Applied Mathematics for Business, Economics, Life Science and Social Sciences [Solution Manual ed.] 0130858897, 9780130858894

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- Student syions Manual

Nhlensti(5 for Business, Economics, LtSciences andSocial SCIences

‘Seventh

Edition

‘

Raymond A. Barnett / Michael R. Ziegler / Karl E.Byleen

Digitized by the Internet Archive in 2021 with funding from Kahle/Austin Foundation

https://archive.org/details/studentsolutionsOO0Obarn_y5i3

Student Solutions Manual

Applied ~ Mathematics for Business, Economics, Life Sciences, and Social Sciences

Seventh Edition

Raymond A. Barnett / Michael R. Ziegler / Karl E. Byleen

PRENTICE HALL, Upper Saddle River, NJ 07458

Acquisitions Editor: Kathy Boothby Sestak Supplement Editor: Joanne Wendelken Special Projects Manager: Barbara A. Murray Production Editor: Maria T. Molinari Supplement Cover Manager: Paul Gourhan

Supplement Cover Designer:

Manufacturing Buyer:

Liz Nemeth Alan Fischer

© 2000 by Prentice Hall Upper Saddle River, NJ 07458

All rights reserved. No part of this book may be reproduced, in any form or by any means, without permission in writing from the publisher

Printed in the United States of America

LORS EeSi) "On, Sta meme il

TSBNE

OG = 13> 0658657,

Prentice-Hall International (UK) Limited, London Prentice-Hall of Australia Pty. Limited, Sydney Prentice-Hall Canada, Inc., Toronto Prentice-Hall Hispanoamericana, S.A., Mexico Prentice-Hall of India Private Limited, New Delhi

Prentice-Hall (Singapore) Pte. Ltd. Prentice-Hall of Japan, Inc., Tokyo Editora Prentice-Hall do Brazil, Ltda., Rio de Janeiro

CONTENTS CHAPTER 1 A

A BEGINNING ES ea eR

LIBRARY OF ELEMENTARY

FUNCTIONS

ii

stercsceasesa asses csecesovoctesesesedesosacotomeeeres featcorer aree ete tebenesssossacossincsee odorett mm 1

EMEA ASR gNe BARMMMae

ermaticrrnancbsaxeneasecsanesco+sh-n

3.5;

y intercept:

-4.2

10 25)

(Oa

(Ses,

.co)

-10 29.

Using line,

3 with a = 3 for we find that the

equation

16

CHAPTER

1

of

A

the

the vertical line and b = -5 for equation of the vertical line is

horizontal

BEGINNING

LIBRARY

line

OF

is y =

ELEMENTARY

-5.

FUNCTIONS

the horizontal x = 3 and the

31.

Using 3 with a = -1 for the vertical line and b = -3 line, we find that the equation of the vertical line equation of the horizontal line is y = -3.

Sore

m=

-3

For

the

point

(4,

and y, = -1. Re

fo

a

-1),

x, =

4

te:

Using 6, we get: Ste

A)

ye ob fSq-32ci+ 012 y=

-3x

+

-

2

3

For

the point

and

Lg

we

Pals

pogy,

2091

39.

%9035

y=

The

are

(1,

3)

AR

OT

points

ee SS Re

x, = 7,

and y, =

-5

5.

4ijiLet

a 43.

45. 47.

=

Using

4,

3x

+

we

© get:

6)

4

OU

:

-5,

yy, =

-2,

X,

=

5,

and

Using 4, we get:

_-4-

(-2)

m5 024(=5)

~

-4+2

=-2

21

G45

“So 0eemae

See

> M63 -3

-

Bes

Gio

(2

SS

7

-10

the

,

slope

is

using

4:

not

:

defined;

the

;

line

through

(2,

7)

and

(ibs

3)

Vertical

393-203 2 80 S*

Se ai First,

find

he

the

slope

eaSey ee X, Then, or

y -

-

X,

Tak

by using

6,

(7,

5),

3 =

we

6 y -

3

bee

m(xX

-

x,),

where

m

= 2 and

(Gc,

y,)

=.

get:

F(x —Ss)h)

iOrietaya—

These two equations are these, we obtain: =X + Sie =) 4 OF 9 X= eS

49.

x;

Y, = -4.

we get: 2

=

= and

Zi

ie

6,

(=O

> 3 (x +

5 =

y tas

or

Skee

5

Using

fy? 45158 Ee 4

y+ 5 =0

y=0

i

iad

11 yr

“the

for the horizontal is x = -1 and the

5 = 5 (x == olf)

equivalent.

After

simplifying

either

one

of

EXERCISE

1-3

ye=n-8

First, find the slope using 4: Paes ca Wastin1,Wetye et, "OS Se ey S SeeiteaS) anSeae 5% 7a 5 By using 6, and either one of these

points,

we

obtain:

y - (-2) = -2 [x - (-5)] fusing (-5, -2)1 Vaastu

=

QE (x +

S(y +22)2=15%.— Sy el =e x + Sy =) -15

>|)

5 5

17

Biss, 37)’ and) (2iaes3)) Since each point has the same x coordinate, the graph of the vertical line. Then, using 3, of

the

formed a = 2,

by these two points will bea we have x = 2 as the equation

line.

B31. ¥(2, 3) manda (= 5a) Since each point has the same y coordinate, the graph of the horizontal

equation 55.A

line with

line.

of

Then,

the

using

line formed by these two points will 3, with b = 3, we have y = 3 as the

line.

linear function Yy

57.

Not

a

10

function Yy

59.

A

x

10

-10

-10

0

x

10

-10

-10

The graphs of y = mx + 2, m any real number, all intercept (0, 2); for each real number m, y = mx line that passes through the point (0, 2).

65.

The graph

=

|mx + b| coincides

for all x satisfying mx + b = 0. graph of f in the x axis for all is never a linear function.

We (A)

are

given

A =

100(0.06)t

PAG

NE l=

Sm

wemhianiouA

ING

ME

20,

we

have

A

+

100

0

with

the graph

The graph of x satisfying

=

6t

+

have the same y + 2 is a non-vertical

of

g is the mx + b


(1 + r)3

= 1500 5 3

1+eo

i

Vi S= 11447 0.1447

(ibd,

CHAPTER

1 REVIEW

b=)

33

87.

(A)

R(130) = 2085) -R2(50)) =) 80 l s20BEESO 228). 1.6

Slope?

730, 2150 ta eon)

Equation:

38.

R =

R(1L20)eept

(C)

L76h=

(D)

1.6; The slope the cost.

1i6Ge

= Gueme,—

(A)

ep

2

7%6(Cc

-

50)

@rm R.=\126C

e=25192 Si

gives

the

change

in

retail

5.

a0

isn

an.

2 oom

SS

6mmm2a69. 1e2 52%)

2234

price

per

0

caon is 09smi2iOne

AES)

3031

2S

20"

The

year

1995

£(25)) =F3103 34 The year 2000 (3.0) (= 3 08n4) (D)

39.

(A)

corresponds

Ge)

to

x

=

10)

AS;

approximately

17

CHAPTER

1

120

25

0:49

“for

024408

“ for

OLS 9x

“for

consumption

is

dropping

A BEGINNING

eggs (1-3)

(OPS

) ee mo OY. 27117, corresponds to x = 30 3 466 Ooo 6 501, 2.00

The per capita egg every five years.

COX)

change

"(B)

5

(C)

unit

3246x

x Consumo

=

6 (120)

(B)

£(x))

80

LIBRARY

OF ELEMENTARY

FUNCTIONS

36

and (1-2)

40.

(A)

Let

=

number

of

CE)

x

=

84,000

+

IR (ote)

ee

OSs

video

tapes

produced.

15x

R(x)

C(x)

@ 200000

(B)

Re (se)

= |CS)

©

5Oret=

S45 000) +

B5oa=) S=)

847.0100 2, 400, unites

REeGucomsOususc xt 27200

15x

41.

xis,

0

2, 400%) Rp

= =

Rie)

= xp(x)

00s

Garor

27200

p(x) C(x)

=

Oss

(1-3)

50 - 1.25x Price-demand function 160l¢ 20x 1 Cost’ function x(50

-

1.25x)

Revenue

function

R(x) C(x)

(B)

R=C x(50

=

DLOZ25Sc)e"=b16ORs

31,25" 50x08, -1.25x* + 40x = ~1.25 (x4 1 3 2x08S256) = -1.25(x - 16)? = (x #006)2% = x =

iO,

160 4 10x 160 160! tape -160 128 16 4° V¥i28 = 27 318 16 - V128 =~ 4.686

R= C at x = 42 686u thousand units (47,686 units) and x = 27.314 thousand units (ATP Biley wtaalies) RVC LOre lee Aen OO Iams 4 a exe S104 Ri> C £or 4a6S86e

1,

‘then

through

OF

b%

f(x)

(0,1);

=

p° =

curves;

b%,

1 for

there

as

x

increases.

f(x)

(0,1) x

48

CHAPTER

2

ADDITIONAL

ELEMENTARY

tices

(=. bi

FUNCTIONS

isb i> 7.

0,

any

are

asymptote.

increases

Graphvota

b>

no

RANGE

b#¥1

base

b.

holes

or

jumps.

e.

If

0’
1

i 4.

