ

Assume QC > 0 is such that D(QC ) = S(QC )

or D(QC ) − S(QC ) = 0

(this is the perfectly competitive market equilibrium). Assume QM > 0 is such that D′ (QM )QM + D(QM ) = S(QM )

or

D′ (QM )QM + D(QM ) − S(QM ) = 0

(this is the monopolized market equilibrium) Show that QM < QC and PM = D(QM ) > PC = D(QC ). Answers to Chapter. 8, 8.6 Exercises #1 We must first derive the market supply function so we can solve it with the market demand function for equilibrium price and quantity. To derive the market supply function we first find a typical firm’s supply equation and then aggregate over 300 firms. The first step in the process of deriving a firm’s supply function is to determine the shut-down price, which is the minimum of the average variable cost, AVC, function. AVC =

VC Q3 − 10Q2 + 45Q = = Q2 − 10Q + 45 Q Q

FOC : AVC′ = 2Q − 10 = 0

−→

Q=5

8 Additional Topics in Perfect and Imperfect Competition

SOC :

157

AVC′′ = 2 > 0 Minimum

Psd = AVC(5) = 25 − 50 + 45 = 15 Using the firm’s profit maximizing condition MC = P, we write 3Q2 − 20Q + 45 = P

−→

3Q2 − 20Q + (45 − P) = 0

We solve this equation for Q in terms of P Q=

20 ±

√

20 ± 400 − 12(45 − P) = 6

√ √ 10 ± 3P − 35 12P − 140 = 6 3

Since there must be a positive relations between the quantity supplied and price, then √ 10 − 3P − 35 we must discard . A typical firm’s supply function is 3 ⎧ ⎨0 for P < 15 √ i = 1, 2, . . . , 300 Qi = 10 + 3P − 35 ⎩ for P ≥ 15 3

We arrive at the market supply function by aggregating the supply functions of 300 firms Qs =

300  i=1

Qi =

300  10 + i=1

√

√ 3P − 35 = 1000 + 100 3P − 35 3

We obtain the market equilibrium price by invoking the equilibrium condition Qd = Qs √ 4000 − 20P = 1000 + 100 3P − 35

−→

√ 3P − 35 = 30 − 0.2P

After squaring both sides of this equation, we have 3P − 35 = 900 + 0.04P2 − 12P

−→

0.04P2 − 15P + 865 = 0

This equation has two solutions: P = 71.1761 and P = 303.824. The second solution is not acceptable because it leads to a negative market quantity when substituted in the market demand equation. The equilibrium market quantity is then Q = 4000 − 20 ∗ 71.1761 = 2567.478 units. #2 In problem #1 the total market supply is 2567.478 units. Given that there are 300 firms in the market, then the output of an individual firm is 2567.478 ÷ 300 = 8.5883 units. With the price/output combination of (71.1761, 8.5883) each firm’s profit is

158

8 Additional Topics in Perfect and Imperfect Competition

π = 71.1761 ∗ 8.5883 − 8.58833 + 10(8.58832 ) − 45 ∗ 8.5883 − 100 = 228.9346 and this induces other firms to enter the market, expand the supply pool and lower the price. The long run price in a competitive market is established at the lowest level of the average total cost, ATC, curve. We must, therefore, find the minimum of the ATC function. ATC =

100 Q3 − 10Q2 + 45Q + 100 = Q2 − 10Q + 45 + Q Q

The FOC for a minimum is ATC′ = 2Q − 10 −

100 =0 Q2

−→

2Q3 − 10Q2 − 100 = 0

This cubic function has a real solution Q = 6.2713. The SOC for a minimum is satisfied ATC′′ = 2 +

200 >0 ∀ Q>0 Q3

The long-run price is PLR = ATC(6.2713) = 37.5619 And from the demand equation we determine the long-run market quantity QLR = 4000 − 20(37.5619) = 3248.762 Given the long run market quantity is 3248.762 units and each firm produces 6.2713 units, then the number of firms in the market must be N=

3248.762 = 518.04 ≈ 518 6.2713

which means 518 − 300 = 218 new firms have entered the market. #3 We derive the trace function of the firm by first determining the shutdown price. Here f (Q) = Q3 − 20Q2 + 150Q and g(Q) = −2Q

8 Additional Topics in Perfect and Imperfect Competition

159

Using Eq. (8.19), we must form and solve f ′ (Q)Q − f (Q) − g ′ (Q)Q2 = 0 Thus (3Q2 − 40Q + 150)Q − Q3 + 20Q2 − 150Q − (−2)Q2 = 0

−→

2Q3 − 18Q2 = 0

leading to a real solution Q = 9. The shutdown price is P˜ = AVC(9) = 51 Using Eq. (8.23), we write f ′ (Q) − g ′ (Q)Q − P = 0

−→

3Q2 − 40Q + 150 − (−2)Q + P = 0

3Q2 − 38Q + 150 − P = 0 After solving this equation for Q in terms of P, we have Q=

√ 1 (19 + 3P − 89) 3

We can now express the trace function of the firm as ⎧ ⎨0 √ Q= 1 ⎩ (19 + 3P − 89) 3 #4

for

P < 51

for

P ≥ 51

We derive the market supply by following the familiar steps. AVC = 2Q2 − 60Q + 950 AVC′ = 4Q − 60 = 0

−→

Q = 15

Psd = AVC(15) = 450 − 900 + 950 = 500 MC − P = 0

−→

6Q2 − 120Q + 950 − P = 0

√ 14400 − 24(950 − P) Q= 12 √ 60 + 6P − 2100 Q= 6 120 ±

160

8 Additional Topics in Perfect and Imperfect Competition

Thus the ith firm’s supply function is ⎧ ⎨0 √ Qi = 1 ⎩ (60 + 6P − 2100) 6

for P < 500 i = 1, 2, . . . , 150

for P ≥ 500

The market supply function is Qs =

150  i=1

Qi =

150  1 i=1

6

(60 +

√

√ 6P − 35) = 1500 + 25 6P − 2100

After invoking the market equilibrium condition, we have √ 1500 + 25 6P − 2100 = 4568.2 − 3.5P

→

√ 6P − 2100 = 122.728 − 0.14P

6P − 2100 = 15062.162 + 0.0196P2 − 34.3638P 0.0196P2 − 40.3638x + 17162.162 = 0

The acceptable answer to this quadratic equation is P = 600. Subsequently, the market equilibrium quantity is Q = 4568.2 − 3.5 ∗ 600 = 2468.2 units #5

We must first solve the market supply function for P in terms of Q, √ 25 6P − 2100 = Q − 1500 6P − 2100 =

→

√ 1 Q − 60 6P − 2100 = 25

1 2 Q − 4.8Q + 3600 625

leading to the supply function of the newly formed monopolist P=

1 Q2 − 0.8Q + 950 3750

Next we derive the marginal revenue function 3.5P = 4568.2 − Q

→

TR = PQ = 1305.2Q −

P = 1305.2 −

10 2 Q 35

→

10 Q 35

MR = 1305.2 −

20 Q 35

8 Additional Topics in Perfect and Imperfect Competition

161

Fig. 8.1 Demand, MR, and MC curves of the monopolist

The profit maximizing level of out of our newly formed monopoly is determined as 1 20 Q2 − 0.8Q + 950 = 1305.2 − Q 3750 35 simplified to Q2 − 857.143Q − 1332000 = 0 The positive solution to this quadratic equation is Q = 1659.7. Subsequently, the monopoly price is P = 831. Figure 8.1 is a graphic presentation of profit maximizing output/price for the monopolist. #6 Unfortunately, due to aggregation problem it is not possible to provide an answer to this question. The complexity of the problem arises from the fact that we don’t know how to arrive at the total cost and the average total cost functions of the monopolist formed by merger of the 150 existing firms. When I was writing this problem, I thought I knew how to derive the ATC function of a monopolist formed by merger of a large number of firms operating in a competitive market, but later I discovered a flaw in my method. This question should direct attention to another example of some unsettled issues in mathematical economics. #7 (a) The first step is to determine the shutdown price. In this problem

162

8 Additional Topics in Perfect and Imperfect Competition

TC =

1 3 Q − 20Q2 + 350Q + 100 3

and the demand function is P = 350 − 10Q. The shutdown price is the value of the average variable cost AVC at the point of tangency of dislocated demand curve and the AVC curve AVC =

1 2 Q − 20Q + 350 3

dAVC dP = dQ dQ

2 Q − 20 = −10 3

−→

−→

Q = 15

Then the shutdown price, Psd , is Psd = AVC(15) = 125 1 Note that if we write f (Q) = Q3 − 20Q2 + 350Q and g(Q) = −10Q and 3 then form the Eq. (8.19) we get exactly the same answer for the shutdown price. Using Eq. (8.23), f ′ (Q) − g ′ (Q)Q − P = 0, we have, Q2 − 40Q + 350 − (−10)Q − P = 0

−→

Q2 − 30Q + 350 − P = 0

√ √ 30 ± 4P − 500 900 − 4(350 − P) = = 15 ± P − 125 2 2 √ We must discard the solution Q = 15 − P − 125, which cannot be a supply function. Subsequently, the firm’s trace function is Q=

Q=

30 ±



√

0 √ 15 + P − 125

for P < 125 for P ≥ 125

(b) The normal profit occurs when the demand curve is dislocated and becomes tangent to the average total cost curve ATC. At the point of tangency the slope of ATC is the same as the slope of the dislocated demand curve. ATC =

1 2 100 Q − 20Q + 350 + 3 Q

dATC dP = dQ dQ

→

2 100 Q − 20 − 2 = −10 3 Q

This leads to the cubic equation

8 Additional Topics in Perfect and Imperfect Competition

163

2 3 Q − 10Q2 − 100 = 0 simplified to Q3 − 15Q2 − 150 = 0 3 with a real solution Q = 15.615. By evaluating the ATC function at 15.615, we have the normal profit price, Pnp , as 125.38. Alternatively, we can use the supply function (8.1) and determine P for the level of output Q = 15.615, which leads to the same solution P = 125.38. The combination of price and output (125.38, 15.615) is the normal profit combination of price and quantity. The equation of the dislocated demand curve associated with this combination is P = β − 10Q

−→

125.38 = β − 10 ∗ 15.625

−→

β = 281.63

Then the equation is P = 281.63 − 10Q (c) If the price is $206, we determine the firm’s output by using the firm’s trace function √ √ Q = 15 + P − 125 = 15 + 206 − 125 = 24 The equation of the dislocated demand curve associated with combination of price and output (206, 24) is P = β − 10Q

−→

206 = β − 10 ∗ 24

−→ β = 446

P = 446 − 10Q #8 (a) We form the profit function, π, and maximize it. π = TR − TC = 3400Q − 14Q2 − 0.8Q3 + 32Q2 − 1000Q − 32000 = −0.8Q3 + 18Q2 + 2400Q − 32000 FOC :

π ′ = −2.4Q2 + 36Q + 2400 = 0

−→

Q = 40

The second order condition for maximum at Q = 40 is satisfied. The profit maximizing price is P = 3400 − 14 ∗ 40 = 2840 (b) The firm’s profit is π = TR − TC = 2840 ∗ 40 − 113,600 = 41600

164

8 Additional Topics in Perfect and Imperfect Competition

(c) For deriving the firm’s trace, we follow the familiar routine AVC = 0.8Q2 − 32Q + 1000 dAVC dP = dQ dQ

−→

1.6Q − 32 = −14

Q = 11.25

−→

Psd = AVC(11.25) = 741.25 In this problem f (Q) = 0.8Q3 − 32Q2 + 1000Q and g(Q) = −14Q. Using Eq. (8.23), f ′ (Q) − g ′ (Q)Q − P = 0, we have, 2.4Q2 −64Q+1000−(−14)Q−P = 0

−→

2.4Q2 −50Q+1000−P = 0

the right solution to this quadratic equation is Q=

√ √ 1 1 (50 + 9.6P − 7100) = (125 + 60P − 44375) 4.8 12

Subsequently, the firm’s trace function is ⎧ ⎨0 √ Q= 1 ⎩ (125 + 60P − 44375) 12

for

P < 741.25

for P ≥ 741.25

Note that if we plug 2,840 and 741.25 for P in the trace function, we get the corresponding values of 40 and 11.25 for Q, which are exactly the values obtained in parts (a) and (b). (d) As is always the case for noncompetitive firms, the normal profit occurs when the demand curve is dislocated and becomes tangent to the firm’s average total cost curve ATC. At the point of tangency the slope of ATC is the same as the slope of the dislocated demand curve. ATC = 0.8Q2 − 32Q + 1000 + dP dATC = dQ dQ

→

32000 Q

1.6Q − 32 −

32000 = −14 Q2

leading to 1.6Q3 − 18Q2 − 32000 = 0

−→

Q = 31.4589

Subsequently, the normal profit price, Pnp , is

8 Additional Topics in Perfect and Imperfect Competition

165

Pnp = ATC(31.4589) = 1802.245 Note that if we plug this price into the trace function we get the same value 31.4589 for Q. With the normal profit price and output combination, we can determine the dislocated demand as P = β − 14Q β = 2242.6696

−→ −→

1802.245 = β − 14 ∗ 31.4589 P = 2242.6696 − 14Q

(e) Using the firm’s trace function, we find the equation of dislocated demand associated with price of $3,200 as Q=

√ 1 (125 + 60 ∗ 3200 − 44375) = 42.435 12

β = 3200 + 14 ∗ 42.435 = 3794.09 P = 3794.09 − 14Q Problems 9–14 are similar to problems 7 and 8, so without going into detail we provide answers to various parts of problems 9–14. #9

In this problem2 TC = 2Q3 − 20Q2 + 136Q + 100 and P = 1000 − 7Q.

(a) The profit maximizing combination of price and quantity is (899.5, 14.36). (b) The firm’s profit is 9065.71. (c) The firm’s trace function is ⎧ ⎨0 for P < 92.13 √ Q= 1 ⎩ (33 + 24P − 2175) for P ≥ 92.13 12

(d) The normal profit occurs when the demand curve shifts and becomes tangent to the average total cost. At the point of tangency, we have dATC dP = dQ dQ 4Q − 20 −

100 = −7 Q2

−→

4Q3 − 13Q2 − 100 = 0

The real solution to the above cubic equation is Q = 4.49. The normal profit price is 2

Please note the correction made to the total cost function given in the text.

166

8 Additional Topics in Perfect and Imperfect Competition

Pnp = ATC(4.49) = 108.79 Thus, the intercept, β, of the dislocated demand curve associated with the normal profit is 108.79 = β − 7 ∗ 4.49

−→

β = 140.22

and the equation of the dislocated demand is P = 140.22 − 7Q (e) Using the trace function of the firm in part (c), we have Q=

√ 1 (33 + 24 ∗ 3200 − 2175) = 25.51 12

Therefore the intercept of the dislocated demand associate with price of 3200 is 3200 = β − 7 ∗ 25.51

−→

β = 3378.60

P = 3378.60 − 7Q #10 (a) To determine the trace function of the firm we first find the shutdown price dP dAVC = dQ dQ

→

0.16Q − 8.2 = −4

→

Q = 26.25

Psd = AVC(26.25) = 1839.88 f ′ (Q) − g ′ (Q)Q − P = 0

→

0.24Q2 − 12.4Q + 2000 − P = 0

The appropriate solution to the above equation is Q=

√ 10 (31 + 6P − 11039) 12

Thus, the firm’s trace is ⎧ ⎨0 √ Q = 10 ⎩ (31 + 6P − 11039) 12

for P < 1839.88 for P ≥ 1839.88

(b) We write the general equation of the demand as P = a−4Q. At the normal profit we must have TC = TR and MC = MR. Therefore we must solve the following

8 Additional Topics in Perfect and Imperfect Competition

167

system of 2 equations with 2 unknowns: 

0.08Q3 − 8.2Q2 + 2000Q + 32000 = aQ − 4Q2 0.24Q2 − 16.4Q + 2000 = a − 8Q

After eliminating a, this system is reduced to −0.16Q3 + 4.2Q2 + 32000 = 0 The solution to this cubic equation is Q = 68.6667, leading to a = 2554.827. Thus, the equation of dislocated demand curve associated with the normal profit is P = 2554.827 − 4Q (c) Using the trace function of the firm in part (a), we have Q=

√ 10 (31 + 6 ∗ 3200 − 11039) = 101.12 12

Then 3200 = a − 4 ∗ 101.12

−→

a = 3604.48

P = 3604.48 − 4Q #11 The profit maximizing combination of price and quantity is (P, Q) = (1632.18, 37.6819). The firm’s profit is 47303.23. #12 The monopolist new normal profit combination of price and quantity is (P, Q) = (73.27, 4.013). The equation of dislocated demand curve is P = 113.4 − 10Q. #13 1. The level of output that minimizes the monopolist’s average total cost is ATC = 3Q2 − 40Q + 250 + FOC :

900 Q

900 dATC = 6Q − 40 − 2 = 0 dQ Q

6Q3 − 40Q2 − 900 = 0

−→

Q = 8.6646

168

8 Additional Topics in Perfect and Imperfect Competition

2. The level of output that maximizes the total revenue is TR = PQ = 2000Q − 10Q2 TR′ = 2000 − 20Q = 0

−→

Q = 100

100 + 8.6646 = 54.3323 2 (a) The profit maximizing level of output is 17.671 compared to 54.3323. (b) The profit maximizing price is 1823.29 compared to 1456.68.

3. The average of these two levels is

#14 If the monopolist follows the average cost pricing, then the level of output is determined by P = ATC

100 − Q = 0.3Q2 − 8Q + 120

−→

0.3Q2 − 7Q + 20 = 0 → the larger solution is Q = 20

→

P = 80

The profit maximizing level of output is 13.964 with P = 86.04. #15 In exerciser #1 we derived the market supply function of all 300 firms as √ Q = 1000 + 100 3P − 35 and the market demand function was given as Q = 4000 − 20P If this market is monopolized the market supply function becomes the supply function of the monopolist. We must first solve this market supply function for P in terms of Q, √ Q − 1000 = 100 3P − 35 

1 Q − 10 100

P=

2

=

√

−→

3P − 35

2

√ 1 Q − 10 = 3P − 35 100 −→

1 Q2 + 100 − 0.2Q = 3P − 35 10000

1 Q2 − 0.06667Q + 45 30000

Next we derive the marginal revenue function by first solving the demand equation for P in terms of Q,

8 Additional Topics in Perfect and Imperfect Competition

Q = 4000 − 20P

−→

169

P = 200 − 0.05Q

Then TR = PQ = (200 − 0.05Q)Q = 200Q − 0.05Q2 MR =

dTR = 200 − 0.1Q dQ

The monopolist’s profit maximizing level of output is determined by intersecting the MR and the firm’s supply curves, 1 Q2 − 0.06667Q + 45 = 200 − 0.1Q 30000 1 Q2 + 0.03333Q − 155 = 0 30000 Q2 + 1000Q − 4650000 = 0 An acceptable solution to this quadratic equation is Q = 1713.6. We determine the price from the demand function as P = 200 − 0.05 ∗ 1713.6 = 114.32 Comparing these figures with results of exercise #1, it is clear that the monopolization of the market has led to the restriction of supply and increase in price. #16 (a) The firm’s profit function is π = TR − TC = 600Q − 30Q2 − 4Q2 − 30Q − 400 = −34Q2 + 570Q − 400 FOC : π ′ = −68Q + 570 = 0

→

Q = 8.382

SOC : π ′′ = −68 < 0

P = 600 − 30 ∗ 8.382 = 348.54 The firm’s profit is π ≈ 1989. (b) The long-run or normal profit situation for the firm is reached when the demand for the firm’s product declines to the point that the demand curve becomes tangent to the average total cost curve. At the point of tangency dP dATC = dQ dQ

−→

d 400 (4Q + 30 + ) = −30 dQ Q

170

8 Additional Topics in Perfect and Imperfect Competition

4−

400 = −30 Q2

−→

Q2 =

400 34

−→

Q = 3.43

PLR = ATC(3.43) = 160.34 (c) The equation of the dislocated demand curve in part (b) is P = a − 30Q

−→

160.34 = a − 30 ∗ 3.43

−→

a = 263.24

P = 263.24 − 30Q #17

This problem is similar to problem number 16.

(a) The firm’s profit function is π = 100Q − 10Q2 − Q3 + 10Q2 − 45Q − 100 = −Q3 + 55Q − 100   FOC : π ′ = −3Q2 + 55 = 0 → Q = 4.282 SOC : π ′′ = −6Q

Q=4.282

0 for all Q > 0 dQ2 Q The long-run price is PLR = ATC(4.53) = 15.76 And from the demand equation we determine the long-run market quantity QLR = 6000 − 10(15.76) = 5842.4 (e) Given the long-run market quantity is 5842.4 and each firm produces 4.53 units, then the number of firms in the market must be N=

5842.4 = 1289.7 ≈ 1290 4.53

which means 1290 − 500 = 790 new firms must have entered the market. #19 (a) A typical firm’s supply function is determined by MC − P = 0, 2aQi + b − P = 0

−→

Qi =

1 (P − b) 2a

i = 1, 2, . . . , n

The market supply function is Q=

n  i=1

Qi =

n (P − b) 2a

The competitive market equilibrium is reached when n (P − b) = α − βP 2a leading to the competitive market equilibrium price, PC , PC =

2aα + nb 2aβ + n

and the competitive market equilibrium quantity, QC , QC = α − β



2aα + nb 2aβ + n



=

nα − nbβ 2aβ + n

174

8 Additional Topics in Perfect and Imperfect Competition

(b) To find the solution to the monopolized market, we must first solve the market supply function for P. P=

2a Q+b n

Next we determine the marginal revenue function, Q = α − βP TR = PQ =

−→

P=

α 1 − Q β β

α 1 Q − Q2 β β

2 α − Q β β

MR =

The profit maximizing level of output for the monopoly, QM , is determined by α 2 2a Q+b= − Q n β β leading to QM =

nα − nbβ 2aβ + 2n

and the monopoly price, PM PM =

1 α − β β



nα − nbβ 2aβ + 2n



(c) It is clear that QM is smaller than QC ; its numerator is the same while its denominator is larger than that of QC . But it is not easy to show that PM is larger than PC . This was clearly a motivation for stating and proving the theorem related to monopolization of a competitive market in its most general form in Sect. 8.3 of Chap. 8 in the text. #20 This problem is a restatement of the monopolization of the competitive market theorem couched in direct economic terms. Chapter 8 Supplementary Exercises 1. Assume all 350 firms in a competitive market have the same cost function given by TC =

1 3 Q − 15Q2 + 260Q + 400 3

8 Additional Topics in Perfect and Imperfect Competition

175

If the market demand function is Q = 6750 − 20P determine the short-run market equilibrium price and quantity. What is each firm’s level of output and profit or loss? 2. In problem #1, what is the long-run price established in the market? In the longrun how many firms enter into or exit from the market? 3. Assume all firms in problem #1 merge and form a giant monopoly. Determine the market equilibrium price and quantity, and the monopolist’s profit or loss. 4. All 250 firms in a competitive industry have the same cost function given by TC = 1.25Q3 − 20Q2 + 125Q + 500 If the market demand function is P = 450 − 0.1Q (a) What is the market equilibrium price and quantity? (b) What is each firm’s level of output? (c) What is each firm’s profit or loss? 5. In problem #4 determine the market long-run price and output and the number of firms in the market. 6. Assume through merger and acquisitions 250 firms in problem #4 form a monopoly. Determine the market equilibrium price and quantity. 7. A natural monopolist has the following total cost function TC = 0.05Q3 − 6Q2 + 520Q + 5000 Assume the market demand for the firm’s output is given by P = 6500 − 20 Q

(a) Determine the firm’s profit maximizing level of output and price. (b) Find the firm’s larger break-even level of output and price. 8. In problem #7, find the firm’s normal profit level of output and price and the equation of the dislocated demand. 9. Determine the level of output and price if the natural monopolist in problem #7 follows the average cost pricing. Show that the result is the same as part (b) in problem #7, that is, the same as the larger break-even output and the associated price.

176

8 Additional Topics in Perfect and Imperfect Competition

10. Assume a noncompetitive firm with the total cost function TC = Q3 − 10Q2 + 50Q + 150 faces the following demand function for its product Q = 4000 − 25P Derive the trace function of the firm. 11. The total cost function for a noncompetitive firm is given as TC = 2Q3 − 25Q2 + 150Q + 400 Assume the market demand function is P = 5000 − 30Q + 0.2Q2

Q < 62.5

(a) Determine the short-run profit maximizing level of output for the firm and the market price. (b) Find the normal profit level of output for the firm and the market price. (c) Find the shut-down price. 12. Assume the market demand and supply functions for a good are given by 1000 Qd = √ P2 + 5 Qs = −10 + P2

P≥

√

10

Graph both functions. Visually approximate the point of intersection between the supply and demand curves. Linear approximate both functions around that point and find the approximate market equilibrium price and quantity.

Chapter 9

Logarithmic and Exponential Functions

Chapter 9, Exercises Page 288–290 1. Find the value of each expression without using a calculator (a) log 10000 (d) log8 2 1 (g) log5 5

(b) lg 32

(c) log7 49

2

(f) ln e−20

(e) ln e

(j) log 0.001

1 36 √ (K) log 10

(m) lg 23 25

(n) eln 2

(i) log1 23 + log1 248

(h) log6

1 (l) log √ 10 1 (o) ln( √ ) e

2. If 101.8 = 63.095, then find the value of log 63.095. What is log 6309.5? √ 3. If y = log x what is log x in terms of y? 4. Solve the following equations (a) 20(2)2x = 80

(c) 8e0.15x = 20(1.5)x (e) lg(x − 2) + lg(2x + 3) = 2 (g) 5(3.4)x+2 = (2.5)2x

(i) log(3x − 1) = 5 − log x

√ (k) ln(2 z + 3) = 5

(m) log6 (2x + 4)2 = 8

(b) e3x+1 = 65 (d) log(x + 1) − log(3x) = 0.2 (f) exp(t + 2) exp(2t) = 125 (h) ln(5 − 3x) = 4 (j) e2−6x = 150

3e2x−1 = 10 2e1−2x (n) e4x+6 = log 125 (l)

© Atlantis Press and the authors 2015 S. Vali, Principles of Mathematical Economics II, Mathematics Textbooks for Science and Engineering 4, DOI 10.2991/978-94-6239-088-1_9

177

178

9 Logarithmic and Exponential Functions

5. Doubling Time: The population of a city at a certain time is P. If the annual rate of growth of the population is 2 %, after how many years does this city’s population doubles? What if the rate is 2.5 %? Write a general expression for the doubling time assuming the rate of growth is g percent. Show that the doubling time is only a function of g. 6. Half Life: A company has 25 million dollars’ worth of equipment. This company depreciates the value of the equipment at the rate of 10 % per year. How long does it takes for the value of the equipment to drop to 50 % of its original value? What would be the value of equipment after 25 years? What would be the value of equipment after 25 years as proportion of the original value? 7. Use either methods described in Example 3 and write LP and CD sales, discussed in Example 4, as exponential function of time. 8. The sales of a department store grow at the rate of 4.5 % annually. In the year 2000 the store reported $125 million in sales. Write an exponential function for sales. What is the volume of sales in year 2012? 9. Assume that the population of a country grows exponentially. Suppose N5 = 20 million and N10 = 22 million. Write the expression for Nt . What is the average annual rate of growth of the population in this country? Estimate the population for year 20. 10. The real rate of growth of the US economy is, on average, about 2.5 % annually. The Chinese economy grows, on average, at a rate of 8.5 %. According to World Bank data in 2011 the GDP of the United States and China was 14.99 and 7.31 trillion dollar, respectively. How long will it take for China to reach the current size of the US economy? Is it possible for China to overtake the United States in 21 century? 11. The number of cell phones (CP) in the United States has been growing according to the following model CPt = CP0 (1 + g)t where g is the growth rate and CP0 is the number of cell phones (in thousands) in 1995. After linearizing this model by a log transformation, we have ln CPt = ln CP0 + t ln(1 + g) and we obtain the following result as regression estimate ln CPt = 8.5 + 0.23t

t = 1 for 1995

(a) What is the rate of growth of cell phones? (b) What was the number of cell phones in use in the year 1995? (c) Write the exponential equation and estimate the number of cell phones in use for the year 2012.