PROPERTIES

OF

= 10

LOGARITHMIC

FUNCTIONS

If b, M, and N are positive real numbers, then:

numbers,

ar

log, 1 =

0

e.

log, MN =

De

log, b =

ib

fe

log, 5 =

Cc. log, b* pe

Gr

log, MP

may

Log Me = log, N

bos 5.

real

ee Bt MeSxqe

LOGARITHMIC

Common

ae

NOTATION;

logarithm

logarithm

log*x*= dresses

¥ ey

linGea—

is equivalent is equivalent

and

log, M +

log, N

log, M =

log, N

if

and

56

CHAPTER

2

ADDITIONAL

Ss

only

2é

M=N

log,) x

log, x

to to

em.

ELEMENTARY

x are

RELATIONSHIPS

pee 1 Oe x = er

EEE n nnn nn

Ee 2) = 8? (emg 3)

p and

= p log,M

LOGARITHMIC-EXPONENTIAL

logusa=

Natural

b #1,

ileal

FUNCTIONS

ese ty

SSS

log, AS)

a=A2

aaa log, 8 rea

pe A

log,,1

= y is equivalent

pO

log,e

17.

log, ,0.2 = y is equivalent

a9.

log,,10° = 3

9.

= y is equivalent

to 10” = 1; y= to e” = e; y=

log, 5 =

log,P

29.

een

=

log,p

-

=

log,p

-

(log, a +

=

log,p

-

log,¢g -

-

log,2

log,x = 2 x=

m =u

log,A

0. 1.

(0.2)¥ = 0.2;

2" 43

y=

1.

23 slog, 1000

3%

(using

4f)

27.

log, ars log,r

+

log ,r -

log, s)

log, L

(using

3)

log.,,7?

=

5 log, L

(using

4f)

(using

4e)

a

fous

ee

This

tel

b =

4x ae yes

39.1 °

09473

oFshe

dem pat?

41.1 ‘

6

9 = (3)

x= 2

(since

x” 3 log, by

aoe

(37jy

B=

3°Y

45.

2

=

log

x =

=

5 log) x 3

log,y°

Nate ui

are

implies the the

same).

ae

iy

log,10° = 2 3 1og,10

“= =

log,10 = :

This inequality implies that 2= VOL y= =2.

43.

OS)

p-4

equality 10

exponents

L OG

4g)

35. log,10°4 = -4

ay

BS ee

(using

log, s

33. log,49 = y

x=9

37.

=-log,,10° = 3

2a)

25.

31.

to

ai. Log,

(using

Li

10 = pi/2 Square both sides: MOOR esbDitele Gan — sl Oe

3 log,o VN = log,n'/3 b = 3 log,N

3 log,y

47.

log, (x Vy) =

log, x? +

lege”

=

2 log, x + F log,y

49.

Wee i20° 20 *") = 1og,50 +) 10g,2/°:2" = log,50 - 0.2t log.2

EXERCISE

2-3

57

SL.

log, P(1

+ r)' =

log,P

+ log,(1 + r)* =

Dist.

log,100e°°°F* = 1og.100 + Yeglem log 100

=

0.01t

log,P +

tlog,(1

+ 2)

ae

log.e

= log

100

=)

0). Oané

/ 55.

2 log,x = 3 1log,8 =

log,4

log,x =

log,2

x= 57.

+

3 5 1og,4

log, x = =

=

log,6

log,6

=

=

1og,8*/°

+

egyor '2

2 3 109,8

-

8

log,4

(using

Therefore,

x(x

Seay. (se AMoS hau Pe

+

+

2 log, 2 =

log) 4 =

log ,4°/? -

tog 87/4 +

x

4) 4)

=

log,21

-

4)

=

21

eon

-3

is

log, (-3)

log,21

-

=3

not

is

61.

log, (x -

1)

Therefore,

z 3 Alike

not

x

1) =

There

is

no

not

not

solution, 1) =

defined.

it

FO

1 =40(xoe

a

defined. ]

jae

Ol

= 1 = = 95en=

solution

TOs

ae

vi

bee

since 20

log, (-%5 )

Similarly,

1)re = Log, 4(-2 5]

defined.

Y

Y =slog. (x =z)

Gee 9, VO

graphiok

ahi

CHAPTER

10?

x-

0

a

1)

Losi o( 5n

is

58

log, , (x +

b, legal

Log) o(-aL 9 +

The

-

ys«|

ees) el:

=

=

is

65.

log, 2°

log, 8

ala og, (-"5 -

Mo aE

log,6

2e)

log, (x log, x(x

63.

-

4-3 log, &

2e)

log, 8 -

log, x +

since

-

log, 8

x=

[Note:

log,3

e(using

log,x =

59.

aly + 5 log,9

y =

log, (x =

2)

teethne

Graph

Zenash

2

ADDITIONAL

ELEMENTARY

FUNCTIONS

OL

y =

1S.420

log,

x shifted

to

the

67.

Since

logarithmic

functions

must have x + 1 > y= 1+ In(x + 1)

69.

are

defined

only

for

(A) 3.54743 (B) -2.16032 (©) 45 62629 (D) -3.19704

Divs

10* =

12

(Take

common

log 10* = log12 wie

de OT92

logarithms

of

both

sides)

4dogi0”

=

x log 10

=

x

logil0

+fad fac

1.03%

Jog(1.03)*

7005)7"

=

log 2.475

Tog 2,03' =

3

(Take here

sides)

logarithms

of

both

sides;

30.6589

either common or natural logarithms we'll use natural logarithms. )

of

both

sides;

= 1n 3

A 2ZE= ey

fem)

=i)

(Take either common or natural we use common logarithms)

In 1.00514°

81.

4. gh ~ 3. g

= -1og.2.475

x=

79.

OnE S54

= 1.0792

e~ = 4.304 (Take natural logarithms of both In e* = 1n 4.304 = 1.4595 xm u 1.4595 (lne*. = xlne =,x; ine= 2) 2.475

we

of

Woge = i.1285 KeeelsiwA 43:1) (B) logx = -2.0497 x =) 020089 (C) In xt="26:7763 x = 16.0595 (Dipole 18.879

75.

=

"inputs",

range

(A)

Xie

Wes.

positive

0 or x > -1; domain: (-1, ©). The is the set of all real numbers.

In

ua, 1.005 1.005

” =

B20 e273

ais eySissy)

83>

Xx, xO

y=) imei

x > G Y 3

increasing

(0,

©)

decreasing

(0,

1)

increasing

[1,

)

EXERCISE

2-3

59

85.

y=

21n(x

oe Been

x

>

-2

BTUs.

y

=925) -1 0

Wea

Aa

eS

| Xe

0

Yy

|=e 0 Ee 1K 33S)

1 5

a. =

2 389

10

=)

Aeon

increasing

(-2,

©) increasing

89.

The

calculator

109 (7)

calculate

109 (7"}= or

91.

calculate

Hom

= and

y =

log, oY ces

log

log

13

0.

-

bust

Thus,

log

=

as

the

common

not

(0,

)

13

as

og (77) To

logarithm

of

the

;

find

result:

0.2688453123

7 to

Ole Dy) tick, log, 1 =

log 13

7

take

log(1.8571...)

anya numbernwb,s

implies 93.

interprets

13

get

the

same

result.

log, 1 = y is

0 for

any

equivalent

permissible

base

to

b”

=

1 which

b.

log, 9¢ ="058x =

0.8x

Therefore,

© =

(using

1)

and y = c-10°-8*, 95. A £unction £ 1s) "larger than" a function gryon haleraicewl [[cl, Jolll alse ie((S%)) Ss Vep(s@) aeCha 2) eS be S Jo). 2g((9) S38 eilSid) SS jollss) steke Al ES Be Ss ANG, eles SiS x > Vx > In x for 1< x < 16

an

1

PMT = $600, Ds

ik

2s 29

In CIGOay

PV = $9,000, =

=T2

= 2 = 0.75

=

PV

,

a O01

f(r oly 2? =n (0275) =n in(1.01) =°1n(0.75) woe (0 275) 11.

_

Giion

= -2OM000TI— pee eins

)

i

ieee

= 200

40,000

= 68.25843856 ~ *°86-01

Pur = 200

0.01,

Pv = $5000/"a'= We

_40,000

1i-. (1 +.Qnor

496]0 .0075

3010750504 7 %199-29 a.

0. BOT

7. PMT = 40,000

1 - (1 + 0.01)

GL

n = 20,

i=?

+) —n

xn

1.)

i Substituting

9000 Sys

Graph The Ma

600

= r

Bert

ES

15i

+

ee

Y,

=

15x

+

1

given values into e +, -20 (1° + 1)

(faI

Cade

eer

a

(1 +2 43)?°

bi 54290

=)

=

me ee

S4000,

+ ote,

Yo

Date XC)

m=

i

Et

this

formula

gives

il

aeCea ial

(1 ea)

%

1

2

curves intersect at x = nOCO29 . Thuse2 = 0.029%

Homer

PV

the

104)

=

=e

-05

0 and

"40

O.,0S 0.02 gO =

Present

=

pr

=

PMTa>,;

3

4000aFA

il

value —

a

1) 1

see

02

= 4000 (27.35547924) = $109,421.92

EXERCISE

3-4

89

15.

This

is

a present

PMT’ =

$350,

Hence,

PV

mee =

They should $16,800.00. 17.

(A)

value

4{12)

=

problem.

48,

i = S57. =O

=

deposit

= 350(40.18478189) = $14,064.67 The child will receive $14,064.67.

2Vj=esco0,

n=

35047819

OOM

PMTa>;

18,

i =

0075

350(48)

=

0.0L

Monthly payment = PMT = Py ——————— So) a Oe a ae 1 =e PV

600

“2;

Aglo.01

The amount paid in 18 payments = Thus, the interest paid = 658.62 (B) PMT

=

000 @7B]0.015

“1

600

16-39826858

= $36.59 per month 36.59(18) = $658.62. - 600 = $58.62.

(4 = 0.015)

600

= $38.28 per month ~ 15.67256089 For 18 payments, the total amount = 38.28(18) = $689.04. Thus, the interest paid = 689.04 - 600 = $89.04. 19.

Amortized amount = 16,000 - (16,000) (0.25) = $12,000 ABOHEKS), WEAVE ee SEO ONO) Miall a a ((SUA)) es TA) ah (0) 0S

PMT

= monthly

The total Thus, the

21.

ee I NOE aa; "1 A7F0.015

ee ANE 43 .84466677

PV

PMT

=

The

amortization

Totals

CHAPTER

3

$273.69

per

amount paid in 72 months = 273.69(72) = $19,705.68. interest paid = 19,705.68 - 12,000 = $7705.68.

First, we compute the required quarterly i = 0.045, and’ n= 98;"as follows:

Payment number 0 al 2 3 4 5 6 yi 8

90

payment

cae

4

al2 a

5000

i schedule

payment

See ES

i

$758.05 is as

Ci

Unpaid balance reduction

$758.05 758.05 158205 758.05 758.05 758.05 158205 PS 03

$225.00 ZOd On S95 ASRS T2238 OB, 63.88 Bi2ni64

S5sen05 BIS) 7]5 OH SKF 5 1K) 608.30 635.67 664.28 694.17 T2589

6064.38

1064.38

5000.00

FINANCE

222

per quarter follows:

Interest

OF

PV = $5000,

0 045 )oc ne = (0d) 2°

Payment

MATHEMATICS

for

Unpaid balance $5000.00 4466.95 3909.91 38200,8a PTTALS) Sal 2083.84 AESS16 WAGES, 0.00

month

23.

First, ihe =

we 12

compute

72(100)

the

OO,

required

12, =

3.(12)

i

ee

ON gen eee a ee — pas™

Vie

ie

a)?

Now, compute the unpaid payments: PV =

for

PV =

0.01 = (1 e0OsORiC Men 1)-

()()()()

unpaid balance PMT = $199.29,

S=he

payment

$6000,

VErsiGe

Je~ = $199.29

arated Z + 2)

PMT

monthly

60 (LO

after 12 payments by considering i= 0.01, and n = 24.

ee

5L99..29

= 19, 929(10—

24

Ey8ieee

0.01

(1.01)~**}

= $4233459

Thus, the amount of the loan paid in 12 months is 6000 - 4233.59 = $1766.41, and the amount of total payment made during 12 months is 12(199.29) = $2391.48. The interest paid during the first 12 months (first year) is:

2391.48

-

1766.41

Similarly, the considering 12

=

$625.07

unpaid unpaid

PV = 199.29 cE

balance as he

after PMT

eo

two years Sol OOo

can be computed by eto OOdee” ANC Tl —seleze

mulOpGg0( te 01)87) = $2243.02

Thus, the amount of the loan paid during 24 months is 6000 - 2243.02 = $3756.98, and the amount of the loan paid during the second year is 3756.98 - 1766.41 = $1990.57. The amount of total payment during the second year is 12(199.29) = $2391.48. The interest paid during the second year is:

2391.48

-

1990.57

=

$400.91

The total amount paid in 36 months is total interest paid is 7174.44 - 6000 during

= 25.

the

third

year

is

1174.44

-

(625.07

+

= $7174.44. Thus, the and the interest paid

400.91)

=

1174.44

-

1025.98

"$148.46.

PMT = monthly payment = $525, Thus, PV

the

1=

PMT

Hence,

present

UST

selling

value

of

epee |a i

price

=

n = 30(12)

all

(SVAls)

loan

payments

1 n(

= The The Pa

The

total amount paid interest paid is: esO.0 00m

Dieaaee

(2)

total

amount

owed

= 360,

i = a

+ down +

eee

at

the

$60,846.38

payment 25,000

$85,846.38 in 30 years 189,000

(360 months) = 525(360) 60,846.38 = $128,153.62

-

aS

end

of

= 0.0081667.

is:

dee 0. 0081667)22°° .=. 0.0081667

60,846.38

27.

199.29(36) = $1174.44

=

the

=

$189,000.

0.0029167

two

years

is:

A = P(1 + i)” = 6000(1 + 0.0029167)74 = 6000(1.0029167)24 = 6434.39 Now,

the

monthly

payment

is:

EXERCISE

3-4

OL

PMT

=

where

29.

ope: ie

PV

1

es

a)? PV

e035

48,

6434.39

0.0029167 (1 + 0, 0029167.

=

The the

total amount paid in 48 payments is 143.85(48) interest paid is 6904.80 - 6000 = $904.80.

‘

=

aymnent.

Monthl

eo

Se

al

(tee

Ina,

(A)

OROTSE

ee

1 =) (E00 1s

US

CSC

Similarly, loan to be

Wee

841.39 =

(C)

Tie

841.39

(1acOl alywees

='$55,

Om Omel:

oe

after

ee

sige. >HALEN MgWhy

841.39

noo

of

payment

PMT

=

PV

aCe

20

after

years

=

12(10)

(with

25

12(5)

n =

remainder

(with

years

=

0125 7 11="20(12)"=*

sama deny

60]

240%

ad (de togt)

ip

0.0125 = 30,000 ——— 1 = (ee OO rebar oo 236,000 The

total

amount

paid

in

240

Ts

3

MATHEMATICS

OF

FINANCE

GTO a8

(IOUS ine!

payments

395.04(240) = $94,809.60 Thus, the interest paid is: $94,809.60 - $30,000 = $64,809.60

CHAPTER

remainder

of

120]

$36,813 .32

; OR 2 = oie

to

loan

909502

[Note:

=

OROWE:

$30,000,)

PV =

02011)°° OnOLst

Monthly

92

a

arrel 4rere OL) -240 3= $70,952.33

n =

[Note:

The balance of the loan paid in 5 years) is:

= (A)

OO OO

Sais39

31.

the balance of the loan paid in 10 years) is:

OF Only

=*s

12

(with balance to compute the balance after 10 years Now, arene. Olas Oe $841.39, = PMT use years), 20 in paid be Sa o = 240. Heese 202) PMT Se Balance after 10 years

= 841.39 fi die = 841.39 (B)

ORAS2

Oe OL 1) ees

ek

1,

eee

21750000

;

(0). @atal

000

PV ss

Thus,

$6904.80.

=

a=

S75, O00,

=

PV

payment:

monthly

compute the First, = Stools Myce, 3012)

month

per

$143.85

=

PMT

fe

Thus,

0.0029167.

=

12

=

1

$6434.39,

=

nli=—4gjm—

is:

2 S795. 04

of

loan

to

be

(B)

New a=

payment 0.0125,

=

PMT

ee 1

-

(1

=

$395.04

+

$100.00

=

$495.04.

+.2)

495.04 = 30,000

$30,000,

375

:

0.0125

4 =

PV =

(1 + 0. 0f359e2%

4 =

(1.0125)

Therefore,

Bis

ha) O105)

“hrs es 0.7575 495.04 (£00125) 7 re 1 - 0.7575 = 052425 in(170125)"" ‘= In(o.2425) -n 1n(1.0125) = 1n(0.2425) =

Se

ante

=

114.047

The total amount paid in 114 payments 495204 (114)e = 9556, 43.4. 56 Thus, the interest paid is: $56,434.56 - $30,000 = $26,434.56 The savings on interest is: $64,809.60 - $26,434.56 = $38,375.04 33.

PV = $500,000,

PMT

=

months

$495.04

or

9.5

years

is:

2 5243.00625)52 PMT 0.00625 | © = 1 - (1.00625) ., it _ B002000(0.00625) ah

$5,000

(1.00625)

Oh)

=

(1.00625)" = ae Thus,

114

i = pone = 0.00625

G 500,000 = 500,000(0.00625) Sur = ae00625)0 =n.= (A)

of

=

157

1

#4

500,000(0.00625) 5000

0.375 in(0.375) (0.375)

in

'

E>.

eG OOEASy withdrawals.

(B) PMT = $4,000 500,000(0.00625) 4000 =.0.21875

(1.00625)" = 1 = (1,00625)." TG)

Thus,

(C)

The

243

1n(0.21875) 1n(1.00625)

than

per

month

$3,000. 12

the

owner

can

on

For

500, 000("45=] a Thus,

_ ~ ~243

withdrawals.

interest

greater

a

$500,000

example,

at

the

7.5%

compounded

interest

in

monthly

the

first

is

month

is

3-4

2)

So teo

withdraw

$3,000

per

month

forever.

EXERCISE

35.

(A)

= 11

4

i 1 + )" Raed

i

FV =e

=

a = a

$7007,

=

PMT

.