9 Logarithmic and Exponential Functions

179

12. Go to the site of BEA (Bureau of Economic Analysis, US Department of Commerce) www.bea.gov and collect the US nominal GDP figures for years 2008 to 2012. Now you can extend Table 9.1 in Example 3 to include more recent data, specially the years that the economy was in crisis. Use Method One and the extended data and estimate a new exponential function for US GDP similar to (9.5). Is there an appreciable change in the rate of growth of the nominal GDP? 13. [For those familiar with regression] Use the extended GDP series from Exercise 12. Implement Method Two of Example 3 and estimate the parameters of the GDP growth function. 14. Use your estimated models from Exercises 12 and 13 and forecast the US nominal GDP for 2012. Which model is more accurate? 15. [For those familiar with regression] In Method Two of Example 3, we offered a procedure for determining c and a in the exponential model of GDP, GDPt = cat , fitted to the data in Table 9.1. In the text we denoted t to take the value of 1 for 1978, 2 for 1979, …, and 30 for 2007. Re-estimate the model by assuming t takes the value of 0 for 1978, 1 for 1979, …, and 29 for 2007. Do you find any difference between these two models? Answers to Chapter 9, Exercises Page 288 − −290 #1 (a) (b) (c) (d) (e) (f) (g) (h) (i)

log 1000 = log 103 = 3 log 10 = 3 lg 32 = lg 25 = 5 lg 2 = 5 log7 49 = log7 72 = 2 log7 7 = 2 √ 1 1 1 log8 2 = log8 3 8 = log8 8 3 = log8 8 = 3 3 ln e2 = 2 ln e = 2 ln e−20 = −20 1 log5 = log5 1 − log5 5 = 0 − 1 = −1 5 1 = log6 1 − log6 36 = 0 − log6 62 = −2 log6 36 log1 23 + log1 248 does not exist. We use this exercise and amend our discussion about restriction on the base of the logarithm on page 274. The base of logarithm, b, cannot be 1, 0, or a negative number. It could be a positive number, as long as it is not 1, that is, {b ∈ R  b > 0 and b = 1, }. Let us, for example, find log0.5 10. By denoting this number by x, we have log0.5 10 = x

−→

10 = (0.5)x

By taking log at base 10 from both sides, we have

180

9 Logarithmic and Exponential Functions

log 10 = log(0.5)x

−→

1 = x log(0.5) = −0.30103 x

1 = −3.3219. In this case x is simply the reciprocal of 0.30103 logarithm of the base. If the base is 1, then the logarithm of 1 is zero and its reciprocal is undefined. In general, based on the definition, log1 a = x means a = 1x . But there is no x to make this possible, unless a is 1, which in that case there is an infinite number of answers for x. 1 (j) log 0.001 = log = log 1 − log 1000 = 0 − 3 = −3 1000 √ 1 1 1 (k) log 10 = log(10) 2 = log 10 = 2 2 √ (l) log( √1 ) = log 1 − log 10 = −0.5 and x = −

(m)

10 3 5 lg 2 2 =

lg 28 = 8

(n) Let’s assume eln 2 = x. By taking the natural logarithm of both sides, we have ln(eln 2 ) = ln x

→

ln 2 ln e = ln 2 = ln x then x = 2. Therefore eln 2 = 2

In general, eln x = x. √ 1 1 (o) ln( √ ) = ln(e−0.5 ) = −0.5. Or ln( √ ) = ln 1 − ln e = 0 − 0.5 = −0.5. e e #2 If 101.8 = 63.095 then log 63.095 = log 101.8 = 1.8. Since 6309.5 = 100 ∗ 101.8 then log 6309.5 = log 100 + 1.8 = 3.8 #3 If y = log x then x = 10y and

√ √ x = (10y )0.5 = 100.5y → log x = 0.5y.

#4 (a) 20(2)2x = 80 −→ 22x = 4 = 22 then 2x = 2 and x = 1. (b) By taking the natural logarithm of both sides of e3x+1 = 65, we have ln(e3x+1 ) = ln 65

−→

(3x + 1) ln e = 3x + 1 = 4.1744

−→

x = 1.0583

(c) By taking the natural logarithm of both sides of 8e0.15x = 20(1.5)x , we have ln 8 + 0.15x = ln 20 + x ln 1.5 0.25547x = −0.91629 (d) log(x + 1) − log(3x) = 0.2 is 1.58489, therefore

−→

−→

2.07944 + 0.15x = 2.99573 + 0.40547x

x = −3.58668   x+1 = 0.2. The anti-log of 0.2 −→ log 3x

9 Logarithmic and Exponential Functions

x+1 = 1.58489 3x

−→

181

x + 1 = 4.75467x

−→

x = 0.26633

(e) lg(x − 2) + lg(2x + 3) = lg(x − 2)(2x + 3) = 2 then (x − 2)(2x + 3) = 22 = 4 leading to the quadratic equation 2x 2 −x−10 = 0 with two solutions: x = −2, and x = 2.5. (f) Noting that exp(x) is the notational equivalent of ex , we write the expression in part ( f ) as et+2 e2t = 125

−→

et+2+2t = e3t+2 = 125

By taking the natural log of both sides, we have ln e3t+2 = ln 125

−→

3t + 2 = 4.8283

−→

t = 0.9428

(g) After taking the natural log of both sides in (g), we have ln 5+(x +2) ln 3.4 = 2x ln 2.5

−→

1.60944+1.22378(x +2) = 1.83258x

4.057 + 1.22378x = 1.83258x

−→

0.6088x = 4.057

−→

x = 6.6639

(h) If ln(5 − 3x) = 4 then 5 − 3x = e4 = 54.5982 leading to 3x = −49.5982

−→

x = −16.5327

5 5 −→ 3x − 1 = and (3x − 1)x = 5. This x x leads to the quadratic equation 3x 2 − x − 5 = 0 with an acceptable solution 1.4684. (j) After taking the natural log of both sides, the equation is reduced to 2 − 6x = 5.01064 with the solution x = −0.50177. √ √ (k) If ln(2 z + 3) = 5 then 2 z + 3 = e5 = 148.4132, which leads to √ z = 72.7066 −→ z = 5286.2467. 3e2x−1 (l) We want to solve = 10. We first write the denominator as 2e−(2x−1) 2e1−2x and simplify the left had side as (i) log(3x − 1) = 5 − log x = log

3 2x−1+2x−1 e = 1.5e4x−2 = 10 2 4x − 2 = ln 6.6667 = 1.8971

−→ −→

e4x−2 = 6.6667 4x = 3.8971 and x = 0.9743

(m) We rewrite log6 (2x + 4)2 = 8 as (2x + 4)2 = 68 . By taking the square root of both sides, we have 2x + 4 = 64 = 1296, leading to solution x = 646. (n) Since log 125 = 2.09691 then the natural log of both sides leads to 4x + 6 = ln 2.09691 = 0.74046 and x = −1.3149

182

#5

9 Logarithmic and Exponential Functions

If the initial population is P then the population at time t, Pt , is Pt = P(1 + 0.02)t = P(1.02)t

When the city population is doubled then Pt = 2P and we have, 2P = P(1.02)t

(1.02)t = 2

−→

To solve for t, we take the log of both sides of the equation t ln(1.02) = ln 2

−→

t=

ln 2 = 35 ln 1.02

That is, the population of the city doubles in 35 years. If the rate of growth is 2.5 % ln 2 then the population doubles in t = ≈ 28 years. The general expression for ln 1.025 ln 2 . the doubling time when rate of growth is g is t = ln(1 + g) #6 If we denote the initial value of the equipment by K and the rate of decline in value, depreciation, by ρ, then the value of company’s equipment after t periods is Kt = K(1 − ρ)t . With ρ = 0.1 and Kt = 0.5K, we have 0.5K = K(1 − 0.1)t = (0.9)t

−→

(0.9)t = 0.5

ln(0.9)t = ln 0.5 −→ t ln 0.9 = ln 0.5 −→ t =

ln 0.5 = 6.58 ln 0.9

That is, 50 % of the value of equipment depreciates over approximately six and a half years. The value of equipment after 25 years will be K25 = K(0.9)25 = 0.0718K = 0.0718 ∗ 25 = 1.795 million The value of equipment after 25 years is only 7.18 % of the original value. #7 Given the data in Table 9.2 in the text, we want to estimate the models LPt = cat and CDt = bd t . Designating 1982 and 1993 as the year number 1 and 12, we have (1) LP1982 = LP1 = ca

and

(2) LP1993 = LP12 = ca12

By dividing (2) by (1), we have LP12 ca12 1.2 = = LP1 244 ca

−→ a11 = 0.004918

ln a11 = ln(0.004918) → 11 ln a = −5.31485 → ln a = −0.48317 and

9 Logarithmic and Exponential Functions

183

a = e−0.48317 = 0.6168 Using LP1 = 244 = ca = c(0.6168), we determine c as 395.574. Then the exponential model for sales of LP is LPt = 395.574(0.6168)t . Similarly, we determine d and b for exponential model for sales of CD. The only twist here is that since sale of CD in 1982 is 0 we have to use 1983 as the base year 1. (1) CD1983 = CD1 = bd;

(2) CD1993 = CD11 = bd 11

By dividing (2) by (1), we have cd 11 CD1993 495 = = CD1983 0.8 cd d = 1.9017 and b =

→

d 10 = 618.75 which leads to

0.8 = 0.4207 1.9017

Subsequently, the model for CD is CDt = 0.4207(1.9017)t . Both of these models perform rather poorly compared to the model developed in the text, relating sales of LP to CD. #8 Let S0 denote the sales of our department store at initial period and St the volume of sales at time t. If we denote the growth rate of sales by g then the general expression for sales is St = S0 (1 + g)t . With g = 0.045 and S0 = S2000 = 125, we have St = 125(1.045)t and the sales volume for 2012 is S2012 = 125(1.045)12 = 211.985 million #9

The general exponential model for the population is Nt = cat . Our task is to

find the estimate of c and a. Using the information N5 = 20 and N10 = 22, we write N5 = ca5 = 20 and N10 = ca10 = 22 N10 ca10 22 −→ a5 = 1.1 −→ 5 ln a = ln 1.1 = = N5 20 ca5 ln a = 0.019062 −→ a = e0.019062 = 1.01924 20 = 18.182. Then the 1.019245 t model is Nt = 18.182(1.01924) . The annual rate of growth is 1.924 % and the estimate of the population after 20 years is N20 = 18.182(1.01924)20 = 26.617 million. Next from N5 = ca5 = 20 we determine c as c =

184

9 Logarithmic and Exponential Functions

#10 Using the Internet domain extension of the United States and China, we denote their respective GDP at time t by GDPUSt and GDPCNt . Given the growth rates of the economy, the exponential model of GDP of US and China can be expressed as GDPUSt = GDPUS0 (1.025)t

for

t = 1, 2, . . .

GDPCNt = GDPCN0 (1.085)t

for

t = 1, 2, . . .

and

where the GDP at time 0 denotes the base year GDP. Using 2011 figure as the base year, we have GDPUSt = 14.99(1.025)t and GDPCNt = 7.31(1.085)t

t = 0, 1, 2, . . .

We determine the answer to the first question by writing GDPCNt = 7.31(1.085)t = 14.99 and solving for t. (1.085)t =

14.99 = 2.05062 −→ t ln(1.085) = ln(2.05062) −→ t = 8.8 7.31

That is, after 8.8 years (around year 2020) China’s GDP will be equal to the GDP of the United States in 2011. For the second question we must solve the following equation for t 7.31(1.085)t = 14.99(1.025)t −→ (1.085)t = 2.05062(1.025)t After taking the log from both sides, we have t ln(1.085) = 0.718142 + t ln(1.025) −→ t =

0.718142 = 12.6 0.056887

Which means after 12 and a half years (about year 2023/24) China’s GDP would be equal to the GDP of the US (roughly 20.44 trillion dollars). #11

By comparing the linearized model after a log transformation

ln CPt = ln CP0 + t ln(1 + g) with the estimated model ln CPt = 8.5 + 0.23t it is clear that 8.5 is the estimate of ln CP0 and 0.23 is the estimate of ln(1 + g). (a) The growth rate is determined by solving for g in ln(1 + g) = 0.23.

9 Logarithmic and Exponential Functions

1 + g = e0.23 = 1.2586

−→

185

g = 1.2586 − 1 = 0.2586

which means the number of cell phones in the US grow 25.86 % annually. (b) The number of cell phones in use in 1995, CP0 , is ln CP0 = 8.5

−→

CP0 = e8.5 = 4914.769

That is, in 1995 there were about 4,914,769 cell phones in use in the United States. (c) Using the estimated values of CP0 and g, the exponential equation for cell phone is CPt = 4914.769 (1.2586)t

t = 0, 1, 2, . . .

And the estimate of the number of cell phones in use in 2012 is CP2012 = CP17 = 4917.769 (1.2586)17 = 245241.789 There were about 245 millions cell phones in use in 2012. #12 On July 31, 2009, the Bureau of Economic Analysis (BEA) released the results of the comprehensive, or benchmark, revision of the national income and product accounts (NIPAs). The comprehensive revision incorporated the results of the 2002 benchmark input-output (I-O) accounts as well as changes in definitions, classifications, statistical methods, source data, and presentation. On July 27, 2012 BEA published and posted a report “Results of the 2012 Annual Revision of the National Income and Product Accounts”. These revisions indicate that the GDP of the United States is a number in flux and the past values of the series are revised periodically. These revisions, of course, do not lead to substantial changes in the numbers. To be consistent, I decided to use the revised GDP series from 1978 to 2012. The revised series is reproduced in the following Table 9.1 (figures are in billion dollars). In order to have a consistent basis for assessing the impact of recent economic crisis, we must first estimate c and a in the exponential model GDPt = cat using the revised data from 1978 to 2007. We then re-estimate these parameters with the extended data and compare the results. Denoting 1978 as time period 1 and 2007 as time period 30, we have GDP1978 = GDP1 = ca1 = 2293.8 GDP2007 = GDP30 = ca30 = 14028.7 Subsequently, ca30 14028.7 GDP30 = 6.11592 = = a29 = GDP1 ca1 2293.8

186

9 Logarithmic and Exponential Functions

Table 9.1 US GDP from 1978 to 2012, revised (figures in billion)

a29 = 6.11592

GDP

Year

GDP

1978 1979 1980 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990 1991 1992 1993 1994 1995

2293.80 2562.20 2788.10 3126.80 3253.20 3534.60 3930.90 4217.50 4460.10 4736.40 5100.40 5482.10 5800.50 5992.10 6342.30 6667.40 7085.20 7414.70

1996 1997 1998 1999 2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012

7838.50 8332.40 8793.50 9353.50 9951.50 10286.20 10642.30 11142.20 11853.30 12623.00 13377.20 14028.70 14291.50 13973.70 14498.90 15075.70 15681.50

ln a29 = ln(6.11592) = 1.81089

−→

29 ln a = 1.81089

Year

→

ln a = 0.06244

→

a = e0.06244 = 1.0644

and value of c is determined by substituting for a in the first GDP equation c(1.0644) = 2293.8

−→

c=

2293.8 = 2155.02 1.0644

The estimated exponential model for the nominal GDP from 1978 to 2007 based on the revised data is GDPt = 2155.02 (1.0644)t We now estimate the model for the extended data. Denoting 1978 as time period 1 and 2012 as time period 35, we have GDP1 = ca1 = 2293.8 GDP35 = ca35 = 15681.5

9 Logarithmic and Exponential Functions

187

GDP35 ca35 15681.5 = 6.83647 = = a34 = GDP1 ca1 2293.8 a34 = 6.83647

ln a34 = ln(6.8365) = 1.92227

−→

34 ln a = 1.92227

→

c(1.0582) = 2293.8

ln a = 0.056537

−→

c=

→

a = e0.056537 = 1.0582

2293.8 = 2167.64 1.0582

The estimated exponential model for the nominal GDP from 1978 to 2012 is GDPt = 2167.64 (1.0582)t The growth factor of 1.0582 implies an annual rate of growth of 5.82 % for the nominal GDP from 1978 to 2012, which is less than 6.44 % estimated for data from 1978 to 2007. #13 To implement Method Two using extended series, we must first re-estimate the model for time period 1978–2007 with the revised data. We first transform the exponential function GDPt = cat to a linear function by taking the logarithm of both sides, ln GDPt = ln cat = ln c + t ln a

(9.1SM)

By renaming ln GDPt = yt , ln c = β0 , and ln a = β1 we have the familiar simple linear regression (trend) model yt = β0 + β1 t

t = 1, 2, . . . , 30

The estimated equation is yˆ t = 7.823367 + 0.05933 t

t = 1, 2, . . . , 30

Comparing this estimated equation with equation (9.1SM) it is clear that 7.823367 is an estimate of ln c and 0.05933 is an estimate of ln a. Therefore, ln c = 7.823367 ln a = 0.05933

→ →

a = e7.823367 = 2498.30 and a = e0.05933 = 1.06113

We can now write the model as GDPt = 2498.30 (1.06113)t , which implies a growth rate of 6.11 %. Next we estimate the model for the extended data. Denoting 1978 as time period 1 and 2012 as time period 35, we have

188

9 Logarithmic and Exponential Functions

yˆ t = 7.868877 + 0.055326 t

t = 1, 2, . . . , 35

Therefore, ln c = 7.868877

→

a = e7.868877 = 2614.63 and

ln a = 0.055326

→

a = e0.055326 = 1.0569

We write the model as GDPt = 2614.63 (1.0569)t , which implies a growth rate of 5.69, smaller than 6.11 % for data from 1978 to 2007. #14 By plugging 35 for t in estimated models from Exercises #12 and #13, we obtain GDP2012 = 2167.64 (1.0582)35 = 15698.52 and GDP2012 = 2614.63 (1.0569)35 = 18138.3 Comparing these two estimates with 15681.5, the actual value of GDP for 2012, it is clear that the first method generates more accurate estimates. #15 Again, for consistency we estimate the models with the revised data from 1978 to 2007 given in Table 9.1. From the first part of Exercise #13 our estimate for the first model, which designates the initial period 1978 as 1 and increments the subsequent periods by one to 30 for 2007, is GDPt = 2498.30 (1.06113)t

t = 1, 2, 3, . . . , 30

Our estimate for the alternative model, which designates the initial period 1987 as 0 and increments the subsequent periods by one to 29 for 2007, is GDPt = 2651.02 (1.06113)t

t = 0, 1, 2, . . . , 29

Both functions have exactly the same base a, 1.06113, but different constant c. But both generate exactly the same results. For estimate of the initial period in the first model we set t = 1 and obtain GDP1 = 2498.30(1.06113) = 2651.02 which is the same as the initial estimate from the second model when t = 0. Similarly, in the estimate of GDP for year 1979 from the first model, t assumes value 2 leading to GDP1979 = 2498.30(1.06113)2 = 2813.08. 1979 GDP estimate from the second model is GDP1979 = 2651.02(1.06113)1 = 2813.08, which is exactly the same. Note that we can write the first model GDPt = 2498.30 (1.06113)t = 2498.30(1.06113)(1.06113)t−1 GDPt = 2651.02(1.06113)t−1

t = 1, 2, 3, . . . , 30

9 Logarithmic and Exponential Functions

189

This modified version of the first model has exactly the same base and constant as the second model, but the initial period is designated as 1. Chapter 9, 9.6 Exercises 1. Given the following market demand and supply functions, determine the market equilibrium price and quantity Qd = Qs =

1200 0.2P + 0.5 P3 e0.02P+0.1

= P3 e−(0.02P+0.1)

2. Assume the following market supply and demand functions. Find the market equilibrium price and quantity. Dd =

1200 0.2P

Qs = P2 e−0.02(P+0.5) 3. Consider a noncompetitive firm with the cubic cost function1 TC =

1 3 Q − 20Q2 + 450Q + 100 3

facing the log linear demand function ln P = 7.5 − 0.6 ln Q

4. 5. 6. 7. 8.

in the market. What is the firm’s profit maximizing combination of price and quantity? In problem (3) determine the firm’s break-even price/quantity combination. In problem (3) determine the firm’s normal profit price/quantity combination. In problem (3) find the dislocated demand function for shutdown and normal profit situation. Derive the trace function for the firm in problem (3). Assume a noncompetitive firm with cubic cost function Q3 − 20Q2 + 150Q + 200 facing a semi-log demand function P = 300 − 3 ln Q

1

Note that typographical error in the text is corrected here.

190

9 Logarithmic and Exponential Functions

9. 10. 11.

in the market. What is the firm’s profit maximizing combination of price and quantity? In problem (8) determine the firm’s break-even price/quantity combination. In problem (8) determine the firm’s normal profit price/quantity combination. In problem (8) find the dislocated demand function for shutdown and normal profit situation. Derive the trace function for the firm in problem (8).

12.

Answers to Chapter 9, Exercise 9.6 #1 By invoking the equilibrium condition, we have Qd = Qs

−→

1200 = P3 e−(0.02P+0.1) 0.2P + 0.5

By taking the natural log of both sides, we get ln 1200 − ln(0.2P + 0.5) = 3 ln P − (0.02P + 0.1) simplified to the following equation 3 ln P + ln(0.2P + 0.5) − 0.02P − 7.19 = 0 WolframAlpha gives two possible solutions to this mixed logarithmic function: P = 8.86476 and P = 926.431. The second solution is not acceptable. The reason lies in the fact that for Qs to be a legitimate supply function the range of values that P can take (the domain of the function) must be limited to 0 < P < 150. As the graph of this function, Fig. 9.1, shows this function has a maximum at P = 150, so the acceptable segment of the curve for representing a supply curve must be for values of P less than 150. Algebraically, the extreme point of the function is determined by

Fig. 9.1 Graph of P3 0.02P+0.1 e

9 Logarithmic and Exponential Functions

191

Fig. 9.2 Graph of P2 0.02(P+0.5) e

FOC :

dQs = 3P2 e−(0.02P+0.1) − 0.02P3 e−(0.02P+0.1) = 0 dP

e−(0.02P+0.1) (3P2 − 0.02P3 ) = 0 Since e−(0.02P+0.1) > 0 for all values of P, then we have 3P2 − 0.2P3 = 0, with the non-trivial solution P = 150. Students should be able to check that the SOC for a maximum is satisfied. #2 In this problem the legitimate values of P for the supply function are P < 100, as indicated in Fig. 9.2. The equilibrium condition leads to 1200 = 0.2P3 e−0.02(P+0.5) which after taking the log of both sides and simplifying leads to the following mixed logarithmic equation 3 ln P − 0.02P − 8.709516 = 0 The legitimate answer is P = 20.961 with the equilibrium quantity Q = 286.1626. #3

We rewrite the log linear demand function ln P = 7.5 − 0.6 ln Q as P = e7.5−0.6 ln Q =

e7.5 e0.6 ln Q

=

1808.0424 Q0.6

Subsequently, the TR and the profit π functions are TR = PQ =



1808.0424 Q0.6



Q = 1808.0424Q0.4

1 π = TR − TC = 1808.0424Q0.4 − Q3 + 20Q2 − 450Q − 100 3

(9.2SM)

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9 Logarithmic and Exponential Functions

the FOC for a maximum is FOC :

dπ = 723.21696Q−0.6 − Q2 + 40Q − 450 = 0 dQ

−Q2.6 + 40Q1.6 − 450Q0.6 + 723.21696 = 0 The real solution to this equation, using WolframAlpha, is Q = 27.068, where the second order condition for a maximum is satisfied. At this level of output the profit maximizing price is P = 249.88 and the firm’s profit is 2525.99. #4

The break-even level of output is found by TR = TC

−→

1808.0424Q0.4 =

1 3 Q − 20Q2 + 450Q + 100 3

leading to the equation 1 3 Q − 20Q2 + 450Q − 1808.0424Q0.4 + 100 = 0 3 There are two break-even level of outputs Q = 0.00073 and Q = 41.197 with associated prices 137789.11 and 194.22. #5 If due to deterioration of the business environment, demand for output of this firm declines to the point that the firm earns only a normal profit then the following conditions must simultaneously satisfied 

TC = TR or alternatively MC = MR

ATC = P

(9.3SM)

The decline in demand is exhibited by a leftward shift in the demand curve, i.e. the demand dislocation. Using Eq. (9.2SM) from Exercise #3, we express this dislocated demand equation and the total revenue as P=

β Q0.6

and

TR = PQ = β Q0.4

Then using (9.3SM), we have the following system of two equations with two unknowns, Q and β. 

1 3 2 3 Q − 20Q + 450Q + 100 = β Q2 − 40Q + 450 = 0.4 β Q−0.6

Q0.4

(9.4SM)

We can eliminate β in (9.4SM) by multiplying both sides of the second equation by −2.5Q and adding it to the first equation.

9 Logarithmic and Exponential Functions



193

1 3 3Q

− 20Q2 + 450Q + 100 = β Q0.4 −2.5Q3 + 100Q2 − 1125Q = − β Q0.4

−2.1667Q3 + 80Q2 − 675Q + 100 = 0 This cubic equation has 3 real solutions. The first value is very small and we ignore it. The other two solutions are Q = 12.7244 and Q = 24.0473. For reason that we will see in problem #7, where we will derive the firm’s trace function, we must discard the first solution. From either equations in (9.4SM) we calculate β = 1118.462. Subsequently, the normal profit price is Pnp =

1118.462 = 165.95 24.04730.6

Note: As it was mentioned in the text (page 153), if the demand function is non-linear dP dATC = , the tangency between the ATC and then the method of equating dQ dQ the demand curves at the normal profit level of output, may fail. This is an instant of this failure. Here the equation of dislocated demand curve, a log-linear function, 0.6β dP β = − 1.6 . Equating this derivative with the is P = 0.6 and its derivative is Q dQ Q derivative of the average cost function leads to an equation with two unknowns, β and Q, which cannot be solved. #6 In Exercise # 5 we determined β associated with dislocated demand functions for normal profit as 1118.462. Therefore, the equations of the demand function after shift leading to normal profit is P=

1118.462 Q0.6

To find the dislocated demand associated with the shutdown price, we must solve the following system 

TVC = TR or alternatively MC = MR

AVC = P

By eliminating the fixed cost of 100 from the first equation of (9.4SM), we have 

1 3 2 0.4 3 Q − 20Q + 450Q = β Q 2 −0.6 Q − 40Q + 450 = 0.4 β Q

We eliminate β in (9.5SM) and arrive at the following cubic equation

(9.5SM)

194

9 Logarithmic and Exponential Functions

−2.1667Q3 + 80Q2 − 675Q = 0 which can be reduced to a quadratic equation −2.1667Q2 + 80Q − 675 = 0 There are two solutions Q = 13.0498 and Q = 23.8737. For reason that will be offered in problem #7, where we will derive the firm’s trace function, we must discard the first solution. We use each of the equations in (9.5SM) and determine β associated with Q = 23.8737, namely 1090.542. The shutdown price is then Psd = 162.51. Therefore, the equations of demand function after shift creating the shutdown situation is P=

1090.542 Q0.6

#7 To derive the trace function of the firm, we utilize Eq. (9.22) from the text reproduce below f ′ (Q)g(Q) − Pg(Q) + PQg ′ (Q) = 0 In this problem f (Q) =

1 3 Q − 20Q2 + 450Q and g(Q) = Q0.6 . Thus, 3

(Q2 − 40Q + 450)Q0.6 − PQ0.6 + PQ(0.6Q−0.4 ) = 0 After simplifying this equation, we have Q2 − 40Q + 450 − 0.4P = 0 Solving for Q in terms of P, we have √ Q = 20 ± 0.63246 P − 125

√ The acceptable solution is Q = 20+ P − 125 . We already computed the shutdown price in problem #6 as 162.51. Thus, the trace function of the firm is Q=



0 √ 20 + 0.63246 P − 125

for P < 162.51 for P ≥ 162.51

In problem #5 we rejected Q = 12.7244 as a normal profit solution. The reason is that the normal profit price associated with this level of output is 257.34. In order √ to arrive at the value 12.7244 for Q from the trace function, we must use Q = 20− P − 125, which is obviously not a supply relation. The same reasoning lead to the rejection of Q = 13.0498 as the legitimate shutdown level of output in problem #6.

9 Logarithmic and Exponential Functions

#8

195

The total revenue function is TR = PQ = 300Q − 3Q ln Q and the profit function is π = TR − TC = 300Q − 3Q ln Q − Q3 + 20Q2 − 150Q − 200 π = −Q3 + 20Q2 + 150Q − 3Q ln Q − 200 FOC :

dπ = −3Q2 +40Q+150−3 ln Q−3 = −3Q2 +40Q−3 ln Q+147 = 0 dQ

This mix of quadratic and logarithmic equation has a real solution Q = 16.1882, which satisfies the SOC SOC :

3  d2π = −57.3145 < 0 = −6Q + 40 −  dQ2 Q Q=16.1882

for a maximum. The profit maximizing price is P = 300 − 3 ln(16.1882) = 291.6472 and the maximum profit is 3091.9244. #9

For the break-even, we have TR = TC

−→

300Q − 3Q ln Q = Q3 − 20Q2 + 150Q + 200

Q3 − 20Q2 − 150Q + 3Q ln Q + 200 = 0 The break-even levels of outputs are Q = 1.16617 and Q = 25.2444. #10 We first write the dislocated demand equation and the associated total revenue function as P = β − 3 ln Q

−→

TR = βQ − 3Q ln Q

and then similar to problem #5, we solve the system of equations setting TC = TR and MC = MR.  Q3 − 20Q2 + 150Q + 200 = β Q − 3Q ln Q 3Q2 − 40Q + 150 = β − 3 ln Q − 3 We eliminate β by multiplying both sides of the second equation with −Q and adding it to the first equation.

196



9 Logarithmic and Exponential Functions

Q3 − 20Q2 + 150Q + 200 = β Q − 3Q ln Q −3Q3 + 40Q2 − 150Q = − β Q + 3Q ln Q + 3Q

−2Q3 + 20Q2 + 200 = 3Q leading to

− 2Q3 + 20Q2 − 3Q + 200 = 0

The real solution to this cubic equation, Qnp = 10.729, provides the normal profit level of output. From the second equation, we calculate β as β = 3Q2 − 40Q + 3 ln Q + 153 β = 3(10.7292 ) − 40 ∗ 10.729 + 3 ln(10.729) + 153 = 76.2932 Subsequently, the normal profit price is Pnp = 76.2932 − 3 ∗ ln(10.729) = 69.174. #11 The dislocated demand function for normal profit is P = 76.2932 − 3 ln Q To find the dislocated demand function for shutdown situation, we must solve a system of equations setting TVC = TR and MC = MR. We can also use Eq. (9.16) on page 305 of the text to derive the shutdown level of output. 

Q3 − 20Q2 + 150Q = β Q − 3Q ln Q 3Q2 − 40Q + 150 = β − 3 ln Q − 3

After eliminating β by multiplying both sides of the second equation with -Q and adding it to the first, we end up with −2Q3 + 20Q2 − 3Q = 0

−→

−2Q2 + 20Q − 3 = 0

This equation has an acceptable solutions Q = 9.8477. We then determine β from either equations as β = 3Q2 − 40Q + 3 ln Q + 153 = 3(9.84772 ) − 40 ∗ 9.8477 + 3 ln(9.8477) + 153 = 56.8853 and find the shutdown price as Psd = 56.8853 − 3 ln(9.8477) = 50.02. The dislocated demand function associated with the shutdown situation is then, P = 56.8853 − 3 ln Q #12 To drive the trace function for a firm facing a semi-log demand function, we utilize Eq. (9.17) on page 305 of the text

9 Logarithmic and Exponential Functions

197

f ′ (Q)g(Q) + bQg ′ (Q) − Pg(Q) = 0 In this problem f (Q) = Q3 − 20Q2 + 150Q, b = 3, and g(Q) = Q. Thus, (3Q2 − 40Q + 150)Q + 3Q − PQ = 3Q2 − 40Q + 153 − P = 0 In problem #11 we computed the shutdown price as 50.02. Therefore, by solving this quadratic equation for Q in terms of P, we determine the trace function as ⎧ ⎨0 √

Q= 1 ⎩ 20 + 3P − 59 3

for P < 50.02 for P ≥ 50.02

Chapter 9 Supplementary Exercises 1. Evaluate the following without using a calculator. (a) log2 64

1 256

(b) log2 1024

(c) log2

(d) log2

(e) log4 2

(f) log2 0.5

(j) log16 128

(h) log16 4 √ (k) log4 64

(i) log4 8 √ 3 (l) log4 64

8 64 (g) log8 4

2. Without using a computer determine which of the two numbers 3300 or 28100 is larger. 3. Assume f (x) = 2x−1 and g(x) = 3x 2 −2x+1. Determine the following functions (a) f [g(x)]

(b) f [g(−x)]

(c) g[f (x)]

(d) g[f (−x)]

(c) g[f (1)]

(d) f [g(−1)]

4. In Exercise #3 evaluate (a) f [g(1)]

(b) f [g(−1)]

5. In Exercise #3 evaluate (a) log2 f (5)

(b) log2 f (−2)

(c) log3 g[f (2)]

(d) log2 f [g(−1)]

6. Determine value of x such that (a) log2 x = −2

(b) log4 x = 0.5

(c) log3 x 2 = 2

7. Solve the following equations: (a) log2 (log3 x) = 1

(b) ln(log x) = 0

(c) ln(ln x) = −1

198

9 Logarithmic and Exponential Functions

8. Solve the following equations: (a) 42x + 4x = 6

6x − 1 = 0.5 6x + 2 (e) log √4 (7x + 2) = 2 ( (f) 5 x − 2) = 25

(d)

(b) 84x − 82x − 20 = 0 √ (c) 9x + 9x − 12 = 0

9. Differentiate the following functions: (a) f (x) = 5 ln(2x + 1) (b) g(x) = 32x+1 + log(x 2 − 1) (c) f (y) = log(3y3 − 2y2 + 5) (d) g(x) = 5x 3 (5)3x−1

(e) f (z) = (5z + 4)e2z−1 3 (f) g(w) = log √ 6 (aw − 2bw + c) (g) f (x) = 2 x 2 + 1 ln x (h) g(x) = ln(1 − x 3 )

10. The Southern California Northridge earthquake in 1994 was registered 6.7 on the Richter scale. The 1995 Kobe earthquake in Japan was registered 7.2 on the Richter scale. How many times more severe was Kobe earthquake compared to Northridge? 11. What is the Richter scale difference between earthquake A which is 100 times more severe than earthquake B? 12. If log 4.83 = 0.68395, what is (a) log 483

(b) log 48300

(c) log 0.483

(d) log 0.00483

13. CFO of a company claims that their annual revenue, Rev, (in $1,000) over time grows according to the following model: Revt = Rev0 (1 + g)t She used the revenue data from the last 10 years and estimated the model as ˆ t = 5.521 + 0.03t ln Rev (a) Determine the growth rate of the revenue. (b) What is the estimated value of Rev0 ? (c) Predict the firms revenue for period 11. 14. The resale value, RV , of a car is a function of age of the car, t. Assume the resale value is expressed as RVt = Pe−0.12t + 0.3

9 Logarithmic and Exponential Functions

199

where P is the original price paid for the car. If P = 35000, determine the selling price of the car after 5 years. What is the rate of depreciation of the car after 7 years? 15. Many colleges are experiencing financial difficulties because of increasing costs of computer maintenance and services. The dean of a college uses data from 1992 to 2011 and estimates the exponential trend model Costt = aebt from the data ln Costt = 8.87 + 0.09 t (a) What is the computer services cost for 2012 and 2013 based on this model? (b) What is the model’s estimate of computer services cost for year 1994 [this is called Back casting (as opposed to forecasting)]? (c) If actual computer services costs for 2012 and 2013 are $44,700 and $46,800 respectively, what is the average of the percentage forecasting errors of the model. 16. In January 2012, the US M1 money supply was $2,200.1 billion. In December 2012 the M1 money supply was $2,440.2 billion. (a) Use Method One and find an exponential function M1t = cat for the monthly US money supply. (b) Use the model and estimate the US money supply for months of August and September of 2012. (c) What is the annual rate of growth of M1 money supply? (Note: you must annualize the monthly rate) (d) If the actual figure of the money supply for August and September were 2,339.0 and 2,373.8, what is the average percentage error of the model? 17. Use the date from the the previous problem, (a) Use method One and find a natural exponential function M1t = ceβ t for the US money supply. (b) Use the model and estimate the US money supply for months of August and September of 2012. (c) Should these estimates be the same as the estimates in the last problem? (d) Convert this natural exponential model to an exponential model. 18. The following data in Table 9.2 show the US population from 2001 to 2012 (a) Use Method One and find an exponential function Pt = cat for the US population. (b) What is the annual rate of growth of the population? (c) Use the model and estimate the US population for 2011 and 2012. (d) Given the actual figure for year 2011 and 2012 in the table, what is the average percentage error of the model?

200

9 Logarithmic and Exponential Functions

Table 9.2 US population from 2000 to 2012 Year Population (1000) 2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012

282162 284969 287625 290108 292805 295517 298380 301231 304094 306772 309350 311592 313914

Source Department of commerce (bureau of the census)

19. Assume monthly demand for 5-pound bags of potatoes is given by P = 20 − 2 ln Q Assume each bag cost a farmer $2 (including the fixed cost) to produce. Determine the profit maximizing number of bags and price per bag. 20. Assume a noncompetitive firm with the following cost function TC =

1 3 Q − 4Q2 + 150Q + 200 3

faces a semi-log demand function P = 210 − 10 ln Q in the market. Determine the firm’s profit maximizing price and quantity. 21. In problem #20, what is the firm break-even price/quantity combination? 22. In problem #20 determine the firm’s normal profit price/quantity combination. What is the firm’s dislocated demand function associated with the normal profit? 23. Derive the trace function for the firm in problem # 20. 24. Consider a firm with the cost function TC =

1 3 Q − 2Q2 ln(Q2 + 10) + 200Q + 200 3

9 Logarithmic and Exponential Functions

201

Fig. 9.3 Graph of MC and MR

Fig. 9.4 Graph of TC and TR curves

with the following demand function for its product P = 500e−0.02Q Linear approximate the firm’s profit maximizing level of output and price. Use the Fig. 9.3 as guide. 25. In problem # 24, linear approximate the firm’s break-even level of output. Use the Fig. 9.4 as guide. 26. Assume all firms in a competitive market have the following cost function: TC = 100Q − 100 ln Q + 500 What is the long-run price in this market?

202

9 Logarithmic and Exponential Functions

Table 9.3 National health expenditure 2003–2012 Year NHE 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012

1778.00 1905.70 2035.40 2166.70 2302.90 2411.70 2504.20 2599.00 2692.80 2793.40

Source Center for medicare & medicaid services, www.cms.gov

27. The demand for product of a firm is given by P = 10e3 − 0.25Q where Q is in 1000 units. What is the firm’s revenue maximizing level of output and price? 28. In problem #26 determine the price elasticity of demand at prices 10 and 20. 29. Assume monthly demand for a certain product is given by P = 32 − 4 ln Q determine the price elasticity of demand at prices 10 and 20. At what price is demand unit elastic? 30. Table 9.3 provides the National Health Expenditure (NHE) of the United States for 10 years from 2003 to 2012 (figures are in billions). (a) Use Method One and find an exponential function NHEt = cat for the US health expenditure. (b) What is the annual rate of growth of the expenditure? (c) Use the model and estimate the US health expenditure for 2013 and 2014. 31. The demand function for the product of a monopolistically competitive firm is given by P = 600e−0.08Q (a) Write the demand elasticity function ǫ(P). (b) Determine the price elasticity at prices 200 and 400. (c) At what price the demand is unit elastic?

9 Logarithmic and Exponential Functions

203

32. Show that in a semi-log demand function Q = a − b ln P a we need the restriction P < e b . 33. A noncompetitive firm with the total cost function TC = 50Q − 50 ln Q + 200 faces the following demand function Q=



5000 − 2P2

in the market. Determine the firm’s profit maximizing price and quantity. 34. The salvage value, Sal, of a $300,000 piece of equipment as a function of time t is given by Sal(t) = 300000e−0.2t Determine the salvage value of the equipment after 6 years. What is the rate of depreciation of the equipment after 6 years? 35. In problem #34, after how many years the value of the equipment drops to only 5 % of its original value?

Chapter 10

Production Function, Least-Cost Combination of Resources, and Profit Maximizing Level of Output

Chapter 10, 10.7 Exercises 1. Assume a firm has the production function Q = 2.5K 0.4 L 0.5 . If this firm employs 15 units of labor and 20 units of capital and the price of the good in the market is $100 per unit, find the following: (a) Marginal product of labor and capital. (b) Marginal revenue product of labor and capital. (c) Average product of labor and capital. Should the fact that in part (a) the marginal product of labor is larger than the marginal product of capital encourage this firm to substitute labor for capital? 2. Assume a firm has the production function Q = 2.5K 0.4 L 0.5 . (a) Write the firm’s Isoquant equation associated with the production of 20 units of output. (b) Calculate MRTS for labor input L = 6 and L = 8. (c) Assume labor costs $80 per unit and capital rental costs $200 a unit. What is the cost of producing 20 units of output when the firm employs 6 units of labor? What about 8 units of labor? Which one costs more? Could this be anticipated by the MRTSs you calculated in part(b)? (d) What is the firm’s least-cost combination of labor and capital for producing 20 units of output? (e) What is the firm’s total cost at the least-cost combination of labor and capital? (f) What is the firm’s profit if the price of this good in the market is $200 per unit? (g) Write the firm’s Isocost equation. 3. A firm has the following short run production function Q = 20L 2 − 2L 3 © Atlantis Press and the authors 2015 S. Vali, Principles of Mathematical Economics II, Mathematics Textbooks for Science and Engineering 4, DOI 10.2991/978-94-6239-088-1_10

205

206

10 Production Function, Least-Cost Combination of Resources …

(a) Write the firm’s marginal product of labor. (b) At what level of labor input does the MPPL reach its maximum? At what level does it become zero? (c) Show that the output reaches its maximum at the level of labor input where MPPL is equal zero (use the graph as a guide).

4. A firm with a CD production function Q = 2K 0.5 L 0.4 plans to produce 500 units of output per day. Assume the price of labor and capital are w = $50 and r = $100 a day. (a) What is the firm’s optimal combination of factors that minimizes its cost? (b) What is the market price of this good if this firm makes a $2,500 profit each day? 5. Consider a firm with the production function Q = 1.5K 0.5 L 0.5 . (a) Write equations of three isoquants associated with 30, 40, and 60 units of output. (b) If the prices of labor and capital are $40 and $160 per unit, respectively, find the least-cost combination of labor and capital for 30, 40, and 60 units of output by setting MRTS equal to the factor price ratio. (c) Show that the three combinations in part (b) fall on a straight line (use the following graph as a guide). In general, if the production function is homogeneous then the least-cost combinations of all different levels of output fall on a straight line. This line is called the Output Expansion Path. Note 1 that the capital-labor ratio in all three cases in this problem is , which is 4 10 equal to the ratio of prices of labor and capital: 40 = 41 . Equation (16) in Sect. 10.4, which relates K to L by K =

αw L βr

is indeed the equation of the linear output expansion path.

10 Production Function, Least-Cost Combination of Resources …

207

(d) Write the equation of the linear expansion path for this production function.

6. For the production function in problem (3), Q = 20L 2 − 2L 3 , write the equation for the average product of labor. At what level of labor input does the A P of labor reach its peak? At what level does it drop to zero? 7. A farmer uses land and labor to grow soybeans. He has the following production function Q = 126.2 A0.6 L 0.4 where Q is the amount of soybeans in bushels, A is land in acres and L is number of farm hands. (a) If this farmer has 100 acres of land, how many farm workers must he employ in order to produce 10,000 bushels of soybeans? What is the marginal product of labor at this level of employment? (b) Assume this farmer pays each farm worker $1,000 over the growing season and spends $2,000 for seed and fertilizer. If the price of soybeans is $14.75 a bushel, what is the farmer’s profit? (c) Is this farmer paying his workers the value of their marginal product? (d) If this farmer doubles the amount of land and labor, how many bushels of soybeans he can produce? 8. A firm operates with the production function Q = 5.4K 0.3 L 0.5 . (a) If the price of labor and capital are $10 and $20, respectively, and the firm’s management has decided to spent only $7,000 daily on production costs, how many units of output can the firm produce? (Use the substitution method). (b) How many units of labor and capital does firm employ? (c) If the price of the good is $20 per unit, what is the daily profit for this firm? 9. Redo part (a) of problem (8) using the method of Lagrange Multiplier. 10. Assume the firm in problem (8) has a fix 213.71 units of capital. (a) Write the firm’s short run production function.

208

10 Production Function, Least-Cost Combination of Resources …

(b) If the price of the good, labor, and capital are $20, $10, and $20 respectively, what is the profit maximizing level of output and employment? (c) What is this firm’s daily profit? 11. Consider a firm with the production function Q = KL. (a) Is this production function homogeneous? If yes, of what degree? (b) If labor and capital cost $100 and $200 per unit, what is the equation of the output expansion path? (c) If the firm wants to spend $8,000 on labor and capital, what is the maximum output it can produce? How many units of L and K does it employ? 12. For the production function Q = K 0.35 L 0.5 write the marginal product of K and L and determine whether they are downward sloping. Let K = 50 and write the equations of total, average, and marginal product of L and graph them. 13. Assume a firm’s output is a function of capital, labor, and energy (E) given by Q = AK α L β E γ (a) Show that this function is homogeneous and determine the degree of homogeneity of the function. (b) What should the relations between α, β, and γ be in order to have a constant returns to scale? Increasing or decreasing returns to scale? 14. In an interesting paper published in the journal American Economic Review in September 1970, Professor Martin Weitzman from Yale University estimated a production function for Russia (then the Soviet Union) based on a large amount of macroeconomic data from 1950 to 1969. The production function he used in his estimate − 1  Q = γ δK −ρ + (1 − δ)L −ρ ρ is known as the Constant Elasticity of Substitution (CES) production function. Weitzman estimated the parameters of the function and expressed it as Y = 0.8(0.64K −1.5 + 0.36L −1.5 )−0.67 (where Y is output) (a) Is this function homogeneous? (b) Write the marginal physical product of labor MPP L and capital MPP K . (c) If K = 5000 units and L = 8000 units, what is Y , MPP L , and MPP K ? 15. Assume a firm with the production function Q = AK α L β . Show that α is the output elasticity of labor, ǫ Q L , and β is the output elasticity of capital, ǫ Q K .

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209

Answers to Chapter 10, Exercises 10.7 #1 (a) The marginal products of labor and capital are MPP L =

∂Q = 1.25K 0.4 L −0.5 = 1.25(20)0.4 (15)−0.5 = 1.07 ∂L

MPP K =

∂Q = K −0.6 L 0.5 = (20)−0.6 (15)0.5 = 0.642 ∂K

(b) The marginal revenue products of labor and capital are MRP L = P ∗ MPP L = 100 ∗ 1.07 = $107 MRP K = P ∗ MPP K = 100 ∗ 0.642 = $64.2 (c) The average products of labor and capital are AP L =

Q 2.5K 0.4 L 0.5 = = 2.5K 0.4 L −0.5 L L

AP L = 2.5(20)0.4 (15)−0.5 = 2.14 AP K =

2.5K 0.4 L 0.5 Q = = 2.5K −0.6 L 0.5 K K

AP K = 2.5(20)−0.4 (15)0.5 = 1.605 The fact that the marginal product of labor is greater than the marginal product of capital is necessary but not sufficient condition for factor substitution. A definite answer to this question requires additional information about the prices of labor and capital. Note that with 20 units of capital and 15 units of labor this firm produces Q = 2.5(20)0.4 (15)0.5 = 32.092 units of output. The equation of isoquant for this level of output is 32.092 = 2.5K 0.4 L 0.5

→

K 0.4 =

32.092 = 12.8368L −0.5 2.5L 0.5

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K 0.4

2.5

2.5  = 12.8368L −0.5

K = 590.397L −1.25

→

Next we find the marginal rate of technical substitution MRTS when the labor input is 15 units, by differentiating the equation of this isoquant and evaluating it at L = 15 dK = −738.0L −2.25 dL

→

  MRTSLK = 738.0L −2.25 

L=15

= 1.667

w As long as the MRTSLK is greater than labor/capital price ratio the substitution r of labor for capital reduces the cost of production (this issue is directly addressed in part (c) of the next exercise). The next step is to connect M RST and the marginal products of labor and capital. In exercise #2 of the Supplementary Exercises for this chapter you are asked to show that for Cobb-Douglas production functions MPP L dK =− dL MPP K

−→

MRTSLK =

MPP L MPP K

That is, MRTS is the ratio of marginal products. In part (a) we computed the marginal 1.07 = products of labor and capital as 1.07 and 0.642, respectively. The ratio 0.642 1.667 is exactly the value of MRTS that we calculated directly. #2 (a) We derive the equation of isoquant for 20 units of output as 20 = 2.5K 0.4 L 0.5

−→

K 0.4 =

20 = 8L −0.5 2.5L 0.5

2.5  K = 8L −0.5 = 181.02L −1.25 (b) By differentiating the equation of isoquant, we have dK = (−1.25)181.02L −2.25 dL

→

M RSTL K = 226.275L −2.25

Subsequently, the MRTS for 6 and 8 units of labor are   226.275L −2.25 

L=6

= 4.016

and

  226.275L −2.25 

L=8

= 2.102

(c) We must first determine the number of units of capital needed to mix with 6 units of labor to produce 20 units of output. Using the equation of isoquant for 20 units of output from part (a), we have

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211

K = 181.02(6)−1.25 = 19.277 units Similarly, the number of units of capital needed for production of 20 units of output when the firm employs 8 units of labor is K = 181.02(8)−1.25 = 13.454 units With the given prices of labor and capital, $80 and $200, the variable cost of production for combination (L , K ) = (6, 19.277) is T V C1 = 6 ∗ 80 + 200 ∗ 19.277 = 4335.4 and for combination (L , K ) = (8, 13.454) is T V C2 = 8 ∗ 80 + 200 ∗ 13.454 = 3330.8 Obviously, the first combination costs more than the second. This could be anticipated by the fact that MRTS for 6 units of labor, 4.016, is greater than the 80 w = = 0.4, which implies substituting labor for capital factor price ratio r 200 would reduce the firm’s total variable cost. (d) The firm achieves the least cost combination of labor and capital when MRTSLK w = 0.4. By using MRTS from part (b), we have is equal to the price ration r M RSTL K = 226.275L −2.25 = 0.4

−→

L 2.25 = 565.6875

resulting in L = 16.725 units. By substituting for L in the equation of the isoquant in part (a), we find K as K =

181.02 = 5.352 units 16.7251.25

(e) The combination (L , K ) = (16.725, 5.352) has a total variable cost of only 80 ∗ 16.725 + 200 ∗ 5.352 = $2,408.4. (f) The total revenue of the firm is 200 ∗ 20 = $4,000. Given the total fixed cost of the firm, then the firm’s profit is π = 4000 − 2408.4 − TFC = 1591.6 − TFC (g) The firm’s isocost associated with the total variable cost of 2408.4, is K =

80 2408.4 − L = 12.042 − 0.4L 200 200

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Fig. 10.1 Isocost and isoquant

Figure 10.1 depicts both isoquant and isocost. The point of tangency of the two curves is the least cost combination of labor and capital.