0.0062,

annuity:

ordinary

the

of

value

future

the

calculate

First,

.= 360

m=3012)

- 11 2 3608°" 1.006062) (1.0

100

0.0062

=) SISSploy The interest earned during the 30 year period is: = 133,137 = Se70007 = $97,137 i3ees7 -\360(100) determine Next, using $133,137 as the present value,

the

monthly

payment:

PV = $133,137, i = 0.0062, nm = 15(12) = 180 = 1337137 —___0.0062 ___— ey per 1 - (1.0062) dee: (14 2) = ~ EO?= 2000

(eyoPer = $2000, m=1, pur

S97

(t shih) 458% ne

m=1,

r=

eat he = Sl O73

37

(3-3)

i, n=3

gy

=.12)9038,28(1 + i)? ieee Ma = {5993799 ~ 1-3475775

3 In(1 + i) = 1n(1.3475775) In(1 +1) 1 +a 30.

= 0.0994362

= ef -9994362 ~ 1.1045 i = 0.1045 or 10.45%

(3-2)

P = $1500, I = $100, Seng med. 7m 3: ¥ 8% From

fae

31.

= a

Problem

wnewn =

pete

24,

oe

51,500,

We

want

PV

=

to

PMT

(20

2-=

find

wal i

Or

20S

—77v=-0.02,..m

the

present

a

i

=

value,

= e200

Te

2(4) PV,

0.02

= of

Cay

8 this

annuity.

$10,988.22 32.

The

committee

(A)

The present value of an annuity which provides for quarterly withdrawals of $5000 for 10 years at 12% interest compounded quarterly is given by: Quand PV

=

for hi

Il

deposit

i as i)-"

iS

will

PMT = $5000,

and n = 10(4)

(Gh se OBOE 0.03

166 66626711 amount

$10,988.22.

with

= 5000

This

should

= have

= 0.03;

= 40

-40

(1203)07°] to

i = Tg

be

in

=2S115,573 86 the

account

when

he

retires.

CHAPTER

3 REVIEW

101

(B)

To

determine

(A),

we

the

use

quarterly

the

si

deposit

to

accumulate

x|

where

FV’=

$115,573.86,

PMT | SSR dye 2 epand pry==442Qmneme0

(C)

=

103 eS 5i7 3%. o6 teens

=

a

(1 + 0.03) =

=

$359.64

i =

=

The

amount

of

LS

pees

=

PMT

=

eee

S283 771320

the

loan

O01

a

=

=

is

Seand

ae

(1s

=

in part

1

quarterly

payment

$3000(3] =

m=

=

2000

$2000.

ee Ten

So,

(3-4)

The

monthly

interest

rate

12 (10)\e=eode

79)

$2396.40.

is:

S171, 228es0

= $99.85 per month amount paid in 24 payments

The 99.85(24)

amount

0:03,

The amount collected during the 10-year period is: ($5000)40 = $200,000 _The amount deposited during the 20-year period is: (SS5964))'3 0) =8 S285 77d 120 Thus, the interest earned during the 30-year period SZOOMO00

33%

the

formula:

the

ene

(lO)

is interest

= =o

O25)

paid

1

is

2396.40

(IPOS)

-

2000

=

$396.40.

(3-4) 34.

FV’ =" $507 000Par* =

PMT

=9S=10709, -

———~——_

FV

(a ¢ t)S,

Lint=

92)

0.09 or

OF 00757

7 1 9 Sam

(ie

aye.

ja) saul.

x)

=.

ah))

=

50,000 95.007028

=

$526.28

per

(from

Table

IT)

month

(3-3)

(By

ey

pies OLOies ae 56 wie her

need to obtain:

solve

the

Asay By

ino?

eee)

OnUO027/4

lig 2 In(1.000274)

=

2530.08

days

or

6.93

years.

2 S020 Tauisl

102

we we

2

2 ent

PAWS

same

=, in 2 ¥.

(A)

1256)

raomoars

Sb) ¢ To determine how long it will take money to double, equation 2P = P(1 + i)” for n. From this equation,

da (dete)

="

50,000

-

4

: waee

CHAPTER

rae

3

sql A inti. t) =

MATHEMATICS

OF

7.27

years.

FINANCE

(3-2)

36.

First, we must calculate the future compounded monthly for 2.5 years.

A = P(1

+ i)™ where 0.055

= 8000(2 +

L2

P = $8000,

i = ae

$8000

and n=

at

5.5%

interest

30

30

) = $9176.33 debt

PMT

=

=

PV

ul

=

9176.33

—

where

(1 + 2)

PV

=

Teo

Stet

10, 516.80) =

$8000)

graphs

of

=

y,2 =

are

aa

12

at

5.5%

0.0045833

= 60 =

42 .058179

1 -

=

=

$175.28

(1.0045833)

is:

$2516.80

ere

2%

(3-4)

(beet & 1200 andiiie 222 = 0.005. .

The

.

7

(1005) -29(60),.= $10,516.80 Thus, the interest paid is:

bs

of

Now, we calculate the monthly payment to amortize this interest compounded monthly over 5 years. 0.055

1+

38.

value

will

at

y =

the

right.

120

100,000

$100,000

after

70

payments,

that

is,

after

10 months.

first

find

the

(3-3) monthly

ent = Py an

payment:

PV = $50,000

i = 2:0? = 9.0075

re

12

n = 12(20)

0.0075 50,000 1 =" (1 fom7*° $449.86

per

The present value x years, is given

5

= 240

month

of by

the

$449.86

per

month,

20

year

annuity

at

9%,

after

3 REVIEW

103

1 2 -Gr foo Ts 42S

y = 449.86 0.0075 59,981.33|1 4 (1.0075) ~22(20-» |

CHAPTER

The

graphs

of

50,000

shown

are

The 39.

unpaid

18;

balance

P a SOP From

40.

x =

23,

PV =elOOOFee7 The quarterly

y =

10,000

See

ile

$10,000

after

18

years.

(3-4)

Mim. 360 =

pes 100565

e:,) 0025

(eee

0

below

= 40/3025, (2 = payment is:

=

right.

the

be

tC =

aE

PMT = own = iLoyoyle)

at

will

oO. 08),

Problem

on

0

figure

in the

intersection:

(1.0075)

1-

59,981.33| si Y, = 10,000

“32(20-») |

Ops O25)

=

0.288

or

28.8%

(3-1)

74

: {Ls

25 AenPee ay (AU: OSy),

Payment number 0 1 B 3 4

Payment

$265.82 265.82 265.82 PASS) yishal

S25 00 18.98 2 Sale 6.48

Totals

IMOSS) LAW

63.27

ee

82

Unpaid balance reduction

Interest

$240.82 246.84 PSS) -{0)ih 25933

Unpaid balance $1000.00 759.18 Syl es! 2595585 0.00

1000.00

(3-4) 41.

PMie—asZ2 00," BVee=

S25 00)

a=

ene

=)

0. 00665

By = pur (1=iy = er: 7

215) nels

(1.00665)

i=

f

2500 30,075.188

104

CHAPTER

3

RIMM

take

{52015

Mice a =O

3070755188

-

F(1800665)=

—

19

MOS Sie25

083105 tne. OSsies5 In 1.083125 eae 00GR

n

it will

10).00665)

010

(1 O0e5)2 e = TD) Tye O06S) =

Thus,

(141

One

13

MATHEMATICS

months,

OF

FINANCE

ek

or

12.32

months

1 year

and

1 month.

(3-2)

42.

Pree eee

5e50,000, (iG),

a

=

ry

(Lite)

total

eh)

43.

The

="0.0876,

m=

25

0.0876 1 = 5

eS

="(0F O43ior

ed

———?-0438 —__ _ 37,230

(1

+

amount

a(S 5), 347548)

Thus, foo

&. 76%

92

= 850,000 The

ar =

=

0.0438)

=

invested

is:

af]

(1.0438)

-

S664, 169.76

the interest earned with this annuity 50 7000 .— S664,169.7/6 = S185 .83 0-24

effective

_ _ 555,347.48

1

rate

for

Security

is:

S & L is:

mM

ts (2 + mn) -~ 1 where r = 9.38% = 0.0938 and m = 12 my.

2 2.0938 ) -~ 1 = 0.09794 or 9.794%

= (2+ D2 The

effective

rate

for

West

Lake

S & L is:

m

a e

(2 +

=

-

(2 +

Ves

1 where

r

=

9.35%

=

0.0935

and

m

=

365

365

=

Thus,

44.

45.

A =

West

Lake

$5000, interest

r=

Ve Pt ~

0.09799

is

a better

investment.

t = Be =

0.25

$4899.08,

earned

the

PMT

Pu. ee

eM

1 =

is

I =

or

$5000.00

ee UE ee (4899.08) (0.25) ~ 0.0824

Using

Waste

-

S & L

P =

The

=

|

sinking

=

peo),

fund

9.8%

-

(3-2)

$4899.08

or

8.24%

a =

ae

=

$100.92.

Thus:

(3=f)

formula

ee Tey

S200,

FV

=

sil0) O00}, fand

0.0075

ed

=

0.0075,

we

have:

TD

000 .c apinks AAG AMD Dh Rice.1G Cite O07) > — 2 (GLwOOr7/Sp re: eal

Therefore,

fee 0075)?

om

Tos

(140085) Wk 0075)e% n

The

couple

will

=

OS 75

7k). 023 75ee-? = sim I, 375, Ine Sere

i00nEhe

have

to

1.375

utes 5

make

43

deposits.

CHAPTER

3 REVIEW

105

46.

0.45 29="-25 Pe i

Pv = $80,000, A)

PMT

a

= "25. Suan eee

s ace ay 80, 000

eae ott

0.0125 =

(B)

Now.use PMT = calculate the

!