#3 (a) The firm’s marginal product of labor is MPP L =

dQ = 40L − 6L 2 dL

(b) The first order condition for MPP L local maximum is FOC :

dMPP L = 40 − 12L = 0 dL

SOC :

d 2 MPP L = −12 < 0 d L2

→

L=

40 = 3.333 12

The marginal product of labor reaches its maximum of 40 ∗ 3.333 − 6(3.3332 ) = 66.667 when L = 3.333 units. It becomes zero at MPP L = 40L − 6L 2 = 0

−→

L=

40 = 6.667 6

(c) The output reaches it maximum when FOC :

dQ = 40L − 6L 2 = 0 dL

−→

L = 6.667

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213

which is the value of L when MPP L = 0. The maximum level of output is 296.296 units. #4 (a) The equation of isoquant for 500 units of output is 2K 0.5 L 0.4 = 500

→

L 0.4 = 250K −0.5

→

L = 988211.769K −1.25

(Note that here we are solving for L in terms of K .) The MRTSKL is dL = −1235264.711K −2.25 dK

MRTSKL = 1235264.711K −2.25

−→

The least cost combination of factors occurs when MRTSLK = 1235264.711K −2.25 =

100 =2 50

→

r , then w

K 2.25 = 617632.356 and K = 374.675

Substituting for K in the equation of isoquant, we have L = 988211.769K −1.25 =

988211.769 = 599.49 units 1648.422

(b) Here TR − TC = 2500 , where TR = 500P and TC = 100 ∗ 374.675 + 50 ∗ 599.49 + Fixed cost = 67442 + Fixed cost T R = 500P = TC + 2500 = 67442 + 2500 + Fixed Cost P=

1 1 69942 + Fixed cost = 139.88 + Fixed Cost 500 500 500

If, for example, the fixed cost is $10,000, then the price would be $119.88. #5 (a) the equations of isoquants for 30, 40, and 60 units of outputs are K 0.5 = 20L −0.5

−→

K 0.5 = 26.667L −0.5 K 0.5 = 40L −0.5

−→

K = 400L −1 −→

K = 711.111L −1

K = 1600L −1

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(b) The MRTS of the 3 functions are dK = −400L −2 dL

MRTSLK = 400L −2

−→

dK = −7.11.111L −2 dL dK = −1600L −2 dL

MRTSLK = 711.111L −2

−→

−→

MRTSLK = 1600L −2

w 40 1 The factor price ratio is = = 0.25. The least cost combination of r 160 4 factors for 30, 40, and 60 units of outputs are 400L −2 = 0.25

→

711.111L −2 = 0.25

→

1600L −2 = 0.25

→

L = 40

  K = 400 40−1 = 10

→

L = 53.333

L = 80

→

→

  K = 711.111 53.333−1 = 13.333

  K = 1600 80−1 = 20

(c) We want to show that the three combinations (40, 10), (53.33, 13.333), and (80, 20) fall on a straight line. We write the equation of the line joining (40, 10) and (80, 20) and check whether combination (53.333, 13.333) is on this line. K − 10 =

20 − 10 (L − 40) 80 − 40

→

K = 0.25(L − 40) + 10 = 0.25L

The combination (53.333, 13.333) is indeed on this line, since K = 0.25 ∗ 53.333 = 13.333 (d) The equation of the linear output expansion path is K = 0.25L. #6

The equation of average product of labor is A PL =

20L 2 − 2L 3 Q = = 20L − 2L 2 L L

The maximum of this function occurs at d A PL = 20 − 4L = 0 → L = 5 units; dL

d 2 A PL = −4 < 0 d L2

10 Production Function, Least-Cost Combination of Resources …

The average product drops to zero at L = #7

215

20 = 10 units. 2

Given the production function for soybean Q = 126.6A0.6 L 0.4

(a) 10000 = 126.2(1000.6 )L 0.4 ≈ 2000L 0.4 L 0.4 = 5

−→

L = 55.9 ≈ 56 farm workers

The marginal product of labor at this level of employment is   50.48 1000.6 ∂Q 0.6 −0.6 = 50.48A L MPP L = = ∂L 560.6 MPP L = 71.48 bushels (b) Assuming no other cost, the farmer’s profit is π = TR − TC = 10000 ∗ 14.75 − 1000 ∗ 56 − 2000 = $89, 500 (c) The marginal product of labor from part (a) is 71.48. The value of the marginal product, or the marginal revenue product, of labor is M R PL = P MPP L = 14.75 ∗ 71.48 = 1054.33 Since the wage rate is $1,000, the farm workers receive a slightly less than the value of their marginal product. (d) The farm’s production function is linear homogeneous, 0.6 + 0.4 = 1, therefore by doubling the amount of land and labor the output will doubles to 20000 bushels. #8 (a) This is an example of Type II constraint optimization. Here we want to maximize output subject to a given total cost constraint. Given the prices of labor and capital, the cost constraint is 10L + 20K = 7000. This problem can now be formulated as Maximize

Q = 5.4K 0.3 L 0.5

Subject to

10L + 20K = 7000

We use the constraint and solve for L (or K ) 10L = 7000 − 20K

−→

L = 700 − 2K

(10.1SM)

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10 Production Function, Least-Cost Combination of Resources …

and substitute the result in the production function. Q = 5.4K 0.3 (700 − 2K )0.5

(10.2SM)

This reduces the function to a univariate function that could be optimized.

F OC :

1.62(700 − 2K )0.5 5.4K 0.3 dQ = =0 − 0.7 dK K (700 − 2K )0.5 1.62(700 − 2K ) − 5.4K =0 K 0.7 (700 − 2K )0.5

This leads to 1.62(700−2K )−5.4K = 0 → 1134−3.24K −5.4K = 1134−8.64K = 0 leading to K = 131.25. By substituting this value for K into (10.1SM), we determine L = 437.5 and, subsequently, the maximum level of output as   Q = 5.4 131.250.3 (437.50.5 ) = 487.88 units We now must check that output at K = 131.25 is indeed maximized. So we d2 Q must check the second order condition, which involves evaluating at dK2 K = 131.25. This is not difficult but a little messy. A quicker alternative is the dQ for a K less than 131.25 and first derivative test. In Table 10.1, we evaluated dK a K more than 131.25. As the table shows, values of the first derivative change from positive to zero then to negative, indicating that the function reaches a maximum at K = 131.25. (b) By substituting for K = 131.25 into the constraint (10.1SM), we obtain L = 700 − 2K = 700 − 2 ∗ 131.25 = 437.5 units (c) With 131.25 units of capital and 437.5 units of labor, this firm produces 487.88 units of output. Therefore (assuming no fixed cost), π = TR − TC = 20 ∗ 487.88 − 7000 = $2,757.6 dQ dK for K less than and greater than 131.25

Table 10.1 Values of

K dQ dK

130 5569.6

131.25 0

132 –4127.8

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217

#9 We form the Lagrangian function as Y = 5.4K 0.3 L 0.5 + λ(7000 − 20K − 10L) We partially differentiate Y with respect to K , L , and λ and set them equal to zero. We then solve the system of 3 equations with 3 unknowns to determine the optimal values of K , L , and λ. ∂Y = 1.62K −0.7 L 0.5 − 20λ = 0 ∂K ∂Y = 2.7K 0.3 L −0.5 − 10λ = 0 ∂L ∂Y = 7000 − 20K − 10L = 0 ∂λ We rewrite the first two equations as 1.62K −0.7 L 0.5 = 20λ 2.7K 0.3 L −0.5 = 10λ After dividing the first equation by the second equation, we have 20λ 1.62K −0.7 L 0.5 = 10λ 2.7K 0.3 L −0.5 which is simplified to 1.62L = 2 ∗ 2.7K , leading to L = 3.333K . Substituting for K in the third equation, which is simply the constraint, we have 7000 − 20K − 33.333K = 0 →

53.33K = 7000

→

K = 131.25

This value for K is exactly the same figure we obtained in exercise #8. Using L = 3.333K also leads to exactly the same value for L = 437.45, and subsequently Q = 487.88. The Lagrangian multiplier is λ = 0.0558. Note that if the firm increases its outlay by 5 % from 7000 to 7350, the firm’s output will increase from 487.88 to 507.3, where Q = 19.42 ≈ λ ∗ TC = 0.0558 ∗ 350. #10

The production function in problem #8 is now converted to

Q = 5.4(213.71)0.3 L 0.5 = 27L 0.5 (a) The short run production function is Q = 27L 0.5 . (b) With the capital fixed at 213.71 units, the firm has a fixed cost of 20 ∗ 213.71 = $4,274.2. The profit function is

218

10 Production Function, Least-Cost Combination of Resources …

π = TR − TC = 20(27L 0.5 ) − 10L − 4274.2 = 540L 0.5 − 10L − 4274.2 FOC :

dπ 270 = 0.5 − 10 = 0 dL L

SOC :

−135L −0.5 135 = − 1.5 < 0 L L

→

L 0.5 = 27

→

L = 729

∀L

Then the profit is maximized at the level of labor input 729 units. The firm’s level of output is Q = 27(7290.5 ) = 729 units with the maximum profit of $3,015.8 #11 (a) Function Q = KL is a homogeneous function. It is a second degree homogeneous, or homogeneous of degree 2. Note that if we change capital and labor by a factor of m, the output changes by a factor of m 2 . (mk)(m L) = m 2 KL = m 2 Q (b) Let us assume that the firm wants to produce 200 units of output. This means Q = 200 = KL

→

K =

200 L

The total cost function of the firm is TC = wL + r K + FC = 100L + 200K + FC Substituting for K = TC = 100L +

200 L

in this equation, we have

40000 + FC L

For the least cost combination of labor and capital we must minimize TC FOC :

dT C 40000 = 100 − =0 dL L2

SOC :

d 2 TC 80000 = > 0 d L2 L3

→

L 2 = 400 and L = 20

∀L

The second order condition for a minimum is satisfied. Subsequently, we have K =

200 200 = = 10 L 20

Thus, the combination of 20 units of labor and 10 units of capital is the least cost combination for producing 200 units of output. Note that the capital/labor

10 Production Function, Least-Cost Combination of Resources …

219

ration at the least cost combination is 10 1 w 100 K = = = = L 20 2 r 200 which holds for the least cost combination of factors for producing any level of output. For instance, to produce 800 units of output we need to double our inputs to 40 units of labor and 20 units of capital (remember that Q = KL is a homogeneous function of degree 2, therefore to 4-fold the output from 200 to 800 we need only double our inputs). Since the capital/labor ratio remains the same as 0.5 for the least cot combination of factors producing all levels of outputs, the equation of the expansion path is K = 0.5L. Figure 10.2 depicts the graphs of three isoquant for 100, 200, and 800 units of output and the expansion path. (c) We write the Lagrangian function as Y = KL + λ(8000 − 100L − 200K ) The three partial derivatives, set equal to zero, are ∂Y = L − 200λ = 0 ∂K ∂Y = K − 100λ = 0 ∂L ∂Y = 8000 − 100L − 200K = 0 ∂λ We write the first two equation as L = 200λ K = 100λ Fig. 10.2 3 Isoquants and expansion path

220

10 Production Function, Least-Cost Combination of Resources …

Next we divide the first equation by the second L =2 K

→

L = 2K

We then substitute for L in the third equation 8000 − 100L − 200K = 0. 8000 − 200K − 200K = 8000 − 400K = 0 L = 2K = 2 ∗ 20 = 40

→

→

K = 20

Q = KL = 20 ∗ 40 = 800

Then with $8,000 cost outlay this firm produces 800 units of output and employs 20 units of capital and 40 units of labor. #12

The marginal product of K and L are

MPP K =

0.35L 0.5 ∂Q = 0.35K −0.65 L 0.5 = ∂K K 0.65

MPP L =

∂Q 0.5K 0.35 = 0.5K 0.35 L −0.5 = ∂L L 0.5

For a given L , MPP K declines as K increases. Similarly, for a given K , MPP L declines as L increases. This result is confirmed by ∂MPP K = −0.65 ∗ 0.35K −1.65 L 0.5 = −0.2275K −1.65 L 0.5 ∂K ∂MPP L − 0.5 ∗ 0.5K 0.35 L −1.5 = −0.25K 0.35 L −1.5 ∂L which are both negative. The declining marginal product of labor or capital conforms to the law of diminishing marginal product, which is the root of diminishing return. This is a feature of the class of production functions Q = AK α L β , as long as 0 < α < 1 and 0 < β < 0, ∂MPP K = (1 − α)αK 2−α L β < 0 ∂K ∂MPP L = (1 − β)β K α L 2−β < 0 ∂L

10 Production Function, Least-Cost Combination of Resources …

221

Fig. 10.3 Total, average, and marginal product of labor

By setting K = 50 in the production function, we have Q = (500.35 )L 0.5 = 3.932L 0.5 A PL =

3.932L 0.5 3.932 Q = = 0.5 L L L

MPP L =

1.966 dQ = 0.5 ∗ 3.932L −0.5 = 0.5 dL L

Another feature of the class of Cobb-Douglas production functions is that MPP K = α A PK and MPP L = β A PL Graphs of the three curves are depicted in Fig. 10.3. #13 (a) This production function is homogeneous of degree α + β + γ. Note that if we change the inputs K , L , E by a factor of m, the output changes by a factor of m α+β+γ . A(m K )α (m L)β (m E)γ = m α K α m β L β m γ E γ = m α+β+γ AK α L β = m α+β+γ Q (b) if α + β + γ = 1 we have a constant return to scale; α + β + γ > 1 increasing return to scale; and α + β + γ < 1 decreasing return to scale. #14 (a) The CES production function specified here is a linear homogeneous function. Change of K and L by a factor of m leads to change in output by the same factor m.

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10 Production Function, Least-Cost Combination of Resources …

 − 1 γ δ(m K )−ρ + (1 − δ)(m L)−ρ ρ  − 1 = γ δm −ρ K −ρ + (1 − δ)m −ρ L −ρ ρ   − 1 ρ = γ m −ρ δK −ρ + (1 − δ)L −ρ =γ

  −ρ − 1  −ρ − 1 ρ δK m + (1 − δ)L −ρ ρ

  − 1 = γ m δK −ρ + (1 − δ)L −ρ ρ  − 1 = mγ δK −ρ + (1 − δ)L −ρ ρ = m Q (b) MPP K =

−1.67

∂Y = −0.67∗0.8(−1.5∗0.64K −2.5 ) 0.64K −1.5 + 0.36L −1.5 ∂K

−1.67

= −0.536(−0.96K −2.5 ) 0.64K −1.5 + 0.36L −1.5

−1.67 = 0.5146K −2.5 0.64K −1.5 + 0.36L −1.5 Similarly MPP L is MPP L =

−1.67

∂Y = −0.67∗0.8(−1.5∗0.36L −2.5 ) 0.64K −1.5 + 0.36L −1.5 ∂L

−1.67

= −0.536(−0.54L −2.5 ) 0.64K −1.5 + 0.36L −1.5

−1.67 = 0.2894L −2.5 0.64K −1.5 + 0.36L −1.5 (c) For K = 5000 and L = 8000, Y = 4775.881, MPP K = 0.751, and MPP L = 0.130 #15

The output elasticity of labor for Q = AK α L β is

ǫQ L =

βQ ∂Q L L = β AK α L β−1 = =β ∂L Q Q Q

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223

And similarly, the output elasticity of capital is ǫQ K =

αQ ∂Q L L = α AK α−1 L β = =α ∂K Q Q Q

An Alternative Solution to Problem #28 of Chapter 6, 6.8 Exercises: Based on materials covered in this chapter, if the product and the labor markets are competitive then the workers wage rate w will be the value of the marginal product of labor, that is, the marginal revenue product of labor. w = M R PL = P ∗ MPP L In problem 28 the short-run production function is given as Q = 30L 2 − L 3 and the price of a unit of output as P = $20. The problem states that the firm is using 18 units of labor, therefore MPP L =

dQ = 60L − 3L 2 dL

−→

w = P ∗ MPP L = 20 ∗ 108 = 2160

  MPP L 

L=18

= 60 ∗ 18 − 3 ∗ (182 ) = 108

Chapter 10 Supplementary Exercises 1. For a general Cobb-Douglas production function derive the relationship between the average products and the marginal products of labor and capital. 2. Use the Euler’s theorem given by Eq. (10.3) and derive the relationship between the marginal rate of technical substitution and the marginal products of labor and capital for Cobb-Douglas production functions. 3. A firm operates with the production function Q = 2.5K 0.4 L 0.5 . Assume the price of labor and capital are $10 and $20 and the firm decides to spend only $6,300 on variable production costs (i.e. net of fixed cost). (a) Use the method of Lagrange Multiplier and determine the firm’s level of output. (b) Calculate the Lagrange multiplier λ. (c) Assume that the firm increases its expenditure on variable inputs by $63 (1 %) to $6,363. Show that the level of output increases by λ ∗ 63. 4. Assume the firm in problem #3 decides to increase its output in part (a) by 2 %. By how much, approximately, it must raise its expenditure? 5. A firm with a CD production function Q = 1.5K 0.5 L 0.4 plans to produce 900 units of output per day. Assume the prices of labor and capital are $50 and $100 a day.

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10 Production Function, Least-Cost Combination of Resources …

(a) What is the firm’s least cost combination of factors for production of 900 units of output? (b) Assuming no fixed cost, what is the market price of this good if this firm makes a $2,000 profit each day? 6. In problem # 5, what would be the new optimal combination of labor and capital for production of 900 units if the price of labor is increased to $80? Is there a substantial substitution of capital for labor? 7. A firm has the production function Q = 1.2K 0.55 L 0.5 . (a) Write equations of three isoquants associated with 360, 480, and 600 units of output. (b) If the prices of labor and capital are $20 and $60 per unit, respectively, find the least-cost combination of labor and capital for 360, 480, and 600 units of output. (c) Show that the three combinations in part (b) fall on a straight line 8. Consider a firm with production function Q = 2.2K 0.45 L 0.4 . Assume P, w, r and fixed cost FC are 200, 100, 150, and 200, respectively. Determine the firm’s profit maximizing level of output, employment of resources, and total profit. 9. A firm operates with the production function Q = 1.4K 0.4 L 0.5 . (a) If the prices of labor and capital are $10 and $20, respectively, and the firm’s allocates only $8,000 a day for production costs (net of fixed cost) how many units of output can the firm produce? (Use the substitution method). (b) How many units of labor and capital does firm employ? (c) If the price of the good is $35 per unit, what is the daily profit of this firm? 10. Re-do part (a) of problem #9, using the Lagrange Multiplier method. 11. Use the multiplier from problem #9 and determine the change in the output if the firm decides to increase it expenditure outlay by 1 %. 12. Use the multiplier from problem #9 and estimate by how much the firm’s cost would increase if the firm wants to expand its output by 1 %. 13. A firm has the following short run production function Q = 45L 2.5 − 4L 3 . (a) Write the equations for the marginal and average product of labor. (b) At what level of labor input does the A P of labor reach its peak? (c) At what level of labor input does it drop to zero? 14. A farmer uses land and labor to grow corn. He has the following increasing return to scale production function Q = 130 A0.6 L 0.5 where Q is the amount of corns in bushels, A is land in acres and L is number of farm workers. (a) If this farmer has 120 acres of land, how many farm workers must he employ in order to produce 20,000 bushels of corn? What is the marginal product of labor at this level of employment?

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225

(b) Assume this farmer pays each farm worker $800 over the growing season and spends $3,000 for seed and fertilizer. If the price of corn in the spot market is $4.20 a bushel, what is the farmer’s profit (assume no fixed cost)? (c) Is this farmer paying his workers the value of their marginal product? (d) If this farmer doubles the amount of land and labor, how many bushels of corns he can produce? 15. The yield of corn per acre for the farmer in problem #14 is about 167 bushels. The average yield of corn per acre in Iowa is 187 bushels. If our farmer in problem #14 wants to achieve this yield (a) How many farm workers he must employ? (b) What is the change in his profit? (c) What would be the farmer’s profit if he pays his workers the value of their marginal product? 16. Assuming no fixed costs, show that if a firm with Cobb-Douglas production function pays its labor and capital resources the value of their marginal products then it (a) Earns normal profit if the function is constant return to scale. (b) Earns above normal profit if the function is decreasing return to scale. (c) Incurs losses if the function is increasing return to scale. 17. Given that the partial output elasticity of labor in problem #3 is 0.5 (keeping the level of capital input the same), (a) How many units of additional labor the firm must employ in order to increase its output by 2 %? (b) What is the percentage increase in the variable cost? 18. Given that the partial output elasticity of capital in problem #3 is 0.4 (keeping the level of labor input the same), (a) How many units of additional capital the firm must employ in order to increase its output by 2 %? (b) What is the percentage increase in the variable cost? 19. Show that for production function Q = AK α L β the output elasticity of scale ǫ Q A is 1. 20. Consider a firm with the production function Q = 3.5K 0.45 L 0.4 . Assume the prices of the good, labor, and capital are $150, $80, and $120, respectively. Assuming $2,000 fixed cost, determine the firm’s profit maximizing level of output, the optimal combination of capital and labor, and profit. 21. Use the result of problem #19 and determine the change in the scale factor required so that the firm in problem #20 can increase its output by 5 %. 22. What is the percentage change in the firm’s profit in problem #20 due to 5 % increase in the scale of operation?

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23. An analyst for Davidson Steel Company uses a Cobb-Douglas production function Q = A K α Lβ for forecasting the company’s output. By using quarterly data from 1996 to 2012, she estimates the model as ln Qˆ t = 0.223 + 0.482 ln K t + 0.518 ln L t (a) What is the estimate of A? (b) Is this a constant return to scale production function? (c) What would be the steel company’s output if it employs 150 units of capital and 200 units of labor? 24. Shown that for positive values of a, b, n, and m the following production functions are linear homogeneous: (a) Q = a K + bL (b) Q =

bL m aKn + L n−1 K m−1

25. Find the expansion paths for the following production functions: 1

1

(a) Q = 12K 2 + 15L 2 (b) Q = 4KL − 2K 2 + 2L 2

Chapter 11

Economics Dynamics

Chapter 11, Exercises 11.4 1. Determine the order of the following difference equations. (a) (b) (c) (d) (e) (f) (g)

4yt+2 + 2yt+1 − 5yt = 8 yk+2 − yk+1 − 5yk = k yt+1 = 5yt − 6t 12yt+2 − 5yt = log t yt+2 − 3yt+1 + 2yt = 0 (1 + t) yt+2 + 2t yt+1 − 5yt = 6t + 1 3yt+2 − 2yt+1 + 5yt−1 = 0

2. Classify the difference equations in problem (1) as (a) (b) (c) (d)

linear or nonlinear. with or without constant coefficients. autonomous or non-autonomous. homogeneous or non-homogeneous.

3. Find the general solution of the following equations. Explain the behavior of y. Find the particular solutions for the given initial values. Determine whether y converges to the steady state. (a) (b) (c) (d)

yt+1 = 5yt − 6 4yt+2 + 3yt+1 = 2 yt+1 = 0.34yt + 4 2yt+1 = −0.34yt − 4

y0 = 10 y0 = 4 y0 = 2.5 y0 = 5

4. Let K t denote the market value of capital stock of a firm at the beginning of year t. Assume λ is the rate of firm’s capital depreciation each year. Further assume that the firm makes I dollar of new capital expenditure or investment each year.

© Atlantis Press and the authors 2015 S. Vali, Principles of Mathematical Economics II, Mathematics Textbooks for Science and Engineering 4, DOI 10.2991/978-94-6239-088-1_11

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(a) Write the first order difference equation relating the capital stock of the firm at the beginning of the year K t+1 to the capital stock and investment in year t (b) Solve the equation. What would happen to the firm’s capital stock if (I) Investment is more than depreciation? (II) Investment is only for replacement of the depreciated stock? (III) Investment is less than depreciation? 5. In problem (4) assume that the firm’s initial capital stock is $5 million. Also assume the rate of capital depreciation is 10 % a year. What would be the firm’s capital stock after 5 years if annual investment are (I) $600,000? (II) $400,000? (III) $500,000? 6. Assume the firm in problem (4) has the following production function Q t = 0.45 K t0.5 What is the firm’s level of output after 7 years if annual investment were the same as I, II, or III in problem (5)? 7. Let Pt denote the aggregate price level at time t. The inflation rate from time t − 1 to t, denoted by τ , is defined as τt =

Pt − Pt−1 Pt−1

(a) Write the first order homogeneous difference equation for the price level and solve it. (b) If the annual inflation rate remains stable at 3.5 %, what is the aggregate price index after 10 years? (Hint: the initial price level is the price index at the base year). (c) After how many years the price level doubles? 8. Inflationary Spiral and Hyperinflation When prices increase rapidly then the inflation rate grows from period to period. A rapid growth of inflation rate creates what is known as the inflationary spiral, which if not contained would lead to a hyperinflation. If the inflation rate τ in problem (7) grows at the rate of 100λ percent per period, then τt = τt−1 + λ τt−1 expresses the relations between inflation rates in two consecutive periods.

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(a) Assuming τ0 is the initial inflation rate, write the general solution for the inflation rate. (b) Re-do part (a) in problem (7) after substituting τt by its general solution from part (a). (c) Assume that the initial inflation rate is 35 % and grows at the rate of 10 % each month. What is the aggregate price index after one year? [Note: Solution to the price level in part (b) is a crude approximation; otherwise the actual aggregate price index in period t has a more complicated structure. If P0 and τ0 are the initial values of the price index and inflation rate, and λ is the rate of growth of inflation, then the price index at period t is Pt = P0 (1 + τ0 )[1 + (1 + λ) τ0 ][1 + (1 + λ)2 τ0 ][1 + (1 + λ)3 τ0 ] . . . [1 + (1 + λ)t τ0 ] or Pt = P0

t 

[1 + (1 + λ)i τ0 ]

i=1

  where is product notation, similar to which is summation notation. Students are encouraged to try and drive this expression.] 9. Econometricians/Time Series Analysts refer to linear difference equations with constant coefficient as Autoregressive models. The first and second order difference equations are called the first and second order autoregressive models and generally denoted by AR(1) and AR(2). AR(1) and AR(2) are widely used for estimating economic time series, notably macroeconomic series, with upward or downward trend. The following is an example of AR(1) estimate of US money supply. The narrowest measure of money that the Federal Reserve Bank reports is M1. M1 includes currency, checking account deposits, and the traveler’s checks. A broader measure of monetary aggregate is M2. M2 consists of M1 plus saving deposits, small denomination (less than $100,000) time deposits, money market mutual fund deposits, and several other liquid financial assets. Using monthly data from January 2005 to September 2006, the following AR(1) model for M2 was estimated M2t = 1.0043 M2t−1 − 5.05 (a) Solve the first order difference equation for the money supply and analyze its behavior over time. (b) If in January 2005 the money supply was $6,415.1 billion, what is your estimate of October, November, and December 2006 money supply? (c) If the actual money supply for October, November, and December 2006 are $6,936.2, $6,977.0, and $7,021.0 billion respectively, what is the models average estimation error for the three month? 10. Assume the market supply and demand for a perfectly competitive market are given by the following equations

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Q st = −c + d Pt Q dt = a − b Pt + δ Q dt−1 where a, b, c, d > 0 ; −1 < δ < 1; and Q dt−1 is the quantity demanded at time t − 1. Derive the first order difference equation for P by assuming that the market reaches equilibrium in each period. Find the general solution for the equation and determine the behavior of P over time. Answers to Chapter 11, Exercises 11.4 # 1 and 2 Answers to parts (a) to (g) in problem #1 and parts (a) to (d) in problem #2 are organized in Table 11.1 #3 (a) The general solution to the first order non-homogeneous difference equation with constant coefficients yt+1 = ayt + b is 1 − at yt = a y0 + b 1−a t





In the equation in part (a), we have a = 5 and b = −6, therefore yt = 5t y0 − 6



1 − 5t 1−5



yt = 5t (y0 − 1.5) + 1.5

= 5t y0 + 1.5(1 − 5t ) = 5t y0 − 1.5(5t ) + 1.5 →

for y0 = 10

→

yt = 8.5(5t ) + 1.5

This is a monotonically increasing y diverging to larger and larger values.

Table 11.1 Answers to questions number 1 and 2 Part Autonomous Homogeneous Order (a) (b) (c) (d) (e) (f) (g)

Yes No No No Yes No Yes

No No No No Yes No Yes

Second Second First Second Second Second Third

Linear

With const. coeff.

Yes Yes Yes Yes Yes Yes Yes

Yes Yes Yes Yes Yes No Yes

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(b) We write the equation as 3 yt+2 = − yt+1 + 0.5 = −0.75yt+1 + 0.5 with solution 4   1 − (−0.75)t yt = (−0.75)t y0 + 0.5 1 − (−0.75) yt = (−0.75)t y0 + 0.286[1−(−0.75)t ]. With y0 = 4 the equation becomes yt = 3.714(−0.75)t + 0.286 This equation is damped oscillatory and converges to 0.286. (c) The general solution of the equation yt+1 = 0.3yt + 4 is yt = (0.34)t y0 + 4



1 − 0.34t 1 − 0.34



= 0.34t y0 + 6.061 − 6.061(0.34t )

The particular solution for y0 = 2.5 is yt = 2.5(0.34)t − 6.061(0.34)t + 6.061 = −3.561(0.34)t + 6.061 This function steadily increases and converges to 6.061. (d) The general solution to equation 2yt+1 = −0.34yt − 4 is yt+1 = −0.17yt − 2

→

1 − (−0.17)t yt = (−0.17) y0 − 2 1 − (−0.17) t





yt = (−0.17)t y0 − 1.709[1 − (−0.17)t ] The particular solution for y0 = 5 is yt = 5(−0.17)t − 1.709 + 1.709(−0.17)t = 6.709(−0.17)t − 1.709 This function converges to −1.709. Figure 11.1 depicts the behavior of the function over tine. #4 (a) The difference equation relating the capital stock of the firm to depreciation and capital investment is K t+1 = (1 − λ)K t + It

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Fig. 11.1 Graph of yt = 6.709(−0.17t ) − 1.709

Let’s assume that the investment each year is a fraction, α, of the capital stock of that year, that is 0 1 and the firm’s capital stock grows. Otherwise if it is less than depreciation, that is, if α < λ then (1 − λ + α) < 1 and the stock of capital of the company diminishes over time. In case of equality of investment and depreciation, that is, when α = λ then (1 − λ + α) = 1 and the stock of capital remains unchanged. #5 In this problem, the assumption is that the firm makes a fixed amount of new investment each year. The difference equation relation the capital stock of the firm to depreciation and investment is now K t+1 = (1 − λ)K t + I The solution to this first order non-homogeneous difference equation is 1 − (1 − λ)t K t = (1 − λ) K 0 + I 1 − (1 − λ) t





= (1 − λ)t K 0 +

I 1 − (1 − λ)t λ

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233

Fig. 11.2 Graph of K t = −(0.9)t + 6

By plugging the given values for K 0 , λ, and I in this equation, for part (I) we have K t = (1 − 0.1)t 5 +

0.6 [1 − (1 − 0.1)t ] = 5(0.9)t + 6(1 − 0.9t ) = −(0.9)t + 6 0.1

The firm’s capital stock after 5 years is K 5 = −0.59 + 6 = 5.41. Over a short period of time this function monotonically increases but over a longer period converges to 6. Note that at earlier periods the capital investment of $600,000, or $0.6 million, is more than $0.5 million depreciation, so the firm’s capital accumulates. With larger capital the amount of depreciation becomes greater than $0.5 million. When the firm’s capital grows to $6 million, then the volume of depreciation $0.6 million will be exactly the amount of investment and the firm reaches a steady state of $6 million worth of capital. Figure 11.2 is the graph of the function. In part (II), where the amount of investment is $400,000, or $0.4 million, the difference equation becomes K t = (1 − 0.1)t 5 +

0.4 [1 − (1 − 0.1)t ] = 5(0.9)t + 4(1 − 0.9t ) = (0.9)t + 4 0.1

The amount of firm’s capital after 5 years is $4.49 million. In this case the volume of firm’s capital steadily declines and converges to $4 million. In the third case (III) the capital stock of the company remains unchanged. Note that K t = (1 − 0.1)t 5 +

0.5 [1 − (1 − 0.1)t ] = 5(0.9)t + 5(1 − 0.9t ) = 5 0.1

#6 For case (I), we use K t = −(0.9)t + 6 and determine the level of capital stock for period 7, which then used in the production function to calculate the level of output, K 7 = −(0.9)7 +6 = 5.5217

→

Q 7 = 0.45(K 7 )0.5 = 0.45(5.5217)0.5 = 1.0574

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For case (II), we use K t = (0.9)t + 4 K 7 = (0.9)7 + 4 = 4.4783

Q 7 = 0.45(4.4783)0.5 = 0.9523

→

For case (III), there is no change in the stock of capital, therefore, Q 7 = 0.45(5)0.5 = 1.0062 #7 (a) The first order homogeneous equation is τt Pt−1 = Pt − Pt−1

−→

Pt = (1 + τt )Pt−1 with solution

Pt = (1 + τt )t P0 (b) This means that τt = 0.035 remains the same for all t. Therefore, we have Pt = (1.035)t P0 Given that the initial price level P0 is the base year price index, then its value must be set to 100. Subsequently P10 = (1.035)10 100 = 141.06 (c) When the price level doubles, we have 2P0 = (1.035)t P0 t=

→

(1.035)t = 2

→

t ln(1.035) = ln 2

ln 2 = 20.15 years or approximately 20 years and 2 months ln 1.035

#8 (a) The general solution for the inflation rate is τt = (1 + λ) τt−1

−→

τt = (1 + λ)t τ0

(b) Substituting for τt in the general solution for the price level in part (a) of problem (7), we have

t Pt = 1 + (1 + λ)t τ0 P0 (c) After substituting 0.35 for τ0 , 0.1 for λ, and 100 for P0 in the above equation, we have

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235

Fig. 11.3 Graph of

t Pt = 1 + 0.35(1.1)t 100



t

t Pt = (1 + (1 + 0.1)t 0.35 100 = 1 + 0.35(1.1)t 100 and after 1 year, t = 1, the price index is P1 = (1 + 0.385)100 = 138.5 Although the problem does not ask us to find the level of prices beyond the first year, but it is very instructional to calculate the price index after 4 or 5 years to see the explosive nature of the prices. Figure 11.3, which depicts the graph of the model, clearly displays the rapid increase in the level of prices over time. Hyperinflation is not a thing of very distant past, like Wiemar Republic in Germany after the First World War. As late as 1990s many countries around the world, especially in Latin America, were suffering from hyperinflation. In his speech to the Stanford Institute for Economic Policy Research in February 2005, Ben S. Bernanke, then the chairman of the FED board of governors, remarked A measure of price changes in nine of the most populous Latin American countries shows that inflation in the region averaged nearly 160 percent per year in the 1980s and 235 percent per year in the first half of the 1990s. Indeed, high inflation morphed into hyperinflation – conventionally defined as inflation exceeding 50 percent per month (Cagan 1956) – in a number of Latin countries during the latter part of the 1980s and in the early 1990s. Brazil’s inflation rate, for example, exceeded 1,000 percent per year in four of the five years between 1989 and 1993. Other Latin American countries suffering hyperinflation at about that time included Argentina, Bolivia, Nicaragua, and Peru. A particularly striking aspect of this poor inflation performance is that it occurred while most of the rest of the world was reducing inflation to low levels.

#9 The estimated AR(1) model for M2 money supply is given as M2t = 1.0043 M2t−1 − 5.05

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(a) Solution to the model is 1 − (1.0043)t M2t = (1.0043) M20 − 5.05 1 − 1.0043 

t



M2t = (1.0043)t M20 + 1174.42(1 − 1.0043t ) (b) If we designate January 2005 money supply as M20 then October, November, and December of 2006 are M221 , M222 and M223 , respectively. M221 = (1.0043)21 6415.1 + 1174.42(1 − 1.004321 ) = 6909.25 M222 = (1.0043)22 6415.1 + 1174.42(1 − 1.004322 ) = 6933.91 M223 = (1.0043)23 6415.1 + 1174.42(1 − 1.004323 ) = 6958.67 (c) The model’s average estimation error for the 3 month using Mean Absolute Percentage Error (MAPE) is best calculated by organizing the information in Table 11.2. The estimation error based on MAPE is then MAPE =

0.0039 + 0.0062 + 0.0089 ∗ 100 = 0.63 % 3

Chapter 11 Supplementary Exercises 1. Specify the order of the following difference equation assumed to be defined over the set of t− or k−values 0, 1, 2, . . . . (a) 3yt−1 + 2yt = 10 (b) 0.9yk+2 + 2yk+1 − 5yk = 0 (c) 10yt+1 − 5yt + ln t = 5 (d) (1 + 2k)yk+1 + kyk − 5k + 5 = 0 2 (e) 3xt+1 − 5xt = 25

Table 11.2 Actual and estimated values of M2

Actual Ai

Estimated E i

| Ai −E i | Ai

6936.2 6977.0 7021.0

6909.25 6933.91 6958.67

0.0039 0.0062 0.0089

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( f ) − 5wt+2 − ln wt+1 + wt = 80 (g) 8yt+1 − 2yt + 2yt+1 yt = 10 (h) − 2.5z t−2 + 5z t−1 + 2t z t = 12 2. Classify the difference equations in problem #1 as (a) (b) (c) (d)

Linear or nonlinear. With or without constant coefficients. Homogeneous or non-homogeneous. Autonomous or non-autonomous.

3. Find the general solution of the following equations. Explain the behavior of y. Find the particular solutions for the given initial values. Determine whether y converges to the steady state. (a) (b) (c) (d)

yt+1 = 3yt − 10 2yt+2 + yt+1 = 5 2yt+1 = 0.34yt + 6 4yt+1 = −0.44yt − 4

y0 = 5 y0 = 5 y0 = 2 y0 = 4

4. Assume the value of a piece of machinery at time t, Vt , is expressed as Vt = (1 − λ)Vt−1 − M where λ is the depreciation rate and M is the maintenance costs. Write the general solution of the difference equation. Assume the initial value of the machinery is $500,000, the depreciation rate is 10 % annually, and the maintenance cost is $2,000, (a) What is the value of the equipment after 5 years? (b) What is its value after 10 years? (c) Approximately after how many years the value drops to 0? 5. The British economist Roy Forbes Harrod is an early contributor to the field of economics dynamic. The following is an example of dynamic national income model similar to his earliest work: Yt = Ct + It + G 0 Ct = bYt r It = Yt − Yt−1 where r in the investment equation is called the warranted rate of investment growth, considered to be a rate of growth in investment needed to maintain the full employment in the economy. Rewrite the investment equation as

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It =

1 1 Yt − Yt−1 r r

and write the first order difference equation for the national income. Find the general solution of the equation. 6. In problem #5, assume r = 0.08, b = 0.8, G 0 = 120, and the initial value of Y is Y0 = 2295. Find the values of the endogenous variables after 10 years. 7. In problem #6, change r from 0.08 to 0.10 and recalculate the values of the endogenous variables after 10 years. 8. With values of b, G 0 and Y0 the same as given in problem #6, determine what warranted rate of investment growth, r , is needed so that the size of the economy doubles over 15 years. 9. Consider the following market supply and demand equations Q dt = 250 − 12Pt Q st = −50 + 4.8Pt−1 Determine the equilibrium price and quantity in the market. Graph the path of price to equilibrium. 10. Based on annual data the following first order autoregressive model for the US GDPz is estimated GDPt = 1.0417GDPt−1 + 113.6557 (a) Write the general solution for the GDP. (b) If the initial value of the GDP for year 1978 is $2,294.7 billion, find the estimate of the GDP for year 2000 and 2005. (c) If the actual values for GDP of 2000 and 2005 are $9,817 and $12,433.9 billion, find the estimation error of the model based on MAPE.

Chapter 12

Mathematics of Interest Rates and Finance

Chapter 12, Exercises 12.8 1. How much will you have after 6 years if you invest $15,000 at 6.5 % per year compounded annually? Quarterly? 2. How much will you have at the end of 10 years if you invest $20,000 at 3.5 % per year compounded monthly? 3. How much do you need to invest now at an 8 % interest rate compounded semiannually in order to have $10,000 in 5 years? 4. Find the present value of $12,000 due 4 years from now if the rate of interest is 6.0 % per year and is compounded quarterly. 5. How long does it take for an amount of money invested at 4.5 % interest compounded monthly to triple? 6. What rate of interest compounded quarterly doubles an investment over 10 years? 7. What rate of interest compounded continuously doubles an investment over 10 years? Effective interest rate: The effective interest rate is your true interest rate cost of borrowing, stated as an annual simple rate. Frequent compounding results in a higher effective interest rate. 8. What is the effective interest rate of a nominal interest rate of 6.9 % per year compounded monthly? 9. Assume a $10,000 bond with a coupon rate of 4.5 % and 5 years to maturity is priced at $9,800. (a) What is its approximate YTM? (b) What is the investor’s rate of return?

© Atlantis Press and the authors 2015 S. Vali, Principles of Mathematical Economics II, Mathematics Textbooks for Science and Engineering 4, DOI 10.2991/978-94-6239-088-1_12

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10. Assume a bond with a coupon rate of 5.5 % and 7 years to maturity is traded at $102.50 per $100 face value. (a) What is its approximate YTM? (b) What is the investor’s rate of return? 11. A $20,000 bond with coupon rate of 6 %, maturing in 3 years, is traded at par. (a) What is its approximate YTM? (b) What is the investor’s rate of return? 12. Using results from Exercises 9, 10, and 11, express the relationship between a bond’s YTM and its rate of return when it is traded at par, premium, or discount. 13. What is the market value of a 30-day $10,000,000 commercial paper if the yield of 30-day Treasury bill is 2.5 %? 14. A 60-day $10,000,000 commercial paper is offered at $9,964,000. What is the discount rate? 15. What is the day count of a $5,000,000 commercial paper sold for $4,900,000 if the discount rate is 3.5 %? 16. An investor purchases a property in the year 2000 for $450,000. She sells the property in 2006 for $550,000. During the 6 years that she owns the property, it generates a monthly rental income (net of monthly mortgage payments and maintenance costs) of $400 a month. What is the investor’s rate of return? 17. You get a 3-year $20,000 car loan at 6.5 % interest compounded monthly. What is the amount of your monthly payment? 18. What is the monthly mortgage payment of a 25-year $250,000 fixed loan if the mortgage rate is 4.75 %? 19. What proportion of the 50th monthly payment in Exercise 18 is applied to the principal? 20. What is the accumulated equity in Exercise 18 after the 120th monthly payment? What is the amount of the loan outstanding? 21. You take out a 30-year fixed $500,000 mortgage with the rate of 6.25 %. After 5 years you decide to refinance the balance of the loan with another 30-year fixed mortgage. What is the upper bond of the advantageous rates? 22. A family borrows $350,000 from a mortgage company. The term of the mortgage is 25-year fixed with 5.8 % interest. What is the optimal time horizon for this family to refinance the balance of this loan with a new 4.3 %, 25-year fixed mortgage? 23. You purchase your home with a $200,000, 15-year fixed, 4.5 % mortgage loan. After 7 years, 84 payments, you are offered to refinance the balance of your loan with a 10-year 3.75 % fix-rate mortgage. Is refinancing to your financial advantage?

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24. In Exercise 23, assume that the closing cost of refinancing is $4,000, which you agree to be added to the balance of your loan. What range of mortgage rates makes the refinancing financially to your advantage? 25. A family purchased a house in January 15, 2007 with a 25-year, $350,000, 6.35 % mortgage. In March 15, 2012, when the mortgage rates dropped to historic lows, they refinance the balance of the loan with a 15-year, 4.25 % mortgage. How much does this family save? 26. A family purchased a house in 2008 with a 15-year, $250,000, 5.15 % mortgage. In 2013 after 80 payments they are offered a 10-year loan with the rate that saves the family about $8,000 over the original loan. What is that rate? 27. You borrow $10,000 for one year with the nominal interest rate of 6.9 % per year compounded monthly. You are required to pay the lender the principal and interest after one year. What is the APR of this loan if the lender charges a $350 application and processing fee, and you agree that the $350 should be added to your loan? What is the APR if you instead withdraw $350 from your saving account, which pays 3.5 %, and pay the charge up front? 28. A savings plan is created by depositing $250 a month into an account that pays 4.5 % interest compounded quarterly. What is the value of this fund after 15 years? 29. Parents create a college fund for their newly born child. They deposit $800 every quarter in the fund, which pays 5.5 % interest compounded quarterly. What is the balance in the fund after 18 years? 30. A newlywed couple sets up a fund to save $80,000, to be used as a down payment for buying a house. How much do they have to contribute monthly to the fund over 12 years if the fund pays 4.5 % interest compounded monthly? 31. You contribute $200 a month to an account that pays 3.76 % interest compounded monthly. How long does it take for you to accumulate $32,000 in the account? 32. What rate of interest, compounded monthly, grow the value of a savings fund to $32,000 over 10 years, if the monthly contribution were $200? 33. You buy a $5,000 face value bond with a coupon rate of 4.5 % for $4,750. Three years left to the bond’s maturity, the interest rate on similar instruments with the same risk jumps to 5.5 %. What is the value of your bond now? 34. What would be the value of your bond in Exercise 33 if the interest rate drops to 4.0 %? 35. In our example in Sect. 12.7, the annual cost of a four-year college was assumed to be $30,000, or $120,000 for 4 years. This figure is in the ball park of the national average for private universities. According to Fidelity Investments, a major financial services company who manages 529 plans, the current estimate of the national average amount of money needed to cover 4 years of expenses (tuition, board, books, …) at a private university is $115,000. The average amount

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is $51,000 for public colleges. Fidelity estimates that the cost of college is rising at an average of 5.4 % a year, twice the overall rate of inflation. Use Eq. (12.31) and determine the monthly amount a family needs to contribute to a college fund for 18 years in order to cover the $45,000 annual cost of a 4-year college. Answers to Chap. 12, Exercises 12.8 #1 By using the general formula for compound interest Pt = P(1 +

i mt ) m

we can determine that $15,000 at 6.5 % interest, compounded annually, after 6 years grows to   0.065 6∗1 = $21,887.13 P6 = 15,000 1 + 1 and with the quarterly compounding it grows to   0.065 4∗6 = $22,085.37 P6 = 15,000 1 + 4 #2 We have P10

  0.035 12∗10 = 20,000 1 + = $28,366.90 12

#3 The amount you need is the present value of $10,000 five years from now when the interest rate is 8 %. P V = 10,000

  0.08 −2∗5 1+ = $6,755.64 2

#4 The present value of $12,000 four years from now when the interest rate is 6.0 % compounded quarterly is   0.06 −4∗4 P V = 12,000 1 + = $9,456.37 4 #5 The length of time required for money invested at 4.5 % compounded monthly to triple is calculated by assuming that the initial investment is P and grows to 3P, therefore

12 Mathematics of Interest Rates and Finance

  0.045 12∗t 3P = P 1 + 12

243

(1.00375)12t = 3

→

By taking the log of both sides, we have 12t ln(1.00375) = ln(3) → 0.044916t = 1.0986122

→

t ≈ 24.5 years

#6 Here we must determine the interest rate that doubles an investment of P dollars over 10 years   i 4∗10 2P = P 1 + 4

→



i 1+ 4

40

=2

Taking the logarithm of both sides, we have   i = ln 2 = 0.69315 40 ln 1 + 4 1+

i = e0.17329 = 1.0175 4

→

→

  i ln 1 + = 0.017329 4

i = 0.0175 4

→

i = 0.07

#7 Using the continuous compounding formula Pt = Peit , we write 2P = Pe10i ln e10i = ln 2

→ →

e10i = 2 10i = 0.69315

→

i = 0.069315

#8 Here we want to determine a nominal annual rate equivalent to 6.9 % compounded monthly. In general, if i is the nominal rate compounded m times per year, then the effective rate r is determined by   i m −1 r = 1+ m In this example i = 0.069 and m = 12, therefore   0.069 12 − 1 = 0.0712 or 7.12 % r = 1+ 12 #9 (a) Since this bond is traded at a discount then its BAE is BAE = Coupon Payments + Prorated Capital Gain

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12 Mathematics of Interest Rates and Finance

BAE = 0.045 ∗ 10,000 + AP =

10,000 − 9,800 = 490 5

Face value of Bond + Price of Bond 10,000 + 9,800 = = 9,900 2 2

YTM =

BAE 490 ∗ 100 = ∗ 100 = 4.95 % AP 9,900

As it was noted in the text, in Eq. (12.16) on page 396, the interest rate i that makes the left and right hand sides of the equation equal is the exact YTM. This exercise is not asking us to calculate exact YTM, but it is a good practice to use the numerical method in Excel to solve problems of this nature. We start by generating a series of interest rates in column A. We can put 0.015 (that is 1.5 %) in cell A1 and then increment it by 0.005 from cell A2 to cell A15. This generates numbers 0.015, 0.02, 0.025, . . . , 0.085 (of course, YTM is not going to be widely different from the coupon rate, so we don’t actually need to generate numbers for rate too far from the coupon rate.) Next we enter the right hand side of the formula in cell B1, by entering A1 for i, and then copy it to cells B2 to B15. In the cells under column C, we find the difference between values in column B and the bond price $9,800. When elements in column C change from positive to negative (or negative to positive) we know that the correct interest rate is a number between the two corresponding values in Column A. We then refine our search by choosing smaller increments between the two interest rates. When the numbers in column C get very close to zero, we have approached the correct YTM. For our current problem, this method yields 4.852 % as nearly the exact YTM. Readers should be alerted to the fact that many texts, business calculators, and on-line financial calculators calculate YTM by assuming annual rather than monthly compounding. If in this problem we indeed assume annual compounding, the value of YTM would be 4.962 %, much closer to 4.9 % approximated above. Another twist is whether the coupon is paid annually or semiannually (which is more common). If semi-annually, then the formula (12.10) on page 396 must be modified. Readers are urged to try to derive the modified formula. In this example if the coupon is paid semi-annually, the value of YTM is 5.136 %. (b) We first calculate the investor’s holding period rate of return HPRR =

2,450 5 ∗ 450 + 200 ∗ 100 = ∗ 100 = 25 % 9,800 9,800

and then determine the rate of return, or more accurately, the average annual rate of return as RR =

25 = 5% 5

12 Mathematics of Interest Rates and Finance

245

#10 We solve this problem for $100 face value of the bond (a) This bond is traded at a premium, therefore BAE = Coupon Payments − Prorated Capital Loss BAE = 0.055 ∗ 100 − AP =

2.5 = 5.5 − 0.36 = 5.14 7

102.5 + 100 = 101.25 2

YTM =

BAE 5.14 ∗ 100 = ∗ 100 = 5.079 % AP 101.25

For the record, the exact YTM, assuming annual compounding and an annual coupon payment, is 5.07 %. If compounding is monthly, the bond’s exact YTM is 4.953 %. (b) Investor’s HPRR is HPRR =

5.5 ∗ 7 − 2.5 ∗ 100 = 36 % 102.5

→

RR =

36 = 5.14 % 7

#11 Since the bond is traded at par, there is no capital gain or loss and BAE = the coupon payment. The average price AP is equal to the face value of the bond. Subsequently (a) YTM =

0.06 ∗ 20,000 1,200 ∗ 100 = ∗ 100 = 0.06 or 6 % 20,000 20,000

which is exactly the same as the coupon rate. (b) 3 ∗ 1,200 ∗ 100 20,000 RR = = 6 % which is exactly the coupon rate 3 #12 If a bond is traded at a discount, then RR > YTM. If a is bond traded at a premium, then RR < YTM. If bond is traded at par then RR = YTM. #13 In this problem the face value of the instrument is $10,000,000, the discount rate is 2.5 %, number of compounding is 12, day count is 30, and the annual basis is 360. We plug these values into the expression for the price of the money market instrument that is given on page 401 and reproduced here as Eq. (12.1SM).

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12 Mathematics of Interest Rates and Finance

P=

P=

Face value of instrument   m∗ day count  discount rate annual basis 1+ m 10,000,000  1+

12∗ 30 0.025 360

(12.1SM)

= $9,979,209.97

12

#14 Using Eq. (12.1SM) given above, we determine i from the following equation 60   i −12∗ 360 9,964,000 = 10,000,000 1 + 12 This expression is simplified to   i 2 10,000,000 1+ = 1.003613 = 12 9,964,000   √ i 1+ = 1.003613 = 1.001804 12

→

and i = 12 ∗ 0.001804 = 0.0216 or 2.16 %. #15 Using Eq. (12.1SM) given above, we have 4,900,000 =

5,000,000    12∗ day count 0.035 360 1+ 12

day count 50 30 = 1.02041 (1.00292) = 49 



By taking the logarithm of both sides, we have 

day count 30



ln(1.00292) = ln(1.02041)



day count 30



=

0.020205 = 6.9296 0.002916

day count = 30 ∗ 6.9296 ≈ 208 days

i = 0.001804 12

12 Mathematics of Interest Rates and Finance

247

#16 The stream of income generated by the property is 6 ∗ 12 ∗ 400 = $28,800. The capital gain realized by the investor is 550,000 − 450,000 = $100,000. Then the holding period rate of return is HPRR =

28,800 + 100,000 ∗ 100 = 28.62 % 450,000

28.62 And the rate of return or average annual rate of return is RR = = 4.77 %. 6 This calculation of rate of return is based on the assumption that interest is not compounded. If we take into account compounding, especially monthly compounding which is common in the money market, then we must take a different approach for calculating yield or rate of return. We can treat this investment as purchasing a $450,000 bond which generates a $400 monthly income. This bond matures after 6 years and its value at maturity is $550,000, generating a $100,000 capital gain. The rate, i, that equates the present value of the 6 years stream of $400 monthly income plus the present value of $550,000 with $450,000 is a measure of return similar to the yield to maturity, that is 450,000 =

72  t=1

450,000 = 400

400 550,000 +



12 t i i 12∗6 1 + 12 1 + 12 72  t=1

1

1+

i 12 t 12

550,000 +

i 12∗6 1 + 12

 72   i −12 t 1+ in the above expression is the sum of 72 terms of a The sum 12 t=1

i −12 geometric sequence with 1 + 12 as both the first term and the common ratio. This sum is 

i 72 1 + 12 −1 and the above equation is now



i 72 i 12 1 + 12 −1 1 + 12 

i 72 1 + 12 −1 550,000  + 450,000 =



72 12 i 72 i i 1 + 12 1 + 12 1 + 12 −1 400

which is similar to Eq. (12.16) in the text (page 396). In this expression we cannot analytically solve for i, but by using a simple numerical method in Excel, similar to the method we discussed in problem #9 we can get a fairly accurate estimate of its value, which is 3.429 %.