PV

=

1

PMT

ied =

47.

(0.

=

=

30025))

1000

=e

ieee heor25) $1435.63 monthly

$1435.63, i= 0.0125, unpaid balance.

and

n=

96

;

IWASYS) 4{538}

ge3

-84

(GE sees Ones)

A

0.0125

of

12

unpaid

Deposit:

balance

$10,000

%

Sea oo 0 40! [1

after

the

first

=

84

to

at

8.75%

(PROREZS))

-~84

]

year

Amount of loan paid during the first year: $80,000 - $74,397.48 = $5602.52 Amount of payments during the first year: TONGA S5 5.63) em Slee SO Thus, the interest paid during the first year S$17/,22/ .56e="S5602252 =esilpe2sea04

Certificate

-

payment

(hh FH

$74,397.48 (C)

96

n=48(22Z)5e.

_=,0-0125,

is:

compounded

monthly

for

288

months

Ara

P(

1+)?

P = 10,000

=)

10; 000K 2e00729))

="

SOi) O41

288

:

i=

86

N=

Reduce the principal: Step 1: Find the monthly ai

12

=O

OO 29

A238

payment.

PMP = PV

Day

ORSTS

ee

PV = $60,000

(P)

0.00854 ——————*"— 1 °° = 60,000(0.008961)

Mice0 025 i= > 12 n = 12(30)

= 60,000 = Step

2: Find

$537.66 the

unpaid

present value 288 payments.

PV = =

PMT

per

rege

i

a

CHAPTER

3

Soe

537.66(106.9659599)

MATHEMATICS

OF

after

$537.66

=1S5 7251 4 332

106

= 360

month

balance

of

= 0.00854

FINANCE

72

per

payments,

month

PMT

that

annuity

at

is,

find

10.25%

2553'7..66

i=

ee

n =

288

=

0.00854167

the

for

Step

3:

Reduce

the

required

principal

to

pay

off

by

the

$10,000

and

determine

the

length

of

time

loan.

1 -_ (100854167) eee a) 7” 0.00854167 0.7548005 0.2451995 __In(0.2451995) = 165.3 ~ 1n(1.00854167)

Pe Ei 33 ey ee 1 -

(1.00854167)~" iT}= (1.00854167) ” = _ n ~

The loan will be paid off after 166 more payments. Thus, by reducing the principal after 72 payments, the entire loan will be paid after 72 + 166 = 238 payments.

Step

4: Calculate 8275%

Lor

the

future

360)

—

value

of

a $537.66

2) ee

FV = es

monthly

48.

Use

the

$10,000

to

reduce

A

PMT = 75,000

ie

25s

ero.087s Medias ent the

the total given by:

principal

os, 0 (ora tam

and

invest

the

interest

for

each

of

374)

the

PV = $75,000

n = 2 (30)4=4360

inthe #2

12% mort 12% mortg age:: gage

at

(J=2900-9,

PMT = PVT tite

= Total

annuity

payment.

We find the monthly payment and options. The monthly payment is

=

month

PMT = $537.66

122 .= 547. 66 (1.00729167) a ae pet a537 66 (195.60300TS) = 1054155084 Conclusion:

per

238°="122imonths.

i1 = = ~4,° “75 = 0.01 0.

0.01 ———————+>> Myce: sabtaal th Azo 8

75,000(0.010286)

771.46 per month interest paid = 360(771.46) =—S2 0202 oo0 mortqage:)

PMT = 75,000

1-=

wee

-

75,000

=/05009375

0.009375 ——————— >> tT > (T0093 7S) "

75,000(0.0097126) =

Total The

$728.45

interest

lower

rate

per

month

paid would

= S

360(728.47) - 75,000 Sle), eas. OW save over $15,000 in interest.

(3-4)

CHAPTER

3 REVIEW

107

49.

A-=.

$5000,

erj=e2

P= (1 +2)” 50.

P =

A =

A

= 4476.20(1

=

+

i)

(oe OO oe

Pep

) = 0.0803811

a = ef 0803811 1020837

_ 4 9837

som

r = 10.76%

p = eS

(8.37%

= 0.1076,

first

compute

the

monthly

i=

eS

=. 0.01,

and

nm =

PMT

=

rot ee Che

ess te hee

PMT.

PV

we

calculate

$222.44)" na

Thus,

the 9%

aaa

=

“i

1

unpaid 0.09,

+

r

=

the 1°=

229) 4404

-

9944.73 payment

5(12)

a balance

m= =

unpaid

03.01 *) —n

and?

using

(3-1) PV =

n=.

-36 after

CHAPTER

3

$10,000,

=.60.

balance 60

after -

24°=

24

$2225 44 payments

S

per month

by using

36.

= 22;244[1 - (1.01)~2°) = $6697.11 2 years

is

$6697.11.

(3-4)

12

9b

(2 43

0.09\12 12 }

= (10075)

108

t = = = 0.5

0.01 100 = 10,000 ————————————_ = = 1.— (804-0! 69) 522 = = (i%'91) 76° =°

r=

(3-2)

Se 00 1. +O O76 JiOFS)

ert

10

1n(2.23404)

: dcs

51. A = $5000,

=Gyn

(3:2)

+ i)?°

In(1

Ts =

53.

m= 1,)r

76 14

ts 7)"

> 2020008) > 3.204 4476.20 —

1+

Now,

(nt=vS

er e710

inte

We

$10,000,

Pt

10,000

52.

£00.095,

ee eye 102095) 39) ct. 095)

$4476.20,

10

'=99 75%

MATHEMATICS

OF

FINANCE

=

+7"=11 = 090938

‘or

9-398

(3-2)

54.

(A)

We

first

calculate

compounded

FV

=

(lee

PMT

the

annually

gy

=

4

future

for

1

value

of

an

annuity

of

$2000

at

8%

9 years.

where

PMT

;

="S2000,

7 =

0.08)

andinwang

9

= 2000

oe

aeoe

rete

4.2 95°000[ (1208)

Now, we calculate the future annually for 36 years.

A=

P(1

+ i)™,

where

= 24,975.12(1 This FV

is

=

the

value

of

this

P = $24,975.12,

amount

i = 0.08,

future

value

r

of

a

where

$2000

PMT

=

annuity

$2000,

i =

Rijs 0208)" = 1 = of freee Seek 1 _ 25, 000[(1.08)3© 0.08 The PMT

amount

Sep

ye

of

loan

is

VEY

oe

($100,000)

(0.8)

=

let

a = eo

=a

OUS9S03,

=

$746.79

monthly

Tet

Hee

=

80,000

u

$896.76

(B) To find =

i =

payment =

for

PMT

=

monthly

unpaid

1 an

for

(hisrd)

3746.79,

0.0089583,

for

=

$80,000

n

3

36

(3-3)

and

the

SIIMi,05,

for

after

$41,482.19

serhent

LeOaet

Then

716 .66667

0.7991735

15 10

ees OO

Wes)

years. years,

we

use

mortgage:

i = co

=

0.0089583,

620089593) AGREES

-240

unpaid 15-year

2

balance

for

n =

12(20)

the

30-year

240

iy AD - (1.0089583) ]

mortgage

mortgage:

OOOH,

unpaid

=

te = 83,362.915[1

gi =

Se)

=

Ee

OOSSES 31-0 ow|= 100,103.81[1 896.76 TERE LTE uv

=

- 1] = $374,204

-60

VA

n

years.

=

payment

balance

30-year

eid)

$73,558.78

Mee

and

annually

ao

4

the

I 746.79 PV =

Next,

compounded

—_0.9596687

30

0.0089583

the

PMT

First,

8%

716.66667 Sseae

1 - (1 + 0.0089593)°°

PV

n = 36

0.08,

ie —anbn(s0)

0.0089583 = EES

1, - (1+ 0.0089583) 369

PMT

compounded

6 ae

=) (2X0) O00)

Substitute ete

Thus, 53.

Let

«vy =

the x = y =

2.45

A (aS)

base

into

of of

is

to

(2):

(1):

ai ae ofS Memeo

price

number number

add

—

9.80.)

$17.95;

pounds pounds

of of

17295

the

surcharge

Total

amount

of

Columbian

beans:

132

x 50

amount

of

Brazilian

beans:

132

x 40

=

Columbian

beans

needed:

12 16%

+

Ss 176Y

OF

3G@*

ae Brazilian

beans

needed:

5: 16%

+

10 i¢6Y

SF

igee

we

need

to

(1)

ax + 2y =

(2)

(2)

by

-y

Substitute

Sy eeil) -3

and

=

-9,240

=

6,160

y =

6,160

y

per

pound.

+

6,600

lbs.

5,280

lbs.

3BY

ro) oe

solve:

ox + 2y = 6,600

Multiply

$2.45

robust blend mild blend

Total

Thus,

is

ax + 2 (6,160)

resis

add

into.

to

(1):

(1):

6,600

6,600

—

2,310

=

94,290

SS =—5), 720

Therefore, 5,720 6,160

55.