248

12 Mathematics of Interest Rates and Finance

#17 Using Eq. (12.22) from the text, we calculate the monthly payment as C=

20,000( 0.065 12 )(1 + (1 +

0.065 12∗3 12 ) 0.065 12∗3 −1 12 )

≈ $613

#18 the monthly payment is C=

0.0475 12∗25 12 ) 0.0475 12∗25 −1 12 )

250,000( 0.0475 12 )(1 + (1 +

= $1,425.30

#19 Using Eq. (12.24) from the text, the amount applied to the principal out of 50th payment is Prin50



0.0475 = 1+ 12

This amount is

50−1   0.0475 1425.3 − 2500,000 = $528.77 12

528.77 ∗ 100 = 37.1 % of the monthly payment. 1425.30

#20 The accumulated equity after 120 payments is calculated by using Eq. (12.25) from the text,    120 − 1 (1 + 0.0475 0.0475 12 ) 250,000 EQ120 = 1,425.30 − = $66,761.76 0.0475 12 12 The balance of the loan outstanding is the difference between the amount of loan and accumulated equity, that is 250,000 − 66,761.76 = $183,238.24. #21 We must first determine the loan balance after 5 years, the amount that will be refinanced. The monthly payment of the original loan, C, using Eq. (12.22) from the text, is C=

0.065 12∗30 12 ) 0.065 12∗30 −1 12 )

500,000( 0.065 12 )(1 + (1 +

= $3,078.59

We next use Eq. (12.25) from the text and find the accumulated equity after 60 payments EQ60

  (1 + 0.0625 500,000 = 3,078.59 − 12

0.0625 60 12 ) 0.0625 12

−1



= $33,313.79

The balance of the loan outstanding is then 500,000 − 33,313.79 = $466,686.21. This is the amount the homeowner wants to refinance. Let i ′ denotes the upper bound of the new rate; at or below it the home owner benefits from refinancing. This rate

12 Mathematics of Interest Rates and Finance

249

must be such that the sum of 360 new monthly payments C ′ is less than the sum of the remaining 360 − 5 ∗ 12 = 300 monthly payments from the original loan. That is, 360 ∗ C ′ ≤ 300 ∗ 3079.59

where C ′ =

′

i 466,686.21( 12 )

1 − (1 +

i ′ −360 12 )

Therefore, we must have 360 ∗

′

i 466,686.21( 12 )

1 − (1 +

i ′ −360 12 )



≤ 300 ∗ 3,078.59

This expression is simplified to i′ 12

≤ 0.005497   i ′ −360 1− 1+ 12 Assuming equality between the left and right hand side, we simplify this expression to 0.005497(1 +

i ′ −360 i′ ) = 0.005497 − 12 12

After taking the logarithm of both sides, we have ln(0.005497) − 360 ln(1 + −5.2036 − 360 ln(

i′ i′ ) = ln(0.005497 − ) 12 12

12 + i ′ 0.06596 − i ′ ) = ln( ) 12 12

−5.2036 − 360 ln(12 + i ′ ) + 360 ln(12) = ln(0.06596 − i ′ ) − ln(12) And finally 360 ln(12 + i ′ ) + ln(0.06596 − i ′ ) − 891.848 = 0 The solution to this logarithmic equation, using WolframAlpha, is 0.0521, implying that refinancing the loan at any rate below 5.21 is beneficial to the homeowner. #22 We must first compute components needed for solving Eq. (12.29) in the text. Using (12.22) or its alternative given in the text, the mortgage monthly payment, C, is

250

12 Mathematics of Interest Rates and Finance

C=

350,000( 0.058 12 )(1 + (1 +

0.058 300 12 )

0.058 300 12 )

−1

= $2,212.46

Using (12.23), the balance of the loan after k monthly payments is

(1 + 0.058 k ) − Bk = 350,000(1 + 12 Bk = 350,000(1.004833)k −



0.058 k 12 ) 0.058 12

−1



2,212.46

 1.004833k − 1 2,212.46 0.004833

Bk = 457,781.92 − 107,781.92(1.004833)k Using (12.28) in the text C ′ , the monthly payment of the new loan, is C′ =

Bk ( 0.043 12 )(1 + (1 +

0.043 12∗25 12 ) 0.043 12∗25 −1 12 )

= 0.005445Bk

After substituting for Bk , we have C ′ = 2,492.62 − 586.87(1.004833)k Equation (12.29) in the text can now be expressed as  2212.46k + 300 2492.62 − 586.87(1.004833)k − 300 ∗ 2212.46 = 0 After dividing through by 300 and simplifying, we have 7.37487k − 586.87(1.004833)k + 280.16 = 0 An approximate answer to this equation is k = 78. This means that if the original 5.8 % loan is refinanced by a 4.3 % loan within 78 months of the original loan, the home owner would benefit from the refinancing. #23 The original loan monthly payment is $1,530. After 84 payments the balance of the loan is $123,155. If this amount is refinanced with a 10-year 3.74 % loan, the monthly payment will be $1,232 and the total payment over the life of the loan will be 120 × 1232 = $147,840. If you continue with the original loan you must pay an additional (180 − 84) ∗ 1530 = $147,072, which is almost the same as the total payment for the second loan. Given that refinancing involves substantial closing costs, it is not to your financial advantage. #24 In exercise #23 the balance of the loan after 7 years, or 84 payments, was calculated as $123,155. If this amount is refinanced with an additional closing cost

12 Mathematics of Interest Rates and Finance

251

of $4,000, then your new loan will be $127,155. We want to determine the bound of the advantageous rates for a 10-year loan. This rate, i ′ , must be such that the sum of 120 new monthly payments, C ′ , is less than the sum of the remaining 180 − 84 = 96 monthly payments, $1,530, from the original loan. 120 ∗

′

i 127155( 12 )

1 − (1 +

i ′ −120 12 )



≤ 96 ∗ 1530

This expression is simplified to i′ 12

≤ 0.009626   i ′ −120 1− 1+ 12 Assuming equality between the left and right hand side, we simplify this expression to 0.009626(1 +

i ′ −120 i′ = 0.009626 − ) 12 12

After taking the logarithm of both sides, we have ln(0.009626) − 120 ln(1 + −4.6433 − 120 ln(

i′ i′ ) = ln(0.009626 − ) 12 12

12 + i ′ 0.11551 − i ′ ) = ln( ) 12 12

−4.6433 − 120 ln(12 + i ′ ) + 120 ln(12) = ln(0.11551 − i ′ ) − ln(12) And finally 120 ln(12 + i ′ ) + ln(0.11551 − i ′ ) − 296.03 = 0 The solution to this logarithmic equation, using WolframAlpha, is 0.0296, which means refinancing the balance of your 15-year loan, including an additional $4,000 closing cost, with a 10-year loan at any rate below 2.96 % is financially to your benefit. #25 Using (12.22) in the text, the monthly payment of the original 6.35 % loan is C=

0.0635 12∗25 12 ) 0.0635 12∗25 −1 12 )

350,000( 0.0635 12 )(1 + (1 +

= $2,330.53

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12 Mathematics of Interest Rates and Finance

After 74 payments (6 years from January 15, 2007 to January 15, 2012 plus 2 months), the remaining balance of the loan is (using 12.23 in the text)

B74

  (1 + 0.0635 74 − = 350,000 1 + 12

0.0635 74 12 ) 0.0635 12

−1



∗2,330.52 = $306,800.56

This amount will be refinanced by a 15-year 4.25 % mortgage. The monthly payment of the new loan, C′ , is C′ =

306,800.58( 0.0425 12 )(1 + (1 +

0.0425 12∗15 12 )

0.0425 12∗15 12 )

−1

= $2,308

In spite of the fact that there is no significant difference between the monthly payments, by refinancing to a lower term and rate, the homeowner saves about 300C − (180C ′ + 74C) = 226C − 180C ′ = 526,699.78 − 415,440 = $111,260 #26 The monthly payment and the balance of the loan after 80 payments are C=

B80

250,000( 0.0515 12 )(1 + (1 +

0.0515 12∗15 12 ) 0.0515 12∗15 −1 12 )

= $1,996.57

  (1 + 0.0515 80 − = 250,000 1 + 12

0.0515 80 12 ) 0.0515 12

−1



∗1,996.57 = $162,060

The amount to be refinanced by a 10-year loan is $162,060. The remaining payment of the original loan is 100 ∗ 1,996.57 = $199,657. Then the rate i ′ of the new 10-year loan must be such that the total payment will be 199,657 − 8,000 = 191,657 = $191,657. Assume that the monthly payment of the refinanced loan is 120 $1,597.14, and then using the alternative expression for monthly payment in (12.22) from the text, we have   i ′ 12∗10 1+ 12  12∗10 ′ i 1+ −1 12

162,060 1,597.14 =



i′ 12

    ′  i ′ 120 i ′ 120 i 1+ 1+ − 1 = 101.46887 12 12 12

12 Mathematics of Interest Rates and Finance

253

This expression can be written as    ′   i ′ 120 i 1 − 101.46887 =1 1+ 12 12 After denoting

i′ by x and then taking the logarithm of both sides, we have 12

120 ln(1 + x) + ln(1 − 101.46887x) = 0 The solution to this logarithmic function is x = 0.002857, leading to i ′ = 12x = 0.0343. This means that a rate of 3.42 % in a 10-year loan saves the family $8,000. #27 If a $350 application fee is added to the loan, then the total amount of the loan is $10,350. Since the loan is compounded monthly, then   0.069 12 = 11,087.17 10,350 1 + 12 is the amount you must pay the lender. Your actual loan is $10,000, therefore APR =



11,087.17 − 10,000 10,000



∗ 100 = 10.87 %

In case that you withdraw $350 from an account that pays you 3.5 % to pay for the application fee, then the APR should be calculated as follows: 

0.069 10,000 1 + 12

APR =



12



0.035 + 350 1 + 12

11,074.69 − 10,000 10,000



12

= 10,712.24 + 362.45 = 11,074.69

∗ 100 = 10.75 %

Slightly better than the first case, but by not much. But the lenders do not take into account the opportunity cost of borrowers withdrawing money from interest bearing accounts to pay for the fees. Therefore they calculate APR as   0.069 12 10,000 1 + + 350 = 10,712.24 + 350 = 11,062.24 12 APR =



110,712.24 − 10,000 10,000



∗ 100 = 10.71 %

254

12 Mathematics of Interest Rates and Finance

#28 In Eq. (12.30) developed in the text, we assumed that there are monthly contributions to a saving plan that pays an interest rate i compounded monthly. In this problem contributions are quarterly and the interest is also compounded quarterly. We must first develop a general expression for the total value of a fund when contributions are made n times per year, where n = 1, 2, 4, 12 for annual, semiannual, quarterly and monthly, respectively, and interest is compounded m times, where m = 1, 2, 4, 12 for annual, semiannual, quarterly and monthly. The general formula is

S=

nT  t=1

 mt  i n P 1+ m

for n = 1, 2, 4, 12

and

m = 1, 2, 4, 12

We can rewrite the above equation as  m nT   m(t − 1)  i i n  n 1+ S = P 1+ m m t=1

The term

nT   t=1

i 1+ m

 m(t−1) n

is the sum of nT terms of a geometric series with the

first term equal to 1 and the common ratio of (1 +  nT   i 1+ m

m(t−1) n

t=1

Subsequently, S is

=

i mn m)

. This sum is

m nT

mT 1 + mi n 1 − 1 + mi =

m

m 1 − 1 + mi n 1 − 1 + mi n

1−



⎡

mT ⎤  m i n ⎣ 1 − 1 + mi ⎦ S = P 1+

m m 1 − 1 + mi n

(12.2SM)

Note that if the contributions to a fund are monthly, that is when n = 12, then Eq. (12.2SM) is reduced to Eq. (12.30) given in the text. In this problem, the contributions are monthly, n = 12, but the compounding is quarterly, m = 4. Therefore using (12.2SM), we have ⎡ 4∗15 ⎤   4 1 − 1 + 0.045  ⎥ 4 0.045 12 ⎢ ⎢ ⎥ S = 250 1 + 4  ⎦ = $64,254.50  ⎣ 4 0.045 12 1− 1+ 4

12 Mathematics of Interest Rates and Finance

255

#29 In this problem, the contributions of $800 are quarterly and the interest rate 5.5 % is compounded quarterly. Therefore both n and m are 4. Thus using (12.2SM) the value of the fund after 18 years is ⎡ 4∗18 ⎤   4 1 − 1 + 0.055  ⎥ 4 0.055 4 ⎢ ⎢ ⎥ S = 800 1 + 4   ⎣ ⎦ 4 0.055 4 1− 1+ 4 S = 800(1.01375)



 1 − (1.01375)72 = $98,685 1 − (1.01375)

#30 Since both contributions to the fund and compounding are monthly, then we can use Eq. (12.30) from the text. Here we must solve for P in terms of other given variables: S = 80,000, i = 0.045, m = 12, T = 12. P=

m(1 +

i m)

iS  (1 +

i mT m)

−1

 =

0.045 ∗ 80,000   = $418.44 12(1.00375) (1.00375)144 − 1

#31 In this problem we must determine T , the length of time needed to accumulate $32,000 with $200 monthly contributions earning 3.76 % interest compounded monthly. We can again use Eq. (12.30) from the text   12 ∗ 200(1.003133) (1.003133)12T − 1 32,000 = 0.0376  32,000 = 64029.77 (1.003133)12T − 1 (1.003133)12T − 1 = 0.4998

−→

(1.003133)12T = 1.4998

Taking the logarithm of both sides, we have 12T ln(1.003133) = ln(1.4998)

−→

T = 10.8 years

#32 In this problem we must determine the interest rate i compounded monthly that grows $200 monthly contributions to $32,000 after 10 years.

32,000 =

     i 12∗10 i 1+ −1 12 ∗ 200 1 + 12 12 i

  i 121 i 32,000 i = 1+ −1− 2,400 12 12

→

  i 121 13.41666i + 1 = 1 + 12

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12 Mathematics of Interest Rates and Finance

After taking the logarithm of both sides, we have ln(13.416666i + 1) − 121 ln(1 +

i )=0 12

The solution to this logarithmic equation is i = 0.0547 or i = 5.47% #33 The price of your bond is the present value of coupons 0.045 ∗ 5000 = $225 generated in 3 years plus the present value of the face value of the bond, using 5.5 % as the discount rate. P=

225 (1 +

0.055 12 12 )

+

225 (1 +

0.055 24 12 )

+

225 (1 +

0.055 36 12 )

+

5,000 (1 +

0.055 36 12 )

P = 212.98 + 201.61 + 190.85 + 4,241.07 = $4,846.51 #34 In this case the discount rate is 4.0 %. P=

225 (1 +

0.04 12 12 )

+

225 (1 +

0.04 24 12 )

+

225 (1 +

0.04 36 12 )

+

5,000 (1 +

0.04 36 12 )

P = 216.19 + 207.73 + 199.60 + 4,435.49 = $5,050 #35 This is a problem with massive applications in the real world. A good example is a pension funds. An individual with a pension fund contributes frequently (mostly monthly) to the fund for a long period of time. After retirement the flow of money is reversed; the contribution stops and instead the individual receives payments from the fund over an specified time period, like 10–15 years, when the fund is exhausted. In this exercise we are assuming that the parents contribute to the fund for 18 years and then pay the college expenses, here $45,000 a year, out of the fund such that the fund is exhausted after 4 years. If the amount accumulated after 18 years is S, then after the first year of college this amount is reduced to S − 45,000 which earns interest and becomes (S − 45,000) ∗ (1.005)12 . Out of this amount $45,000 second year expenses are paid, and so on until the final payment depletes the fund. This leads to   1 − 1.0616784 = 0 → 1.19668S−197,349 = 0 (1.061678)3 S−45,000 1 − 1.061678 We now use Eq. (12.30) and find S in terms of P and then use this method for directly calculating the amount of monthly contribution needed. From (12.30) we get S = 389.29P, therefore 1.19668(389.29P) − 197,349 = 0

−→

P = 423.6 ≈ $424 a month

12 Mathematics of Interest Rates and Finance

257

Alternatively, we can modify and use the expression (12.31). To arrive at a general formula, let us assume that $P is contributed to a fund that pays i percent interest compounded m times a year for T years. After T years, $C annually is paid out of the fund such that after T ′ years the fund is totally depleted. By following the strategy we outlined above for determining $424 monthly payments, we arrive at the modified version of Eq. (12.31) as  ′ C( mi ) (1 + mi )mT − 1 P=   ′ (1 + mi )m(T −1)+1 (1 + mi )m − 1 (1 +

i mT m)

 −1

(12.31MOD)

In our college fund problem, i = 6 %, m = 12, T = 18, C = 45,000, and T ′ = 4. Using (12.31MOD), we have P=

450,00(0.005)[(1.005)48 − 1] ≈ $424 (1.005)37 [(1.005)12 − 1][(1.005)216 − 1]

Chapter 12 Supplementary Exercises 1. You invest $20,000 in a money market mutual fund that gives 3.5 % interest annually. How much will you have after 5 years if (a) interest is simple? (b) interest is compounded annually? (c) interest is compounded monthly? 2. How much will you have at the end of 10 years if you invest $10,000 at 3.25 % compounded quarterly? 3. In the previous problem, how much will you have if compounding is continuous? 4. You invest $15,000 at the rate of 2.85 % compounded monthly. After how many years will your investment grow to $20,000? 5. An investment of $25,000 grows to $32,000 over 8 years. What interest rate, compounded monthly, will the investor receive? 6. A zero coupon bond is offered at $25,000 with the promise to give investors $40,000 after 25 years. Assuming monthly compounding, what interest rate will an investor earn? 7. What interest rate compounded quarterly doubles an investment in 15 years? 8. What interest rate compounded monthly doubles an investment in 10 years? 9. What interest rate compounded continuously doubles an investment in 10 years? 10. Show that the interest rate i compounded m times a year that increases an investment n-fold over t years, is given by i = ex p



 1 ln n + ln m − m mt

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12 Mathematics of Interest Rates and Finance

11. In the weekly auction of short term government bonds by the Federal Reserve Bank of New York, a government bond dealer bids $99.10 for a $100 face value of 13 weeks T-bill. If the dealer’s bid is accepted, what is its annualized yield? 12. How long does it take for $10,000 invested at the rate of 3.48 % compounded quarterly to double? 13. How long does it take for $10,000 invested at the rate of 3.48 % compounded continuously to triple? 14. Show that with the interest rate i compounded m times a year, the number of years t that it takes for an investment to increase n-fold is given by t=

ln n   i m ln 1 + m

15. Show that if interest is compounded continuously, the number of years in problem ln n . #14 is simply t = i 16. A $10,000 investment at the rate of 3.5 % compounded grows to $11,903.40 over 5 years. What is the number of yearly compounding? 17. A $5,000 bond is trading for $4,950. The bond’s current yield CY is 0.035. What is the bond’s coupon rate? 18. Assume a bond with $15,000 face value and a coupon rate of 3.2 % has 4 years to its maturity. An investor pays $99 per $100 face value and purchases the bond. (a) What is the bond’s approximate YTM? (b) What is the bond’s exact YTM (assume annual compounding and coupon payment)? (c) What is the bond’s current yield CY? (d) What is the investor’s rate of return? 19. Assume a bond with $10,000 face value and a coupon rate of 4.25 % has 5 years to its maturity. If the price of this bond in the market is $10,500, (a) What is the bond’s approximate YTM? (b) What is the bond’s exact YTM (assume annual compounding and coupon payment)? (c) What is the bond current yield CY? (d) What is the investor’s average annual rate of return? 20. A $5,000 face value bond has 4 years to its maturity. The market price of this bond is $5,100. If the YTM of this bond, calculated using the approximation method, is 2.475 %, what is the coupon rate of this bond? 21. A $10,000 face value bond with a coupon rate of 3.2 % has 5 years to its maturity. If the interest rate in the market for similar instruments is 3.25 %, what is the market value of this bond?

12 Mathematics of Interest Rates and Finance

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22. You buy a property in early 2007 for $350,000. By the end of 2012, when the housing market somewhat recovers, you sell the property for $450,000. During the 7 years that you own the property, it generates a monthly net rental income of $600. What is your rate of return? 23. An individual borrows a 48 month $25,000 car loan at 6.5 % interest compounded monthly. What is the amount of his monthly payment? 24. What is the monthly mortgage payment of a 25-year $200,000 fixed loan if the mortgage rate is 4.25 %? 25. What proportion of the 150th monthly payment in the previous problem is applied to the principal? 26. What is the accumulated equity in Exercise #24 after the 150th monthly payment? What is the amount of the loan outstanding? 27. You purchase a property with a $250,000 mortgage from a bank. This is a 15-year mortgage with a rate of 4.5 %. (a) What is your monthly payment? (b) What is the total amount of money you pay the bank over the life of the mortgage? (c) What is the total amount of interest that you pay? 28. Assume that in the previous problem you decide to make bi-weekly payments instead of a monthly payment (i.e. to pay half of the monthly amount every two weeks). (a) What is the total amount of money you pay the bank over the life of the mortgage? (b) What is the total amount of interest that you pay? 29. A family takes out a 30-year fixed $300,000 mortgage with a rate of 5.25 %. After 8 years they decide to refinance the balance of the loan with another 30-year fixed mortgage. What is the upper bond of the advantageous rates? 30. A family borrows $250,000 from a mortgage company. The term of the mortgage is a 30-year fixed with a 4.85 % interest rate. What is the optimal time horizon for this family to refinance the balance of this loan with a new 30-year fixed mortgage at 3.9 %? 31. You purchase your home with a $300,000, 15-year fixed, 4.25 % mortgage loan. After 6 years, 72 monthly payments, you are offered to refinance the balance of your loan with a 15-year 3.85 % fixed-rate mortgage. Is refinancing to your financial advantage? 32. You purchase your home with a $300,000, 15-year fixed, 4.25 % mortgage loan. After 6 years, 72 monthly payments, you are offered to refinance the balance of your loan with a 10-year 3.65 % fixed-rate mortgage. Is refinancing to your financial advantage?

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33. In the previous problem, assume that the closing cost for refinancing is $4,000, which you agree to be added to the balance of your loan. What range of mortgage rates makes the refinancing financially to your advantage? 34. A family purchased a house in March 15, 2005 with a 25-year, $250,000, 6.35 % mortgage. In June 15, 2012, when the mortgage rates were substantially lower, they refinanced the balance of the loan with a 15-year, 4.0 % mortgage. How much does this family save? 35. A family purchased a house in 2006 with a 15-year, $350,000, 5.25 % mortgage. After 80 payments they are offered a 10-year loan with a rate that saves the family about $10,000 over the original loan. What is that rate? 36. A family pays $250 a month to a savings fund that pays 3.45 % interest compounded monthly. How much money will they have after 12 years? 37. A family sets up a savings fund which pays 4.0 % interest compounded monthly. They contribute $350 a month to this fund. What is the value of the fund after 15 years? 38. A family needs a $50,000 down payment for the purchase of a house. In order to reach their $50,000 target, how much does the family need to contribute monthly to a savings fund, over 10 years, that pays 3.5 % interest compounded quarterly? 39. A family needs a $85,000 for down payment of a house. The family contributes $700 a month to a savings fund that pays 3.25 % interest compounded monthly. How long does it take to have the required amount? 40. Determine the monthly payments a family must contribute to a college fund, paying 3.75 % interest compounded monthly, for 22 years in order to cover the annual college cost of $35,000. 41. In the previous problem, assume that the parents stop their contribution after 18 years (when their child enters a college). Use the modified version of Eq. (12.31), labeled (12.31MOD), and determine the amount they must contribute monthly for 18 years in order to cover the annual college cost of $35,000. 42. The parents of a child set up a college fund for her. They contribute $450 monthly to the fund which pays 4.25 % interest compounded monthly. After 15 years she enters a college, and the parents start to withdraw annually from the fund in order to cover her college expenses. If they continue their contribution of $450 during the 4 years that she is attending college, how much can they withdraw each year so that the fund is depleted in the last year? 43. In the previous problem, assume that the parents stop their contribution after 15 years. Use the modified version of Eq. (12.31), labeled (12.31MOD), and determine the amount they can withdraw annually such that after 4 years the fund would be exhausted. 44. A self-employed individual contributes $5,000 annually for 30 years into an Individual Retirement Account (IRA). At the end of 30 years the account balance

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261

is $230,000. Assuming monthly compounding, determine the interest rate that the IRA paid out. 45. Assuming that the account in the previous problem continues to receive the same interest rate, for how long can this individual withdraw $40,000 a year from the account? 46. Derive the modified formula for price of bond in Eq. (12.16) if its coupon is paid semiannually. 47. An investor has $45,000 to invest. A relatively safe investment offers 3.5 % interest compounded quarterly. A more risky investment offers 5.4 % compounded monthly. The investor allocates $22,500 to each investment. How long will it take for his investment to grow to $56,250? 48. In the last problem, assume that the first investment offers 3.75 % compounded continuously. Determine the time it will take for the investment to grow to $60,000. 49. A student finishes her college after accumulating a $30,000 federally backed student loan over 4 years. The loan charges 3.5 % interest compounded monthly. What is her monthly payment if she wants to repay the loan (amortize) over the next 6 years? 50. A company establishes a sinking fund for replacement of equipment worth $700,000 that depreciates in 8 years. How much should the company deposit semi-annually into an account that pays 4.25 % interest compounded monthly? 51. An individual retires with $L in her pension fund. This money is in an account that pays i percent interest compounded m times a year. She withdraws $P each month from the fund for her expenses. Derive the expression that gives the balance of the fund after t years. 52. In problem #51, assume L = 500,000, i = 4 %, m = 12, and P = 4,200. Find the balance of the pension fund after 8 years. How long will the fund last?

Chapter 13

Matrices and Their Applications

Chapter 13, Exercises 13.8 1. Given A=

(a) (b) (c) (d)





3 18 −2 , 5 4 0

⎤ ⎡ 0 −1 3 B = ⎣4 2 0⎦ , and 9 7 8

⎡

⎤ 2 0 2 C = ⎣−1 5 3⎦ 4 −3 1

Find A′ , B ′ , and C ′ . Find B + C and B − C. Is A + B defined? Find AB, B A′ , A′ A, A A′ and BC. Is B A defined? Show that (AB)′ = B ′ A′ .

2. Evaluate the determinant of the following matrices: ⎡

⎤

⎡

⎤

3 0 −1 0 −5 2 A = ⎣4 1 0⎦ , B = ⎣1 3 2 ⎦ , and 4 −2 5 8 5 9

⎡

2 ⎢−1 C =⎢ ⎣4 0

0 5 0 2

2 3 1 0

⎤ −2 1⎥ ⎥ −4⎦ 0

3. Which one of the matrices in Exercise 2 is not invertible? 4. Find the inverse of the matrices in Exercise 2 that are nonsingular. 5. Find a, b, and c such that matrix S is symmetric, ⎤ 0 −5 a S = ⎣b 1 c ⎦ 8 5 9 ⎡

6. Find a, b, and c such that matrix T is symmetric,

© Atlantis Press and the authors 2015 S. Vali, Principles of Mathematical Economics II, Mathematics Textbooks for Science and Engineering 4, DOI 10.2991/978-94-6239-088-1_13

263

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13 Matrices and Their Applications

⎡

⎤ 4 −a 2b − 4 T = ⎣a 1 c + 2 ⎦ b 5 9 7. For a square matrix A, show that A′ A, A A′ , and A + A′ are symmetric. 8. Write the following system in matrix form and solve it 2x1 − 2x2 + 4.5x3 = 20 −2x1 + 4x3 = 5 3x2 − 2x3 = 2 9. Write the following system in matrix form and solve it 2x1 − 2x2 + 4x3 + 2x4 = 30 −2x1 + 4x3 + 3x4 = 10 3x1 − 2x3 = 2 x2 + 2x3 + 3x4 = 15 10. A chain of grocery stores has 5 stores and 3 warehouses. The following table gives the distances from each warehouse to the five stores.

Warehouse 1 Warehouse 2 Warehouse 3

St1

St2

St3

St4

St5

10 25 30

15 20 22

25 10 15

30 22 10

12 15 30

Assume the cost of shipping one pound of an item per mile is $0.02 or 2 cents. If the weights of items each store must receive per week are given by the row vector W = [10,000 25,000 15,000 30,000 20,000], rank the warehouses in terms of cost effectiveness of shipping items to all stores. 11. In Exercise 10, write a matrix expression that finds the average distance of stores from each warehouse. 12. Find the inverse of the following matrices by method of adjoint, 

 3 2 A= , 5 4

  0 −1 B= , and 4 2

  1 2 C= 5 3

13. Find the inverse of the following matrices using elementary row operations ⎡

⎤ 0 −1 3 A = ⎣4 2 0⎦ , and 9 7 8

⎡

⎤ 2 0 2 B = ⎣−1 5 3⎦ 4 −3 1

13 Matrices and Their Applications

265

14. Assume the following amount of raw materials and labor required for production of a unit of products A and B, Product A Product B Pound of Raw materials Hours of labor

5 3

7 4

If the available amount of raw materials and labor are 13,400 pounds and 7,800 h, determine the outputs of A and B. 15. How do the outputs of A and B in Exercise 14 change if the amount of available raw materials and labor hours are increased to 13,800 and 8,000? 16. In Exercise 14, what amounts of raw materials and labor hours are needed for producing 1,300 units of A and 1,100 units of B? 17. The demand and supply functions of two related goods are given by Q d1 = 30 − 8P1 + 4P2 Q s1 = −60 + 6P1 Q d2 = 200 + 4P1 − 4P2 Q s2 = −40 + 6P2 Write the system in matrix form and solve for the equilibrium prices and quantities. 18. The demand and supply functions of three related goods are given by Q d1 = 60 − 5P1 + 2P2 − 3P3 Q s1 = −10 + 4P1 Q d2 = 70 + 4P1 − 4P2 + 2P3 Q s2 = −20 + 2P2 Q d3 = 50 − 2P1 + 4P2 − 2P3 Q s3 = −70 + 5P3 Write the system in matrix form and find the equilibrium prices and quantities.

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13 Matrices and Their Applications

19. Consider an open economy with the following consumption, investment, and tax functions C = 40 + 0.80Y D T = 10 + 0.10Y I = 38 + 0.15Y Write the model in matrix form. Assume the government expenditure G 0 = 110, export X 0 = 40, and import M0 = 90. What are the equilibrium output, consumption, investment, and taxes in this economy? 20. In Example 9 in Sect. 13.6.2, it was shown that to achieve a balanced budget, cutting government expenditures by 30 % from $125 to $87.5 billion must be accompanied by raising the marginal tax rate from 10 to 14.15 %. Show how the required tax rate of 0.1415 is computed. 21. Consider the following extended version of a national income model presented in Chap. 4: Y = C + I + G0 + N X0

(NX0 is the net export)

C = −228.78 + 0.832Y D I = −41.951 + 0.255 Y D − 11.511 r r = −0.178 + 0.010 Y − 0.012 M0 where r is the corporate bond rate used as proxy for interest rate and M0 is the real money supply. Assume the tax equation is expressed as T = T0 + tY = 135 + 0.15Y and the values (in billions of dollars) of exogenous variables G 0 , N X 0 and M0 are 820.8, −37.5 and 3323.3, respectively. (a) Write the model in matrix form Coe EN = EX. (b) Solve the model for the equilibrium values of the endogenous variables Y, C, I, r and T . Make sure you obtain the following result: ⎤ ⎡ ⎤ ⎡ 4325.837 Y ⎢C ⎥ ⎢2718.132⎥ ⎥ ⎢ ⎥ ⎢ ⎢ I ⎥ = ⎢ 824.405 ⎥ ⎥ ⎢ ⎥ ⎢ ⎣ r ⎦ ⎣ 3.200 ⎦ 783.876 T (c) Calculate the model’s autonomous (exogenous) expenditure, tax, and interest rate multipliers.

13 Matrices and Their Applications

267

22. In Exercise 21, assume the Federal Reserve Bank expands the money supply to $3,500 billion. What would be the new equilibrium values of the endogenous variables? How does the interest rate change? 23. Now assume that, in Exercise 22, simultaneously in conjunction with the expansion of the money supply, the government increases its expenditures by an additional $40 billion dollars. What are the new equilibrium levels of income and the interest rate? Is there evidence of “crowding out”? 24. Go back to the original model in Exercise 21 and change the tax function to T = 100 + 0.13Y That is, lower the marginal tax rate from 15 to 13 % and autonomous taxes from 130 to 100. Redo Exercises 21, 22 and 23. Compare the equilibrium solutions. 25. Below is the inter-sectoral transaction table for the United States, aggregated from data provided by BEA Use Table (after redefinition) 2011. The sectors are Agriculture (AG), Construction and Mining (including utilities)(C & M), Manufacturing (Manu), and Services (Ser) (excluding local, state, and federal government).1 4-Sector Input-Output Table of the U.S. ($1000)

AG C&M Manu Ser

AG

C&M

Manu

Ser

82550 9418 100120 65865

1109 143422 322107 213120

246867 653526 1834966 861584

14595 204375 868271 4863250

The vector of Final Demand (FD) for the 4 sectors is ⎡

⎤ 69980 ⎢ 897154 ⎥ ⎥ FD = ⎢ ⎣1845577⎦ 850000 (a) Use the data and prepare the matrix of inter-sectoral input coefficients. (Hint: use (13.16) and find total sectoral outs Q. Form a diagonal matrix with the sectoral outputs on the main diagonal and zero everywhere else. Multiply inter-sectoral transaction matrix T by the inverse of this matrix. This gives you matrix A) (b) Determine the sectoral gross outputs if the final demand for only the Agricultural sector’s products increases by 3 %. (Hint: create a unit matrix I of size 4. Calculate (I − A) and (I − A)−1 . Note that you must get the original vector of output if you calculate (I − A)−1 F D) 1

Imports and exports are also excluded.

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13 Matrices and Their Applications

(c) What is the growth rate of the economy? (d) Repeat parts (b) and (c) for other sectors. Rank sector based on their impact on the economy. Can this be determined by the column elements of (I − A)−1 that provides the direct and indirect output requirements per unit of final demand? 26. In Exercise 25, assume that the final demand for all sectors of the economy increases by 3 %. (a) Determine the gross outputs of all sectors. (b) What is the growth rate of the economy? Answers to Chapter 13, Exercises 13.8 #1 (a) ⎡

⎤ 3 5 A′ = ⎣ 18 4⎦ , −2 0

⎡

⎤ 0 4 9 B ′ = ⎣−1 2 7⎦ , and 3 0 8

⎡

⎤ 2 −1 4 C ′ = ⎣0 5 −3⎦ 2 3 1

(b) ⎡

⎤ 2 −1 5 B + C = ⎣ 3 7 3⎦ 13 4 9

and

⎡

⎤ −2 −1 1 B − C = ⎣ 5 −3 −3⎦ 5 10 7

A + B is not defined, they are not conformable for addition and subtraction. (c)   54 19 −7 AB = 16 3 15   337 87 AA′ = 87 41

⎤ −24 −4 BA′ = ⎣ 48 28 ⎦ 137 73 ⎡

⎤ 34 74 −6 A′ A = ⎣ 74 340 −36⎦ −6 −36 4 ⎡

⎤ 13 −14 0 BC = ⎣ 6 10 14⎦ 43 11 47 ⎡

B A is not defined. A and B are not conformable for multiplication; the number of columns of B, 3, is not equal to the number of rows of A, 2. ⎡ ⎤ 54 16 (d) From part (c), we have (AB)′ = ⎣ 19 3 ⎦. From part (a), we have −7 15

13 Matrices and Their Applications

269

⎤ ⎤⎡ ⎤ ⎡ 54 16 0 4 9 3 5 B ′ A′ = ⎣−1 2 7⎦ ⎣ 18 4⎦ = ⎣ 19 3 ⎦ −7 15 3 0 8 −2 0 ⎡

which confirms that (AB)′ = A′ B ′ . #2 To find the determinant of matrix A we expand by the first row. The minors and cofactors associated with elements a11 = 0, a12 = −5, and a13 = 2 are:



1 0

= 9 and |C11 | = (−1)1+1 |M11 | = 9 |M11 | =

5 9



4 0

= 36 and |C12 | = (−1)1+2 |M11 | = −36 |M12 | =

8 9





4 1

= 12 and |C13 | = (−1)1+3 |M11 | = 12

|M13 | =

8 5

|A| = a11 |C11 | + a12 |C12 | + a13 |C13 | = 0 ∗ 9 + (−5)(−36) + 2 ∗ 12 = 204

To find determinant of B we expand by the first row.



3 2

= 19 and |C11 | = (−1)1+1 |M11 | = 19 |M11 | =

−2 5



1 2

= −3 and |C12 | = (−1)1+2 |M11 | = 3

|M12 | =

4 5





1 3

= −14 and |C13 | = (−1)1+3 |M11 | = −14

|M13 | =

4 −2

|B| = b11 |C11 | + b12 |C12 | + b13 |C13 | = 3 ∗ 19 + 0 ∗ 3 + (−1)(−14) = 71 For matrix C we expand by the third row with 3 zeros as element. All we have to do is to determine the minor associated with the non-zero element C42 = 2, which is











2 2 −2

−1 3

−1 1

3 1













− 2

−2

|M42 | = −1 3 1 = 2

4 1

4 −4

1 −4

4 1 −4

Since |M42 | = 2(−12 − 1) − 2(4 − 4) − 2(−1 − 12) = 0 then the determinant of C is 0.

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13 Matrices and Their Applications

Matrix C is singular and non-invertible.

#3

#4 Matrices A and B with non-zero determinant are invertible. Given that we have already calculated the determinants of these matrices, then we use method of adjoint to find their inverses. ⎡ ⎤ 9 55 −2 1 ⎣ 1 −36 −16 8 ⎦ Ad j (A) = A−1 = |A| 204 12 −40 20 B

−1

⎡ ⎤ 19 2 3 1 ⎣ 1 3 19 −7⎦ Ad j (B) = = |B| 71 −14 6 9

For matrix S to be symmetric, it must be a = 8, b = −5 and c = 5.

#5

#6 If matrix T is symmetric, then it must be the case that a = −a, which is only possible if a = 0; 2b − 4 = b leading to b = 4; and c + 2 = 5 leading to c = 3. #7 The key to answering this problem is the Eq. (13.2) on page 438, (AB)′ = B ′ A′ , that we specified without proof. We first offer a short proof for (13.2) and then use it to show that for any square matrix A, A′ A, A A′ and A + A′ are all symmetric. Assume A = [ai j ] is an m × n and B = [bi j ] is an n × m matrix conformable for multiplication. Let form AB. A typical i j entry of AB is ai1 b1 j + ai2 b2 j + ai3 b3 j + · · · + aim bm j We now form B ′ A′ . Here column j of B becomes the jth row of B ′ and the ith row of A becomes the ith column of A′ . Then the typical i j entry of B ′ A′ is b1 j ai1 + b2 j ai2 + b3 j ai3 + · · · + bm j aim which is the same as the i jth element of AB above. Thus (AB)′ = B ′ A′ . Recall that a matrix C is symmetric if its transpose is equal to the matrix itself, that is if C ′ = C. This is only possible if C is a square matrix. We want to show that if A is any square matrix then A′ A is symmetric. Let form A′ A. The transpose of this product is (A′ A)′ . By using (AB)′ = B ′ A′ that we established above, we have (A′ A)′ = A′ (A′ )′ = A′ A. The same works for A A′ . Since (A A′ )′ = (A′ )′ A′ = A A′ . Also (A + A′ )′ = A′ + (A′ )′ = A′ + A, which is the same as A + A′ . The system of 3 equations with 3 unknowns in matrix form is

#8

⎤⎡ ⎤ ⎡ ⎤ x1 2 −2 4.5 20 ⎣−2 0 4 ⎦ ⎣x2 ⎦ = ⎣ 5 ⎦ which is in the form of Ax = b 0 3 −2 2 x3 ⎡

13 Matrices and Their Applications

271

Solution to this system is x = A−1 b ⎡ ⎤ ⎡ ⎤−1 ⎡ ⎤ x1 20 2 −2 4.5 ⎣x2 ⎦ = ⎣−2 0 4 ⎦ ⎣ 5 ⎦ 2 0 3 −2 x3 The system’s solution is ⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤ x1 20 0.27907 −0.22093 0.18605 4.8488 ⎣x2 ⎦ = ⎣0.09302 0.09302 0.39535⎦ ⎣ 5 ⎦ = ⎣3.1162⎦ 2 0.13953 0.13953 0.09302 x3 3.6744 The system in matrix form is

#9 ⎡

2 ⎢−2 ⎢ ⎣3 0

−2 0 0 1

4 4 −2 2

⎤⎡ ⎤ ⎡ ⎤ 2 30 x1 ⎢x2 ⎥ ⎢10⎥ 3⎥ ⎥ ⎢ ⎥ = ⎢ ⎥ which is in the form of Ax = b 0⎦ ⎣ x 3 ⎦ ⎣ 2 ⎦ 3 x4 15

the system’s solution is ⎡ ⎤ ⎡ 0.30 x1 ⎢x2 ⎥ ⎢ 0.30 ⎢ ⎥=⎢ ⎣x3 ⎦ ⎣ 0.45 x4 −0.40

−0.8 −1.8 −1.2 1.4

−0.4 −1.4 −1.1 1.2

⎤⎡ ⎤ ⎡ ⎤ 0.6 30 9.2 ⎢ ⎥ ⎢ ⎥ 1.6 ⎥ ⎥ ⎢10⎥ = ⎢ 12.2 ⎥ 0.9 ⎦ ⎣ 2 ⎦ ⎣ 12.8 ⎦ −0.8 15 −7.6

Let D denote the 3 × 5 matrix of distances between warehouses and stores. ⎡ ⎤ 10 15 25 30 12 D = ⎣25 20 10 22 15⎦ 30 22 15 10 30

#10

Then C, the column vector of costs of shipping from each warehouse to the 5 stores, is given by ⎤ 10000 ⎡ ⎤ ⎥ $39,800 10 15 25 30 12 ⎢ ⎢25000⎥ ′ ⎥ ⎣ ⎦ C = DW = ⎣25 20 10 22 15⎦ ⎢ ⎢15000⎥ × 0.02 = $37,200 $39,500 30 22 15 10 30 ⎣30000⎦ 20000 ⎡

⎤

⎡

Then the ranking, in terms of cost effectiveness of shipping, is Warehouse 2, Warehouse 3, and Warehouse 1.

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13 Matrices and Their Applications

#11

 Let S denote the sum vector of all 1. S = 1 1 1 1 1 . Then

⎡ ⎤ 1 ⎤ ⎡ 18.4 1⎥ 10 15 25 30 12 ⎢ ⎥ ⎢ 1 1 ⎦ ⎣ 1⎥ DS ′ = ⎣25 20 10 22 15⎦ ⎢ ⎥ = 18.4 5 5 30 22 15 10 30 ⎢ ⎣1⎦ 21.4 1 ⎤

⎡

is the vector of the average distances of stores from warehouses. #12 We first calculate the determinant of the matrix. Then we form the Adjoint matrix by transposing the matrix of cofactors. Inverse of the matrix is then computed as product of inverse of the determinant and the adjoint matrix. |A| = 12 − 10 = 2, Ad j (A) =



Ad j (B) =



4 −2 −5 3 2 1 −4 0



|B| = 0 − (−4) = 4, 



3 −2 Ad j (C) = −5 1

−→ A−1 =

−→ 

−→

B −1 =

C

−1

|C| = 3 − 10 = −7

    1 4 −2 2 −1 = 2.5 1.5 2 −5 3     1 2 1 0.5 0.25 = −1 0 4 −4 0

    1 3 −2 −0.4286 0.2857 =− ≈ 0.7143 −0.1429 7 −5 1

#13 We form the augmented matrix [A|I ] ⎡

⎤ 0 −1 3 | 1 0 0 ⎣4 2 0 | 0 1 0 ⎦ 9 7 8|0 0 1 and then perform the following row operations (1) Divide the second row by 4. (2) Add the second row to the first row ⎤ 1 −0.5 3 | 1 0.25 0 ⎣1 0.5 0 | 0 0.25 0⎦ 9 7 8|0 0 1 ⎡

(3) Multiply the fist row by −1 and add it to the second row. (4) Multiply the first row by −9 and add it to the third row. ⎡

⎤ 1 −0.5 3 | 1 0.25 0 ⎣0 1 −3 | −1 0 0⎦ 0 11.5 −19 | −9 −2.25 1

13 Matrices and Their Applications

273

(5) Multiply the second row by 0.5 and add it to the first row. (6) Multiply the second row by −11.5 and add it to the third row. ⎡

⎤ 1 0 1.5 | 0.5 0.25 0 ⎣0 1 −3 | −1 0 0⎦ 0 0 15.5 | 2.5 −2.25 1 (7) Divide the third row by 15.5. (8) Multiply the third row by −1.5 and add it to the first row. (9) Multiply the third row by 3 and add it to the second row. ⎤ 1 0 0 | 0.25806 0.46774 −0.09677 ⎣0 1 0 | −0.51613 −0.43548 0.19355 ⎦ 0 0 1 | 0.16129 −0.14516 0.06451 ⎡

Then the inverse of A is ⎤ ⎡ 0.25806 0.46774 −0.09677 A−1 = ⎣−0.51613 −0.43548 0.19355 ⎦ 0.16129 −0.14516 0.06451 #14 Let Q A and Q B denote the levels of outputs of products A and B. With the technical coefficients given in the table and the amount of available material and labor, we can write the system as      5 7 QA 13400 = 3 4 QB 7800  #15  #16 

→



   −1  13400 5 7 QA = QB 7800 3 4

       −4 7 QA 13400 1000 = = QB 3 −5 7800 1200 With the change in the amount of material and labor, the new solution is        −4 7 13800 800 QA = = QB 3 −5 8000 1400 To produce 1,300 units of A and 1,100 units of B, we need        5 7 1300 14200 QA = = QB 3 4 1100 8300

14200 units of materials and 8300 units of labor hours. #17

Using the equilibrium conditions Q d1 = Q s1 , and Q d2 = Q s2 , we have

30 − 8P1 + 4P2 = −60 + 6P1

−→

14P1 − 4P2 = 90

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13 Matrices and Their Applications

200 + 4P1 − 4P2 = −40 + 6P2

−→

−4P1 + 10P2 = 240

This system in matrix form is 

14 −4 −4 10



   P1 90 = P2 240

The vector of equilibrium prices is determined as 

   −1       1 10 4 P1 14 −4 90 90 15 = = = P2 −4 10 240 30 124 4 14 240

We next use the equilibrium values of P1 and P2 and determine the equilibrium levels of outputs Q 1 and Q 2 from the demand or supply function, which are 30 and 140 units, respectively. Alternatively, we can incorporate the outputs of good 1 and 2 as variables into the model and set up the problem in matrix form such that the solution provides a vector of equilibrium prices and outputs. Since at equilibrium the quantity demanded and supplied are the same, we use Q 1 and Q 2 to denote the level of outputs, so the system will be ⎧ ⎪ ⎪8P1 − 4P2 + Q 1 = 30 ⎪ ⎨6P − Q = 60 1 1 ⎪−4P1 + 4P2 + Q 2 = 200 ⎪ ⎪ ⎩ 6P2 − Q 2 = 40 We write this system of 4 equations and 4 unknowns in matrix form as ⎡

8 ⎢6 ⎢ ⎣−4 0

−4 0 4 6

1 −1 0 0

⎤ ⎡ 8 P1 ⎢ P2 ⎥ ⎢ 6 ⎢ ⎥=⎢ ⎣ Q 1 ⎦ ⎣−4 Q2 0 ⎡

⎤⎡ ⎤ ⎡ ⎤ 0 30 P1 ⎢ ⎥ ⎢ ⎥ 0⎥ ⎥ ⎢ P2 ⎥ = ⎢ 60 ⎥ ⎣ ⎦ ⎣ ⎦ Q1 1 200⎦ Q2 −1 40 −4 0 4 6

1 −1 0 0

⎤−1 ⎡ ⎤ 0 30 ⎢ ⎥ 0⎥ ⎥ ⎢ 60 ⎥ 1 ⎦ ⎣200⎦ −1 40

I used R and find the inverse of the matrix equilibrium prices and outputs as ⎡ ⎤ ⎡ 0.080645 0.080645 0.032258 P1 ⎢ P2 ⎥ ⎢0.032258 0.032258 0.112903 ⎢ ⎥=⎢ ⎣ Q 1 ⎦ ⎣0.483871 −0.516129 0.193548 Q2 0.193548 0.193548 0.677419

of coefficients and the vector of ⎤⎡ ⎤ ⎡ ⎤ 0.032258 30 15 ⎢ ⎥ ⎢ ⎥ 0.112903 ⎥ ⎥ ⎢ 60 ⎥ = ⎢ 30 ⎥ ⎣ ⎦ ⎣ ⎦ 0.193548 200 30 ⎦ −0.322581 40 140

13 Matrices and Their Applications

#18 ⎡

275

We can write this problem in matrix form as

5 ⎢4 ⎢ ⎢−4 ⎢ ⎢0 ⎢ ⎣2 0

−2 0 4 2 −4 0

3 0 −2 0 2 5

1 −1 0 0 0 0

0 0 1 −1 0 0

⎤⎡ ⎤ ⎡ ⎤ P1 60 0 ⎢ P2 ⎥ ⎢10⎥ 0⎥ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 0⎥ ⎥ ⎢ P3 ⎥ = ⎢70⎥ ⎥ ⎥ ⎢ ⎥ 0 ⎥⎢ ⎢ Q 1 ⎥ ⎢20⎥ 1 ⎦ ⎣ Q 2 ⎦ ⎣50⎦ 70 −1 Q3

The system is now in the form of Ax = b where A is the matrix of coefficients, x is the vector of unknowns, and b is the vector of constants. Again, with help from R we find the inverse of the matrix of coefficients A−1 as, ⎡ ⎤ 0.125926 0.125926 0.007407 0.007407 −0.05185 −0.05185 ⎢0.088889 0.088889 0.211111 0.211111 0.022222 0.022222 ⎥ ⎥ ⎢ ⎢0.014814 0.014814 0.118518 0.118518 0.170370 0.170370 ⎥ −1 ⎥ ⎢ A =⎢ ⎥ ⎢0.503703 −0.49629 0.029629 0.029629 −0.20740 −0.20740⎥ ⎣0.177777 0.177777 0.422222 −0.57777 0.044444 0.044444 ⎦ 0.074074 0.074074 0.592592 0.592592 0.851851 −0.14814 and the solution to the system as ⎤ ⎡ ⎤ ⎡ ⎤ 3.26 P1 60 ⎢ P2 ⎥ ⎢10⎥ ⎢27.89⎥ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ P3 ⎥ ⎢ ⎥ ⎢ ⎢ ⎥ = A−1 ⎢70⎥ = ⎢32.15⎥ ⎢20⎥ ⎢ 3.04 ⎥ ⎢ Q1⎥ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎣50⎦ ⎣35.78⎦ ⎣ Q2⎦ 90.74 70 Q3 ⎡

#19

After adding Y = C + I + G 0 + X 0 − M0 to the rest of the equations, we

have ⎧ Y = C + I + 110 + 40 − 90 ⎪ ⎪ ⎪ ⎨C = 40 + 0.80(Y − T ) → ⎪ T = 10 + 0.10Y → ⎪ ⎪ ⎩ I = 38 + 0.15Y →

→ Y − C − I = 60 −0.80Y + C + 0.80T = 40 −0.10Y + T = 10 −0.15Y + I = 38

This model in matrix form Coe EN = EX is ⎡ ⎤⎡ ⎤ ⎡ ⎤ 1 −1 0 −1 Y 60 ⎢−0.80 1 0.8 0 ⎥ ⎢C ⎥ ⎢40⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣−0.10 0 1 0 ⎦ ⎣ T ⎦ = ⎣10⎦ −0.15 0 0 1 I 38

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13 Matrices and Their Applications

with solution EN = Coe−1 EX ⎡ ⎤ ⎡ Y 1 ⎢C ⎥ ⎢−0.80 ⎢ ⎥=⎢ ⎣ T ⎦ ⎣−0.10 I −0.15

−1 1 0 0

⎡ ⎤ ⎡ 7.69231 Y ⎢C ⎥ ⎢5.53846 ⎢ ⎥=⎢ ⎣ T ⎦ ⎣0.76923 1.15385 I

0 0.8 1 0

⎤−1 ⎡ ⎤ 60 −1 ⎢40⎥ 0⎥ ⎥ ⎢ ⎥ 0 ⎦ ⎣10⎦ 38 1

7.69231 6.53846 0.76923 1.15385

−6.15385 −5.23077 0.38462 −0.92308

⎤ ⎤⎡ ⎤ ⎡ 1000 60 7.69231 ⎥ ⎢ ⎥ ⎢ 5.53846⎥ ⎥ ⎢40⎥ = ⎢ 752 ⎥ 0.76923⎦ ⎣10⎦ ⎣ 110 ⎦ 188 38 2.15385

Note that the absolute value of the determinant of inverse of the coefficient matrix is the model’s autonomous expenditure multiplier, that is m is equal to abs(|Coe−1 |) = 7.69 Alternatively, we can treat Y D as a separate variable and write the model as ⎧ ⎪ ⎪Y = C + I + 110 + 40 − 90 ⎪ ⎪ ⎪ ⎪ ⎨Y D = Y − T → C = 40 + 0.80Y D → ⎪ ⎪ ⎪ T = 10 + 0.10Y → ⎪ ⎪ ⎪ ⎩ I = 38 + 0.15Y →