Let

x = y =

the manufacturer should produce pounds of the robust blend and pounds of the mild blend

amount amount

of of

mix mix

We want to solve the Oelx +50'..2y = 20 0: 06x + 0. 02y = 6

A, B.

and

following (1) (2)

system

Clear the decimals from (1) and (2) 10 and both sides of (2) by 100. x + 2y = 200 (3) 6x + 2y = 600 (4)

of

equations:

by multiplying

both

sides

EXERCISE

of

4-1

(1)

by

123

Multiply 5x

=

Se

Now 80)

(3)

by

-1

and

add

to

(4):

400

till,

substitute ae

va

2yY y

x =

80

into

(3):

80

grams;

00 120

60 x = mix

Solution:

A =

y = mix

B =

grams

60

pa

57.

p=

200

equation]

[Approach

70

-od + 4

j9) 3 (A)

Avoidance

-3.d + 230 The

[Avoidance

figure

shows

the

equation]

graphs

of

r

the

two

Ng

equations. (B)

a

Setting the two equations other, we have aS15 a+ 70 ~ a 43 d + 230

to

each

S %

Approach 200 d

100

Distance in centimeters

70

-

230

=d =

22 gig

equal

Liles es 15 d'="160 a (C)

The

EXERCISEa 4-2

ae

Things Le

rat

=

141

would

cm

be

(approx:)

very

confused

SR

to

(!);

it

would

ET

vacillate.

IT

ETE

SSDI

DEE

EE IE BLEDEL

EEE

remember:

eMAT RICHES

A MATRIX

is

brackets. matrix mxX

has

nis

a rectangular

Each m

the

number

in

rows

and

n

SIZE;

m

and

array

of

a matrix columns, n

are

numbers is

it the

called is

written an

called

DIMENSIONS.

within

ELEMENT. an

m

If

a

X n MATRIX;

A matrix

with

n

rows and n columns is a SQUARE MATRIX OF ORDER n. A matrix with only one column is a COLUMN MATRIX; a matrix with only one row is a ROW MATRIX. The element in the ith row and jth column of a matrix A is denoted ais: The PRINCIPAL DIAGONAL of a matrix A consists [bo

A system of system if: (a) (b) (c)

124

of

CHAPTER

the

elements

linear

Ay41

equations

AgQ1

is

4337

transformed

two equations ar interchanged; an equation is multiplied by a nonzero a constant multiple of one equation is equation.

4

SYSTEMS

OF

LINEAR

EQUATIONS;

MATRICES

into

an

equivalent

constant; added to another

EE

Jw

Associated

is

with

a,x,

+

a,x,

+ bx,

the

a

b,x,

An

=

b

Lah

linear

system

k,

= k,

AUGMENTED

a, 4.

the

MATRIX

k

3

(I) of

the

system

|

(II)

b,|

augmented

matrix

is

transformed

into

a

row-equivalent

matrix

athe (a)

two

rows

are

(b)

a row

(c)

a constant

is multiplied

(KR; ts (Note: 5.

The

Given

interchanged

—

system

is

added

to

another

row

"replaces.")

linear

(II).

equations

If

(II)

is

(I)

row

and

its

equivalent

to

a matrix

0

i

n

(2)) ie 0

| 0

|. then (I) has infinitely many 0 solutions (consistent and dependent) ;

(3)

[2

"

a

0

0

Pp

2X93 8

(Omens

a

Ae

ia

5 "

(I)

p #0,

then

x3

36

=) > £2

Sra

bale

0

has

and

(I)

: solution;

independent) ;

has

no

solution

(inconsistent).

=65

3

-5

then

(consistent

° Crs,

-4,.a,,

F =

,

E a unique

associated

m

@e=| (B)

row

0

ieeeAgc:

es

one

(KR, => Rik;

1

(a8)

a5

means

of

augmented matrix of the form:

7.

of

constant

R,).

arrow

the

R,)i

by a nonzero

multiple ed

(R; oO

7

|, size

-8

CG

ioe eS

5)

0

2x4

size 3 X 2. F would have to have one more column to be square.

=

—6

EXERCISE

4-2

125

13.

Interchange and row 2.

row

15.

1

R,

1

dhs

=

ak

xX, =

2

eure

IK

e263

3

Z

3

R, OR,

row,

7S 0

(-3)R, +R,

Thus,

ad

£1LO

0

A)

From the last inconsistent.

1.

Te

3

eZ

0

et

-1

5

10

2

R, +R,

(-5)R, +R,

Asie

=:

Sa

@)

Al

-1

ab

0

2

(-2)R,

+R,

>R,

(-3)R, +R,

OR,

conclude

220 2

1

-4

we

i

LOT

R, OR,

2

1

-1

that

3 -5 -10

(-2)R, +R,

there

is

2

1

7>R,

7

R,

= 2

3

2

meet A fae | a0)

i

=5

-3

=5

0

0

0

solution;

the

ij

>R,

no

EXERCISE

system

4-3

is

133

35.]

3°.

-2

1

-7

1

1

-3

1

i:

2

1

=4

0

- | 32

1

-4

Oa

OL

1

hes

al)

Br

=—2

1

R, OR,

Ie

=3

il

-5

10

5R, +R,

From

this

(-3)R,

+R,

AR,

al

Ore

= 1

—1

-2

2

=10

0

0

0

0

matrix,

=6

x,

-

es

4 ,

this -2s

=3

=i

and 26

-3

matrix,

-6

X, + 2x,

+t+1,

Die

5

20

-10 > R,

or

=3

-

x, =

15

and

x,

=t,

sbr | S74] 0

ae

ren rd es 40°

By

0

0

0

0

Get

Let

x,

at

be

any

real

t.

2 -

i

0

0

x, =

s and

0

-4

eve

1

PER

ea

-3]

0

x, 7-t.)

real

i

Then

numbers.

4R, +R,

-10

(-8)FR, hak

1

oly

OR, eae

Ser

3

A

BD - yy

-7

s and

t any

ey

GDR

0

*11G0"

2.

ee:

aL

5

=

x, =

mieile

ies {

~ | oo

2x,

and

OR,

Sh iol Opliga Faeries

| ee

2,

3}

ct Pa

8) F-2

Rote

+

2 | 3R, +R,

oe:

0

fp

5

(-5)85 + TR Ons

-7

ql -42

solution.

Ai

EP 9

ooo | ea) | 20 Se

134

=

X, =

silat

3

From

wat

-2

OR,

then

PO

No

}-5)

1

2

(-1)R,

oy

ee

72ers

O

= }| 0

4 -2

-3°,

OR,

R,

2 1-1.

OE

OL

0

0

MATRICES

64) =19 eet 8

-40

(26)Rn. &4 pio tp ; : : ores

ee

S$

Baie:

5

3

ae

24.3

See

Peo iioes ROR

0 0 system x,

-

0 of

0 equations

9X2

=

-4

X13

2

S

1-3

0

ot

“MT

0

0

“4

hs 0

0

is:

and

x

=

2.5x,

-

4

tee Lee

x, =

t.

x, F

2256

X,

c

a

43.

=

Then =

any

real

2

ali!

ah

Ae

1

-2

4

(-1)R,

+ R, >

-1‘

-2;

9

A

al F

(-2)R,

+

(-4)R,

+R,

bd}

12

= 1 | 7, ag R, -

1

1

0

0

AKON

Fae,

1

0

0

1

TNE on

RR il ey te eee eli

(B) The of

(to simplify

4

R, 3

system is solutions.

OR

3 ~

3 0 0

The

system

of

x

=

1

Lo 3

R, >} R, 0

x.

2

Solution:

of

R,

the arithmetic)

2

0

(A) The

t,

5

“ieee

45.

number

4

-3

:1

for

It

al

0

ail

eee

=

-2

SANS

at

x,

=

A

X,

=

Sin‘

=2,

2}

£2

24 all

-2,

two

parameters

system is dependent solutions.

with

one

parameter

(D)

Impossible.

system

is

independent

with

a unique

(GeVRe

(+i),

2

R, >

4

PRLS

3

R,

RoR

3

R,

1

with

The

1

R, +

is:

dependent

(C)

Al :

x,

=

I:

and

an

and

an

infinite infinite

number number

solution.

EXERCISE

4-3

135

47.

x7

Xoyaee

3x,

+

If

kx,

k = 2

=

-3,

Os

the

system

is

E -1

>

4

2

0

0}

-5

3

-9

-4

Gwe

a a a0

i

7

il}

0

3

-4

-1

7

ee -3

-6

(-2)R,

+R,

OR,

(-2)R, +R,

AR,

(-1)R,

+R,

OR,

(-3)R,

>R,

=

2x,

real

32) -3

0

=

= 2a) a a ORS

ga 1 2

0.2

ONO

R,

XS

7=2

Dk || ENO)

1

=i)

5-2

2

6

0

0

0

0

2X,

te IC Mat,

Al 0

=—=26

et

Be = ise

2R, dep ioy omega) (-3)R, + R, > R, ~

0.6-)'-#12, -& Rage FR,

-2 al 6 | -3.4 2) -0.1

.-32l

xX, =

Xx,

t,

s}

Ss and

3

x,

wheres,

of

ae

Jala 0 1

-1 -4

ail

4 | -2.4 6 | -3.4

0)

SO.

0

5

0

Osa

Tie

Ce

ome).

Ot

&

3" Bye

Ea

=2 = 3

§ 27)

4e-l

1 DS

-=1

C.

t are

>

(-1)R,

+

R, >

2a4

R,

R,

(206

1

=A

IR,

( 2) 3

O

R

3

oes

OF

LINEAR

EQUATIONS;

MATRICES

(to simplify

Ri

0 eee

BR,

+R,

9.3

SiS

SYSTEMS

i 2

a2 aA:

0

4

+R (-2)R,

ge 0

ia.5a '-2) Soe Bi,

AD

al 0

CHAPTER

3" and

3), a

system

=

07

ah Orie ds |eolG

+

-2R,

2

ap

k =

=

ie ial x |e ai

Ry

-4

= 53. | 72

6

si

The system has no solutions. If k #-3, the system has a unique solution.

any

3

aot

aly

raat

Thus,

kx, =

6x,

If

7

alee

3S

51.]