→

Y − C − I = 60 −Y + Y D + T = 0 −0.80Y D + C = 40 −0.10Y + T = 10 −0.15Y + I = 38

This model in matrix form is ⎤⎡ ⎤ ⎡ ⎤ ⎡ Y 60 1 0 −1 0 −1 ⎥ ⎢Y D ⎥ ⎢ 0 ⎥ ⎢ −1 1 0 1 0 ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎢ ⎥ ⎢ ⎥ ⎢ 0 −0.80 1 0 0 ⎥ ⎥ ⎢ C ⎥ = ⎢40⎥ ⎢ ⎦ ⎣−0.10 0 0 1 0 ⎣ T ⎦ ⎣10⎦ 38 −0.15 0 0 0 1 I with solution EN = Coe−1 EX ⎡ ⎤ ⎡ 7.69231 6.15385 Y ⎢Y D ⎥ ⎢6.92308 6.53846 ⎢ ⎥ ⎢ ⎢ C ⎥ = ⎢5.53846 5.23077 ⎢ ⎥ ⎢ ⎣ T ⎦ ⎣0.76923 0.61538 1.15385 0.92308 I #20

The model in Example 9 is

C = 60 + 0.75Y D

7.69231 6.92308 6.53846 0.76923 1.15385

−6.15385 −6.53846 −5.23077 0.38461 −0.92308

⎤⎡ ⎤ ⎡ ⎤ 1000 60 7.69231 ⎢ ⎥ ⎢ ⎥ 6.92308⎥ ⎥ ⎢ 0 ⎥ ⎢ 890 ⎥ ⎢ ⎥ ⎥ ⎢ 5.53846⎥ ⎢40⎥ = ⎢ 752 ⎥ ⎥ 0.76923⎦ ⎣10⎦ ⎣ 110 ⎦ 188 38 2.15385

13 Matrices and Their Applications

277

I = 30 + 0.1Y − 7r0 T = 15 + 0.1Y Originally, the government expenditure G 0 is 125. The country’s parliament decides to cut the budget by 30 % to 87.5 and change the marginal tax rate such that the government has a balanced budget. The question is how they arrived at the marginal tax rate of 14.15 %. If the budget is set at 87.5 and the country must have a balanced budget, then the volume of tax must be 87.5 too. In that case the consumption function would change to C = 60 + 0.75(Y − T ) = 60 + 0.75(Y − 87.5) = 0.75Y − 5.625 After plugging 5 for r0 in the investment function, we have I = 30 + 0.1Y − 7 ∗ 5 = 0.1Y − 5 Substituting for C and I in the national income equation Y = C + I + G 0 , we have Y = 0.75Y − 5.625 + 0.1Y − 5 + 87.5 Y − 0.85Y = 76.875

−→

Y =

−→

Y = 0.85Y + 76.875

78.875 = 512.5 0.15

With the equilibrium value of Y determined, we can use the tax function expressed as T = 15 + tY and find t, the marginal tax rate. 87.5 = 15 + tY → 87.5 − 15 = 515.5t, leading to t =

72.5 = 0.1415 or 14.15 % 512.5

#21 After substituting 820.8, −37.5, and 3323.3 for G 0 , N X 0 , and M0 in the equations we can write the system as ⎧ Y = C + I + 820.8 − 37.5 → Y − C − I = 783.3 ⎪ ⎪ ⎪ ⎪ ⎪ C = −228.78 + 0.832(Y − T ) → 0.832Y − C − 0.832T = 228.78 ⎪ ⎪ ⎪ ⎨ I = −41.951 + 0.255(Y − T ) − 11.511r → ⎪0.255Y − I − 11.551r − 0.255T = 41.951 ⎪ ⎪ ⎪ ⎪ ⎪ r = −0.178 + 0.010Y − 0.012 ∗ 3323.3 → 0.010Y − r = 40.0576 ⎪ ⎪ ⎩ T = 135 + 0.15Y → −0.15Y + T = 135

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13 Matrices and Their Applications

(a) This system in matrix form is ⎡ 1 −1 ⎢ 0.832 −1 ⎢ Coe EN = E X → ⎢ ⎢ 0.255 0 ⎣ 0.010 0 −0.15 0

⎤ ⎤⎡ ⎤ ⎡ 783.3 Y −1 0 0 ⎥ ⎢ ⎥ ⎢ 0 0 −0.832⎥ ⎥ ⎢C ⎥ ⎢ 228.78 ⎥ ⎢ ⎥ ⎢ ⎥ −1 −11.551 −0.255⎥ ⎢ I ⎥ = ⎢ 41.951 ⎥ ⎥ 0 −1 0 ⎦ ⎣ r ⎦ ⎣40.0576⎦ 135 T 0 0 1

(b) The system’s solution is obtained by EN = Coe−1 EX, where the inverse of the matrix of coefficients is ⎡

5.22029651 ⎢3.69179369 ⎢ Coe−1 = ⎢ ⎢0.52850282 ⎣0.05220297 0.78304448

−5.22029651 −4.69179369 −0.52850282 −0.05220297 −0.78304448

−5.22029650 60.2996450 −3.69179369 42.6439090 −1.52850282 17.6557361 −0.05220297 −0.3970035 −0.78304448 9.0449468

⎤ −5.67446231 −4.84497975⎥ ⎥ −0.82948256⎥ ⎥ −0.05674462⎦ 0.14883065

and the vector of equilibrium values of the endogenous variables is ⎤ ⎡ ⎤ ⎡ 4325.169 Y ⎢C ⎥ ⎢2717.659⎥ ⎥ ⎢ ⎥ ⎢ ⎥ ⎥ ⎢ EN = ⎢ ⎢ I ⎥ = ⎢ 824.209 ⎥ ⎣ r ⎦ ⎣ 3.194 ⎦ 783.775 T (c) The absolute value of the determinant of Coe−1 , which is 5.22, is the model’s autonomous spending multiplier, m. The tax multiplier, m ′ , is −bm, where b = 0.832. Hence m ′ = −0.832 ∗ 5.22 = −4.34. The interest rate multiplier, m ′′ , is −km, where k = −11.511. Hence m ′′ = −11.511 ∗ 5.22 = −60.1. #22 Expanding the money supply to $3,500 billion changes E X and leads to a new vector of equilibrium values for the endogenous variables E N . ⎤ 783.300 ⎢228.780⎥ ⎥ ⎢ ⎥ EX = ⎢ ⎢ 41.951 ⎥ ⎣ 42.178 ⎦ 135.000 ⎡

−→

⎤ ⎡ ⎤ ⎡ 4453.028 Y ⎢C ⎥ ⎢2808.082⎥ ⎥ ⎢ ⎥ ⎢ ⎥ ⎥ ⎢ EN = ⎢ ⎢ I ⎥ = ⎢ 861.647 ⎥ ⎣ r ⎦ ⎣ 2.352 ⎦ 803.954 T

It is clear that expansion of the money supply is the source of decline in interest rate form 3.194 to 2.352 %. The rest of the endogenous variables show increases compare to exercise # 21, indicating that declining interest rate had a stimulating effect on the economy.

13 Matrices and Their Applications

279

#23 An additional $40 billion of government expenditure along with the expansion of the money supply, changes E X to a new vector, leading to a new vector of values for the endogenous variables, ⎡

⎤ 823.300 ⎢228.780⎥ ⎢ ⎥ ⎥ EX = ⎢ ⎢ 41.951 ⎥ ⎣ 42.178 ⎦ 135.000

−→

⎤ ⎡ ⎤ ⎡ 4661.840 Y ⎢C ⎥ ⎢2955.753⎥ ⎥ ⎢ ⎥ ⎢ ⎥ ⎥ ⎢ EN = ⎢ ⎢ I ⎥ = ⎢ 882.787 ⎥ ⎣ r ⎦ ⎣ 4.440 ⎦ 834.276 T

The change in the interest rate from 2.352 to 4.449 %, after an about 5 % increase in the government expenditure, is indicative of the crowding out. #24 Change in the tax function changes both the matrix of coefficients Coe and the vector of exogenous values E X . These changes lead to a new matrix formulation of the original system in exercise #21, ⎡

1 ⎢ 0.832 ⎢ Coe EN = EX → ⎢ ⎢ 0.255 ⎣ 0.010 −0.13 The inverse of Coe is ⎡ 5.8885879 ⎢4.2623954 ⎢ Coe−1 = ⎢ ⎢0.6261924 ⎣0.0588858 0.7655164

−1 −1 0 0 0

⎤ ⎤⎡ ⎤ ⎡ −1 0 0 783.3 Y ⎥ ⎢ ⎥ ⎢ 0 0 −0.832⎥ ⎥ ⎢C ⎥ ⎢ 228.78 ⎥ ⎢ I ⎥ = ⎢ 41.951 ⎥ −1 −11.551 −0.255⎥ ⎥ ⎥⎢ ⎥ ⎢ 0 −1 0 ⎦ ⎣ r ⎦ ⎣40.0576⎦ 100 T 0 0 1

−5.8885879 −5.2623954 −0.6261924 −0.0588858 −0.7655164

−5.8885879 −4.2623954 −1.6261924 −0.0588858 −0.7655164

68.019079 49.234930 18.784148 −0.319809 8.842480

⎤ −6.40089507 −5.46522388⎥ ⎥ −0.93567118⎥ ⎥ −0.06400895⎦ 0.16788364

and the vector of equilibrium values of the endogenous variables is ⎤ ⎡ ⎤ ⎡ 5102.899 Y ⎢C ⎥ ⎢3381.703⎥ ⎥ ⎢ ⎥ ⎢ ⎥ ⎥ ⎢ EN = ⎢ ⎢ I ⎥ = ⎢ 937.897 ⎥ ⎣ r ⎦ ⎣ 10.917 ⎦ 763.376 T The impact of reducing the marginal tax rate by 13.3 % and the autonomous taxes by 26 % is evident by comparing this result with that of exercises #21, #22, and #23, presented in Table 13.1, especially a dramatic rise in the interest rate.

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13 Matrices and Their Applications

Table 13.1 Results of problem # 24 EN Original #21 #21 After tax change Y C I r T

4325.169 2718.659 824.209 3.194 783.775

5102.899 3381.703 937.897 10.971 763.376

#22 After tax and M0 change

#23 After tax, M0 and G 0 change

5247.127 3486.100 977.727 10.293 782.126

5482.670 3656.596 1002.774 12.649 812.747

#25 (a) We create the matrix of technical coefficients by using Eq. (13.16). The matrix of inter-sectoral transaction, T , the sum vector of size 4, S, and the vector of final demand, FD, are ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ 69980 1 82550 1109 246867 14595 ⎢ 897154 ⎥ ⎢1⎥ ⎢ 9418 143422 653526 204375 ⎥ ⎢ ⎥ ⎢ ⎥ ⎥ T =⎢ ⎣100120 322107 1834966 868271 ⎦ S = ⎣1⎦ FD = ⎣1845577⎦ 850000 1 65865 213120 861584 4863250 We calculate the vector of sectoral outputs as, ⎡

⎤ 415101 ⎢1907895⎥ ⎥ Q = T S + FD = ⎢ ⎣4971041⎦ 6853819 We form D Q, a diagonal 4 by 4 matrix with the sectoral outputs on the main diagonal, zero elsewhere, ⎡

⎤ 415101 0 0 0 ⎢ 0 ⎥ 1907895 0 0 ⎥ DQ = ⎢ ⎣ 0 ⎦ 0 4971041 0 0 0 0 6853819 Then the matrix of coefficients A is calculated as ⎡ 0.19886726 0.0005812689 ⎢ 0.02268845 0.0751729000 A = T D Q −1 = ⎢ ⎣0.24119431 0.1688284733 0.15867223 0.1117042605

0.04966103 0.13146663 0.36913113 0.17332064

⎤ 0.00212947 0.02981914⎥ ⎥ 0.12668426⎦ 0.70956791

13 Matrices and Their Applications

281

Table 13.2 Results of part(d) problem # 25 Sector Agriculture Const. and mining Ag. C and M Manu. Ser. GDP % Change

417833 1908286 4972722 6856465 14155306 0.053

415975.9 1939503.9 4983914.4 6874136.7 14213531 0.46

Manufacturing

Services

422193.8 1926320.6 5081937.9 6930960.5 14361413 1.51

416854.3 1914706.2 4994721.3 6959328.4 14285610 0.97

(b) After creating I − A and a new vector of final demand FD′ , by changing the first element of FD from 69980 to 1.03 ∗ 69,980 = 72079.4, we find the new sectoral gross outputs, Q ′ , as ⎡

⎤ 417833 ⎢1908286⎥ ⎥ Q ′ = (I − A)−1 FD′ = ⎢ ⎣4972722⎦ 6856465 (c) The sum of sectoral outputs is the total output, GDP, of the economy. The original volume of output in part (a) is $14,147,856, about 14.15 trillion dollars. The total output after 3 % increase in agricultural sector is $14,155,306, about 14.16 trillion dollars. A 3 % increase in the U.S. agricultural activities contribute 0.053 % to the overall national output. (d) Every sector’s final demand are increased by 3 % and then the sectoral outputs and the total national output, GDP are estimated. The result is organized in the following table. Under each sector the result of 3 % increase in the final demand of that sector is listed. The “% Change” in the last row is the percentage change in the GDP from the original model in part (a). As the table indicates, the biggest impact of increase in final demand belongs to the Manufacturing sector, followed by Services, Construction and Mining, and Agriculture (Table 13.2). Following is a short R snippet to do problem #25. # We enter elements of inter-sectoral transaction matrix T by row T = matrix(c(82550,1109,246867,14595,9418,143422, 53526,204375, 100120,322107,1834966,868271,65865,213120 861584,4863250),4,b=T) # We enter element of column vector of final demand FD FD = matrix(c(69980, 897154, 1845577, 850000)) #Next we make the column sum vector S S = matrix(c(1),4)

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13 Matrices and Their Applications

#We now calculate the vector of sectoral outputs Q as Q = T%*%S +FD # Then print Q Q [,1] [1,] 415101 [2,] 1907895 [3,] 4971041 [4,] 6853819 # We add the outputs of 4 sectors to get the original GDP. GDP= t(S)%*%Q #Where t(S) is the transpose of the sum vector S GDP [,1] [1,] 14147856 # To form the diagonal matrix DQ, we first make a 4 by 4 # null matrix DQ = matrix(c(0),4,4) #Note here that ‘‘c(0),4,4’’ tells R that we want a 4 by 4 # matrix all 0. #Next we replace the diagonal elements of DQ with elements of Q for (i in 1:4) { DQ[i,i] = Q[i] } #which is equivalent to DQ[1,1]=Q[1], DQ[2,2]=Q[2],DQ[3,3]=Q[3], # and DQ[4,4]= Q[4] #We can now calculate A A = T%*% solve(DQ) # ‘‘solve(DQ)’’ finds the inverse of DQ # Next we create a 4 by 4 unit matrix I. Here we use ‘‘diag(n)’’ # command that makes a unit matrix of size n I = diag(4) IA = I-A # We now make (I-A) inverse IAinv = solve(IA) # To make sure that we have correctly created the matrix of # technical coefficients A, we form (I-A)ˆ(-1)FD, which must give # us the original vector of sectoral outputs Q

13 Matrices and Their Applications

283

IAinv%*%FD [,1] [1,] 415101 [2,] 1907895 [3,] 4971041 [4,] 6853819 # # # # #

It does. Next we go around a loop, increase the final demand for each sector by 3\,% and call this new vector FDn. We use this vector and find the new vector of sectoral outputs Qn, the new total national output GDPn and calculate the rate of growth RG with respect to the original GDP.

for (i in 1:4) { + FDn=FD + FDn[i]=FD[i]*1.03 + Qn = IAinv%*%FDn + GDPn = t(S)%*%Qn + RG = ((GDPn-GDP)/GDP)*100 + print(Qn) + print(GDPn) + print(RG) } # print(Qn), print(GDPn) and print(RG) prints new sectoral # outputs, new GDP, and rates of growth

#26

After increasing the final demand vector by 3 %

(a) The new sectoral outputs are $427,554, $1,965,132, $5,120,172 and $7,059,434. (b) The national output is $14,572,292, which is exactly 3 % more than the original value of the GDP in problem # 25. This result is a reaffirmation of the linear homogeneity of the Input-Output models. Chapter 13 Supplementary Exercises 1. Given matrices A, B, and C, 

1 7 2 −10 A= −2 6 20 −10

(a) (b) (c) (d)

Find Find Find Find



⎡

⎤ −3 4 ⎢2 3 ⎥ ⎥ B=⎢ ⎣ 0 −10⎦ −1 2

A′ , B ′ , and C ′ . AB and (AB)C. B ′ A′ and show that it is equal to (AB)′ . A A′ and A′ A. Are they the same?



5 −3 C= 12 2



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13 Matrices and Their Applications

2. Find the inverse of the following matrices by method of adjoint.

A=



2 −3 1 2



⎤ 1 4 2 B = ⎣2 6 3 ⎦ 0 8 −2 ⎡

3. Find the determinant of the following matrices. If the matrix is invertible, find its inverse by using elementary row operations. ⎡

⎤ 1 3 5 A = ⎣2 1 4 ⎦ 5 6 2

⎤ 1 2 4 B = ⎣2 1 −3⎦ 5 7 9 ⎡

4. Find the inverse of the following matrices by using elementary row operations. ⎡

⎤

2 3 4 A = ⎣ 5 4 3⎦ 8 9 1

⎡

3 ⎢0 B=⎢ ⎣0 0

−2 −1 0 0

6 4 3 0

⎤ 9 2⎥ ⎥ 6⎦ −4

5. Solve the following system of 3 equations with 3 unknowns x1 , x2 , and x3 . 2x1 + 5x2 − 4x3 = 20 5x1 − 2x2 + 4x3 = 15 3x2 − 2x3 = 2 6. Write the following system in matrix form and solve it. x1 − 2x2 + 2x3 + 2x4 = 30 2x1 − 4x3 + 5x4 = 10 x2 − 2x3 = 8 x1 + 2x3 + 3x4 = 20 7. Solve the following system of 4 equations with 4 unknowns by hand. x1 + x3 = 25 x2 + x3 = 18 x1 + x2 = 23 x3 + x4 = 15 8. Write the system in the previous problem in matrix form and solve it.

13 Matrices and Their Applications

285

Courses

Students

1 2 3 4 5

1

2

3

82 89 65 75 85

95 92 68 79 72

80 97 60 64 70

9. The above table shows grades of 5 students in 3 different courses. (a) Write a matrix operation that determines the grade average for each course. (b) Write a matrix operation that determines the grade average for each student. (c) Write a matrix operation that gives the grade average for all students in all 3 courses. 10. The demand and supply functions of two related goods are given by Q d1 = 40 − 6P1 + 2P2 Q s1 = −40 + 8P1 Q d2 = 250 + 3P1 − 4P2 Q s2 = −60 + 3P2 (a) Use the equilibrium condition and reduce the system to a 2 by 2 matrix system and solve for equilibrium prices. Then determine the equilibrium quantities. (b) Write the system in a full 4 by 4 matrix form and solve for the equilibrium prices and quantities. 11. The demand and supply function of three related goods are given by Q d1 = 120 − 4P1 + 4P3 Q s1 = −40 + 2P1 Q d2 = 140 + 2P1 − 2P2 + 4P3 Q s2 = −60 + 3P2 Q d3 = 150 − 3P1 + 2P2 − 2P3 Q s3 = −90 + 3P3

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Write the system in a full 6 by 6 matrix form and find the equilibrium prices and quantities. 12. An investor has $20,000 to invest. He has two investment opportunities; one relatively low risk with 3 % return, and another relatively high risk with 5 % return. Use matrix formulation and determine the amount he should invest at each rate in order to generate $850 annual interest. 13. Assume an economy with 3 sectors: Agriculture, Manufacturing, and Services. To produce $1 of agricultural output requires $0.10 worth of agricultural products, $0.30 worth of manufacturing products, and $0.20 of services. To produce $1 of manufacturing output requires $0.35 of manufacturing, $0.05 of agricultural, and $0.30 of services worth of input. To produce $1 of services requires $0.15 worth of services and $0.25 of manufacturing products. The annual final demands for outputs of the sectors are $176, $320 and $650 billion, respectively. (a) Determine the value of sectoral and national outputs. (b) What impact an increase of 3 % in final demand for manufacturing products has on the sectoral and national outputs. (c) What impact an increase of 3 % in final demands for all sectors has on the sectoral and national outputs. 14. A vertically integrated steel company owns and operates coal mines, iron ore mines, and a rail road system. Production of a $1 worth of steel requires inputs of $0.10 of steel, $0.05 of coal, $0.02 of iron ore, and $0.08 of rail road (transportation). Production of a $1 worth of coal requires $0.15 of steel, $0.10 of coal, and $0.03 of rail road. Production of $1 worth of iron ore requires $0.10 of steel, $0.12 of coal, and $0.02 of rail road. The input requirements for $1 worth of rail road are $0.35 of steel and $0.05 of coal. (a) Determine the output of steel, coal, iron ore, and rail road companies needed to satisfy a final demand of $550 million of steel, $300 million of coal, $150 million of iron ore, and $120 million worth of rail road service. (b) What is the value of coal and iron ore used in production of steel or sold in the market. 15. Consider an economy with four sectors, agriculture (A), construction and mining (C), manufacturing (M), and services (S). The following table provides the input requirements for $1 worth of output for each sector and the final demand (FD, in million dollars) for each sector. (a) Determine the value of sectoral and national outputs. (b) Assume there is a 2.5 % increase in final demand. Calculate the new sectoral and national outputs.

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Input-Output Table Output

Input

A C M S

A

C

M

S

FD

0.12 0.00 0.08 0.05

0.02 0.06 0.25 0.12

0.05 0.20 0.20 0.18

0.03 0.15 0.10 0.10

1100 2500 3700 4000

16. An electronic company produces 3 different electronic devices in California and in Zhengzhou, Henan province, in China. The following matrix gives the labor hour per unit of each device in California and China. Assume the labor costs, on the average, is $20 per hour in California and $5 per hour in China. Labor Hour per Device Location

Dev. 1

Dev. 2

Dev. 3

California China

0.10 0.30

0.08 0.20

0.05 0.10

(a) Write a matrix expression to calculate the labor cost of producing each device in California and China. (b) Write a matrix expression to calculate the average labor cost of producing these devices in California and China. 17. The original Input-Output table prepared by the U.S. Bureau of Labor Statistics, consisted of 42 industries that comprised the U.S. economy in its entirety and showed the exchange of goods and services in the U.S. for 1947 (this table appeared as Table 2.1 on pages 16–19 of Wassliy Leontiefs book Input-Output Economics.) The following table is an aggregated 3-sector version of the original table (figures in millions of 1947 dollars) (a) Use the table and calculate the matrix of inter-sectoral coefficients. (b) Calculate the sectoral gross outputs if the new vector of final demand, as row vector, for year 1948 is given as  FD′ = 42100 68500 156700 (c) What is the percentage change in the GDP?

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Table of Inter-sectoral Transactions US 1947

Agriculture Manufacturing Services

Agriculture

Manufacturing

Services

Final Demand

34690 5280 10450

4920 61280 25950

5620 22990 42030

39240 60020 130650

18. The following table is an aggregated version of US Department of Commerce, Bureau of Economic Analysis (BEA) 15-sector input-out table estimated for 2012 (See www.bea.gov/industry/io_annual.htm). In this table Mining, Utilities, and Construction industries are aggregated into one sector labeled MUC. Similarly, 9 different service sectors are aggregated into one called Ser. Other sectors are Agriculture (Ag), Manufacturing (Manu) and Government (Gov) (figures in millions of 2012 dollars). Table of Inter-sectoral Transactions US 2012

Ag MUC Manu Ser Gov

Ag.

MUC

Manu

Ser

Gov

Final Demand

86810 8285 80416 69030 37

1502 97899 328561 302979 384

270144 648351 1903440 883549 5597

934112 298770 934112 5132521 76513

2929 114869 370598 667024 9537

74554 989442 2021814 10670276 2619157

(a) Use the table and calculate the matrix of inter-sectoral coefficients. (b) Calculate the sectoral gross outputs if the new final demand vector for year 2013 is given as  FD′ = 74664 998445 2031426 10809234 2867920 (c) What is the percentage change in the GDP? 19. In the previous exercise ignore an eliminate Government and aggregate MUC in the Manufacturing sector. This creates a 3-sector model similar to 1947 table given in exercise #17. (a) Use the table and calculate the matrix of inter-sectoral coefficients. (b) Calculate the sectoral gross outputs if the new final demand vector for year 2013 is given as  FD′ = 74664 3029871 10809234

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(c) What is the percentage change in the GDP? (d) What is the impact of ignoring government as a sector on the GDP? 20. Compare the results of exercise #19 and #17. (a) What sector of the US economy has grown the most over the last 65 years? (b) What is the size of the US economy in 2012 compared to 1947? 21. Consider a close economy with the following consumption and investment functions: C = 250 + 0.67Y I = 150 + 0.07Y Assume the government expenditure G 0 is 80. (a) Write the model in matrix form and determine the equilibrium values of national income, consumption, and investment. (b) From the matrix of coefficients, determine the model’s multiplier. (c) Assume the government expenditure changes to 90. Determine the new equilibrium values of Y, C and I . 22. Consider the following national income determination model for a close economy: C = 220 + 0.84Y D I = 130 + 0.09Y D − 20.5r0 T = 85 + 0.12Y where r0 is the interest rate. Assume the government expenditure G 0 is 100 and r0 = 4.2. (a) Write the model in matrix form and determine the equilibrium values of national income, consumption, investment, and tax. (b) From the matrix of coefficients, determine the model’s government expenditure multiplier and tax multiplier. (c) Assume the government expenditure changes to 90. Determine the new equilibrium values of Y, C, I and T . 23. Assume the central bank of the economy described in the last problem expands the money supply and manages to reduce the interest rate to 3.2 %. Determine the new equilibrium values of the endogenous variables. 24. The following is an extended version of the model in exercise #22. In this model r is treated as an endogenous variable with its value determined by income and the money supply,

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C = 220 + 0.84Y D I = 130 + 0.09Y D − 10.5r r = −5 + 0.016Y − 0.02M0 T = 85 + 0.12Y (a) Treat Y D as a separate variable and write the model in matrix form (the coefficient matrix is now a 6 by 6 matrix). Determine the equilibrium values of the endogenous variables assuming G 0 = 80 and M0 = 800. (b) Calculate the model’s multipliers. 25. What would be the equilibrium values of the exogenous variables in problem #24 if the Federal Reserve decides to cut the money supply by 20 %? 26. Write problem #6 of Chap. 1 Supplementary Exercises in matrix form and find the household’s optimal bundle. 27. Assume a household allocates its $4,500 monthly income to food, housing, transportation, health care, and all other items. According to Table 1.2 in the text, allocation of a typical household’s budget to housing is almost equal to the sum of its allocations to food, transportation, and health care. Allocation to all other items is twice as allocation to transportation plus $500. Allocation to transportation is equal to sum of allocations to food and health care. Finally, allocation to all other items is also equal to one half of allocation to food, plus the allocations to transportation and health care. If the unit price of food, housing, transportation, health care, and other items are $20, $50, $30, $15 and $35 respectively, determine the household optimal bundle. 28. In problem #27, assume that household’s income decreases by 5 %. What is the household’s new optimal bundle? 29. In problem #27, assume that due to inflation prices increase by 5 %. What is the new optimal bundle? 30. Does comparing the results of problems #28 and #29 lead you to believe that inflation is similar to the loss of income?