+

+

X,

3a

ail

x,

2X,

ee

3X

-3R,

49.

7

arithmetic)

oR

4

m& un

SN

CN et

‘i

st

0

St

z

Be:

Coon

| aes a] al

4

ee +st et ot

By es e

:

19) ef com

a

my ~~ |

~ 1s oe

=f ot we la 1

wN o

sie is:

Onesy

~

N

+ +

gl &me oad

st

ot

+

(x ves

&

sO

& ~e

wm

NO

| ~m

XxX, =

od

OF2% wm

Xoo =

coh

equations

-0.5,

oe a a ee fcV2).

AMAR: re

Aaa +perience Ta tenes.

Sant IC

Solution:

of

MO

Te} ! 'CcoOG00

system

8

woON +t

: uw wo moM °

The

(oa) N a)

iZR > Ry wm &

+ & Meee ie ee FON a Ge

“tas fi. ora

_N NN

& ~~

&

Tet

RIL Cot i woman

op woul iS)! elgQh

IN

le YO AN AUN

pe ~m

“i

ina (oe)

th a sho

EXERCISE

4-3

137

1-2) ef

Gmmege

=*| R,

1

2

3

760

AKO)

1

2

AZO)

0)

Oh Bly

Slo)ck

(0.1)R, +R,

330 120

+R,

OR,

(-0.2)R,

+R,

OR,

| See

OR,

tO ome se Os) 05

(-0.6)R,

1

-32

s and

0

One iil

2

0

Ora

10R,

CHAPTER

4

SYSTEMS

=|

1e =2 0. 20,3 0) -0.2

1

Ope

l=t

-80

420

0

Al

2

420

10

0

0

am

100

> R,

LINEAR

EQUATIONS;

MATRICES

760 \etee ues

ab oe (-gy)a a,

R,+R,

(-2)R,

OF

3 =0.6) -0,1)y

-80

(-2)R, + Ry aay

138

ee

t.

760

O76 O52

3

Ot Notas.

boats.

model is: 1.5x, =

+

+ 1,

numbers

boats

x,

ch -5t

2s

one-person

0.5x,

=

1

3

Thee Oy 5 Let

70) tone ae vil oo Oe UNL ee of See "9% onlane

+ 2 5%,= ral x,

Ore

1-2 Ofc. pig peo

+R,

OR,

OR,

if orig aoe 66” | me Oo a Thus,

xX.

person (B)

st

=

20 220 206

20),

boats,

The

Xo) and

+

0.6x,

+

0.9x,

n E 2 1

+

x,

In

Thus,

where (C)

+

x)

=

A202

=

65-280

to

ee

2

i

Hs

Sue

80


R,

ele ~

t is

model

=

380

t < 210

(two-person

x

0

boats,

S\SK0)

-80

+

2 A

one-person

0 2 Be 1 Bh a6

real

mathematical

2R,

20

boats.

(four-person

1

0

1.2x,

0.5x,

OnSe

~10

=)

x,

keep

be

or

is:

1.5x,

(t any

&

order

100,

OR,

=t

i

43

four-person

ie r E 420 Ol

Os

Then,

x,

fee 7 r 209 Be |‘a é is 2 3 | | 330 m6 0.9 “ayo pmez0 0 -0.3 -0.6] -126 1 (-0.6)R, +R, OR, (-5.3)2 7

: 2}!

(-2)R, +R,

x,

and

model

xX, +

ee 1 | 0.6 0.9 1.2] 2R, > R,

Let

100

mathematical

0.5x,

Thus,

220,

an

and

t 2 80.

boats)

integer.

is:

2

760

ONG

ORS

BS

OL 2

OS

120

(-0.6R,)

+R,

OR,

(-0.2R,)

+R,

JR,

il OMe

2

Oe

res

OF

=O

Ja

760 =12'6 | =| -32

(-o.3)®2 >

ik

2

760

70

al

420

OO

-32

0.1R,+R,7>R,

760 | From this matrix, we conclude that there is no solution; 420 | there is no production schedule that will use all the 10

labor-hours

in

all

departments.

EXERCISE

4-3

139

59.

Let

x, =

number

of

tank

cars

gallon

tank

cars

Xy =

number

of

8000

number

of

18,000

3

the mathematical x, + x, +

6000x,

Dividing

+

the

Xj

Hag

3x,

+

The

augmented

eh 4x,

+

Fs 1) 31 3a

model

+

second

equation

x,

=

9x,

=

18, 000x,

="2 5070100

by 2000,

corresponding

O)

t.

24

we

get

the

system

5

2S

x,

cars

24

matrix

xX,

tank

is: Xar=

8000x,,

AR

to

this

Ome

6

system

eA) 2 E 0 4 i

tej fh Sys}

is:

53

(-1)R, +R, Re

(-3)R, +R, > R,

xX, =

gallon

24) " 3 ih

oe

Let

gallon

x Then,

Thus

6000

-

5X3

=

—29

+

6x,

=

156!

Then

Be 5t - 29 and PS) = 53 - 6t Thus, (5t - 29) 6000-gallon tank cars, (53 - 6t) 8000-gallon tank cars and (t) 18000-gallon tank cars should be purchased. Also, since t, 5t - 29 and 53 - 6t must each be non-negative integers, it follows that El= 46), 77 FOrReE

61.

Let

xX,

Then,

iT}

federal

income

tax

x, =

state

income

tax

x,

local

income

tax

=

the mathematical model is: x, = 0.50[7,650,000 —- (x, + x;)]

xX, =

0.20[7,650,000

-

(x, + x3) ]

xX, =

0.10[7,650,000

-

(Xx, + X,) ]

and

The

140

x,

+

0.5x,

+

0.5x,

=

3,825,000

0.2x,

+

X,

+

0.2x,

=

1,530,000

0.1x,

+

0.1x,

+

x,

=

765,000

corresponding

augmented

matrix

is:

a

0). 5

075) 1537825, 000

(-0.2)R,

OZ

AL

OFZ

CMAs.

saver

a

Cote.

CHAPTER

4

SYSTEMS

1715810), 000

765,000

OF

LINEAR

EQUATIONS;

1

+R,

OR

fr) —S R 3 3

MATRICES

a

BHOL5), 000

ae

TES), 000

205

20R,

>

R,

(simplify arithmetic)

382,500

\o On]

.5 | 3,825,000 sa 745,000 |Ry 7,650,000 i iO “OS: ele oe #O"OoO 19 OF Oo

an

+o oo oO

~#

a

0.5

0

at

0

ORS

0.5

| 3,825,000}

(-0.5)R,

+ R,

7,650,000

(=OOK2 DER

+

19 Cipage =9

0

=17"t

>

R

765,000

1 (-a7)%s ma

19 | 7,650,000 Oo:

7 R

R,

~6; 1207000

=] (Sk

=)

0

TS) || Ue SNO! MOOG, 360,000

IR,

+

R, =

R,

(-19)R,

+

R, >

Ry

3,240,000 810,000 ii Sao 360,000

Ore Or ono °o gle} (eos ©

THUS,

ie

Sa), 240); 0100,

itabricty ;

63.

1S

&6,

1

4x4

4,410,000

income

(7*650°000

Let

=

x,

2

_

=

=

X,

4 x

$810,000,

x, =

$4,410,000

By

00 57165

or

$360,000.

which

is

The

57.65%

total

of

tax

the

taxable

57.658).

Taxable

income

of

company

A

Taxable

income

of

company

B

Taxable

income

of

company

C

Taxable

income

of

company

D

PR

Il

ara =

The

taxable

=

income

each

company is given by + 0.03x, + 0.07x,

=

0.12x,

+

0.81(2.6)

Il

0.11x,

+

0.09x,

+

OR 7216 88)))

+

0.02x,

+

0.14x,

et Fae

0.06x,

which

of

OR a3 2)" ih 0.08x,

is

x, —

the

same

0.08x,

+

0.11x, +

+

the

system

of

equations:

0.13x, 0.08x,

0.78(4.4)

as:

-

0.03x,

-

0.07x

-

0.11x,

-

0.13x

Ate.

ORO Ss

-0.12x,

+

X,

-0.11x,

-

0.09x,

Pe

-0.06x,

-

0.02x,

-

0.14x,

+e

ll rw

Dee dls

ns

te

278/856

=)

Sea

PP

EXERCISE

4-3

141

This

is

the

mathematical

1 ~ Oi Delite =0.06

model.

-0.08 -0.03 Qiegal 20.025 Ondee iam &20.024,=0.12

-0.07][ x, R,

all

il

eo

il

2

OG

0

2

0

4

18

(-1)R, +R,

1 | 24

0

0

8

AG)

1

P

10

0

a

-4

-16

-4%3

7,

OR,

(=2)R,.+ R, > R,

1 Se 0

0

0

Hi

2s

0

1

8 4

(-2)R, +R, AR, TAUS 7 (es0i=5 Os xX, = food

(B)

142

B,

The

and

1.

LOG?

2,

4 ounces

mathematical + +

10x, 10x,

= =

340 180

10x,

+

30x,

=

220

4

SYSTEMS

a.

Ow

#2

0

30x, 10x,

CHAPTER

Lio

OF

and of

x,

=

food

model

LINEAR

4 or

8 ounces

Cc.

is:

EQUATIONS;

MATRICES

of

food

A,

2 ounces

of

30

10]

340

207

10;

180

10>

30%

220

i 10

jy]

3

1}

34

Bi

ys

1

a

yto.b

3

a

1 122

I~

(3422

2S

1 7 Fy

R, OR,

oe

10 F2 7 ®2

il

SA

(-3)R, +R,

OR,

(-1)R, +R,

OR,

1

SO

18

1

2alk 2.0

0

,

4

(-2)R,

+R,

OR,

Ae

10 10 %3 7 ¥3

bee

| 1

~lo

18 10 ee

1 (C)

The

From this matrix, we conclude there is no solution.

mathematical

model

30x,

+

10x,

+

20x,

10x,

+

10x,

+

20x,

i 10 10 10

=

is:

340 180

a4 20/_[ 29 201 180) L30

R, © R,

10 10

4 seo] | 20] 340) 13

whe REOER:

elAgy

that

1 1

2 18)_[2 1 ‘4 4 21 34) Lo -2 -4| -20 i (-3)%

(-3)R, +R, OR,

>

ohh apa

i b

1

Omi.

d

ie y E

0

°

21

40

.oga

-2

(-1)R, +R, Let

x3

x, to The x,

67.

Let

=

be

t

0

(t any

real

number).

X%, + 2x, = 10

Then,

x, =

8 ounces

Ofetoodee,

number number

of of

OFS

barrels barrels

of of

of

jt

TOE

92.t

=

-

eLOMES

ftps

Sab

for

Divide each side and each side of equations: XX,+

Xx, +

+ 3X, +

x3 +

3x,

+ 2x,

2x,

A;

x

10

mix mix

+

2x,

=

30

xX, =

30

5x,

=

2t

ounces

of

food

B;

A, B, Gc D

(1) (2) (3)

of equation (1) by 30, each side equation (3) by 10. This yields

2X,

+

food

= R,

solution:is: =

| TUS psAy

of equation (2) by 25, the system of linear

70

EXERCISE

4-3

143

al

Al

qt

2

30

2

3

ul

AL

30m

~ a) @

ff

+i.

3

= 305]

3

2

2

5

70

®

t=.

=).

=2

-20

(-2)R, +R,

7R,

(-3)R,

OR,

al

+R, 0

tC)

0

iif

R, +R,

7R,

1

ill

®

30 73),

il

0

2

0

1

0

0

(-1)R, +R, 5

60

= 3

=3

2,

0 Olen

PAS

5

60

-1

-3

-30

-2

-4

-50

al,

R,+R,>R,

2

= 0

il

(-3)®s

78s

>R, 0

il

lO

1

OPEL

0

0

al

10 -5

2

25

(-2)R, bers en.)

Thus,

x,

peels’ X, x,

Let

=

Be

-

25

each

2t

mix

=—25

number

mix

A,

=

of

number

must

barrels

of

mix

D.

Then

t -

number

of

barrels

2s

be

of

5 =

barrels

of

nonnegative,

mix

5


(-2)R,

hours

(A)

4

10.

10 mix

Also,

‘0 -[2

i (-2)®2

B.

CHAPTER

by

2

Company

6th

20x,

= of

Since

30x,

equation

4

an

the

of

barrels

integer.

system

of

40

E

Thus,

C.

t < 10.

Let

R, OR,

144

2X,

-5

and

Divide linear

71.

+

t = of

of

69.

x,

barrels

Sekai SY

See

Pies ch os

for

i200

ty Xn oe 1000 ope

EQUATIONS;

1300

MATRICES

2

4 deck

ah | alls) +R,

‘Company’

° ia

0

gauss

Si

hourswior

DR,

A ‘and!

(B)

The x,

system

of

equations is: I 1500

st

X, + X, X,

+ X3

ete

=

200

=

1000

=

1300

Ley! e (Opt) wig eal) SAYS OK6) Mra Maro Wazou One ls HOTT LOO0 Om One ets lan L300 (-1)R,

+R,

aeRO

Ore

_{0

1 OS Oo)

Let

Since

7>R,

tt

@r

=

t.

X1

aoorsR,

ales

oe

173

"as }62)

sen.

Se0%

40

a

Ba

29

96.

69 | 89}

a2

865

508

26

ao \S 7 20ene eRe Te | 20.27 “ey 270 Seek we

Thus, Vee

the decoded message is Seen a oie), e238 39 20

which

corresponds

GONE

of

‘ ih

¢ |Pi timid Soak 6’ 823

59.

inverse

ax . 2 | 1 grees: | -1

+R, al

Now

find

WITH

THE

(8°27 ime

Oeec

moma

225.

49

S14)

74

to

WIND

"THE BEST YEARS OF OUR LIVES" corresponds to the sequence Bom £5 27 21-5 19° 9°20 27 25) 5 ae 1s) aoe S Peewee milan 9228 5 19

Gehaks

15

We divide the numbers in the sequence into groups of 5 and use these groups as the columns of a matrix with 5 rows, adding 3 blanks at the end to make the columns come out even. Then we multiply this matrix on the left by the given matrix B.

1

0

il

0

1

1

0

3

Beolg

i

1

5

0 2

A

at

0

0

At

0

2

1

1

1

2

a

AU

50

ORGS epee

The

20

i

208 omsenihy

Ze

505

863

Sy

59

Biome

LOS

27]

WAKOGy Rake)

Od

PS

KEN

70

WA

(9S)

56

ag

123

SS)

SYS,

Si

5S)

eld

ls

5

a

6

82%

G19

28V

274

Tam

27

41S! © ME 12).

LOO.

9

encoded Zio 82 OOO. 99=

5

Dic

AY

Zen

81 2

message

"9 352

is: (S950 wield. AO, 105s alO6e .Sén 97,

g/OS 123 e599 127

50) eehOOM RE Sare/ 27 laps ley hey

S981

EXERCISE

63

4-5

96

163

61.

First,

we

must

find

-2

-1

2

Bei

Now

2r-2.

-4

al

2-44

LP2>e109P

66)

"SS

98.

124)

62

87°

67°

Ie

reek

43

55

41

81

94.

69.

112

HLO9P

896

143

decoded 5) eS)

27) lve

message

7 Saw

corresponds GREATEST

B:

(18: 20

gomets ee

to

Be a

mares 7

15

820"

“23

19

es| 18

5

Bapht

127

8

S27 rt 018.

ae 27

(20°77

is

S.-i) M8

20295 e ome

97.19

«8

15.

935

to

SHOW

ON

EARTH

BASIC

remember: PROPERTIES

Assuming

OF

MATRICES

matrices

all products and sums are A, B, C, I, and O, then

ADDITION

PROPERTIES

ASSOCIATIVE:

(A

COMMUTATIVE:

+

B)

+

defined

C=A+

(B+

ADDITIVE

IDENTITY:

Ag+

OR=30R+

ADDITIVE

INVERSE:

A

(-A)

+

sAne.

=

=I1A=A

MULTIPLICATIVE

If

A

INVERSE:

A(BC)

=

is

exists,

C)

+Az=O

(AB)C

a

square

matrix

then aA!

PROPERTIES

LEFT DISTRIBUTIVE: RIGHT DISTRIBUTIVE:

4

SYSTEMS

indicated

PROPERTIES

AI

CHAPTER

the

A

(-A)

ASSOCIATIVE PROPERTY: MULTIPLICATIVE IDENTITY:

COMBINED

for

A+B=B+H+A

MULTIPLICATION

164

8 5

119,

4-6

Things 1.

327.24

the

VATA

THE

inverse

oa

6

20048"

which

ee

3

B+

Thus,

the

OF

LINEAR

A(B + C) (B+ C)A

EQUATIONS;

=

AB BA

MATRICES

+ AC + CA

ana

a7l

= a-lg = fr.

hogan

EQUALITY ADDITION:

bE

Af=—hBethenyAg+

LEFT

Die

Ag=i

MULTIPLICATION:

RIGHT MULTIPLICATION: 2.

USING

INVERSE

C=

5 anc hienmeAm—

B

47

C-

NCE.

If A = B, then AC = BC.

METHODS

TO

SOLVE

SYSTEMS

OF

EQUATIONS

If the number of equations in a system equals the number of variables and the coefficient matrix has an inverse, then the system will always have a unique solution that can be found by using the inverse of the coefficient matrix to solve the corresponding matrix equation.

Matrix

Equation

Solution

AX = B

ab

5

2

| =| 5]

-14[*,

-4

3.

|

Re + | 7

Thus,

X=aA lB

3x, + x, = 5 2x, - X, = -4

=55

Bae

os

2

0

MO

rei

B

=-2

3

85. | ep

-3x,

i

aX

+

3X,

226 -X.

216 3X,

~

2x,

1

4x,

=

ive

ae

pee 2 =

3

ee

=

sk

-

=

2

2x,

il 5

fe : a 2x,

2

x,

2X, ees) -X, + 3X, - 2x, Thus,

=4

a

XX

Ky

=

3X5)

-2x,

+

3x.

~2x, at

2) 2X,

; =)

and

2

14

=

LX,

2) 5

=a73 =1

tien,

xX, -

=

it 4x,

=

-2

3X, + 2x,

+ 3x, xX

3 =

4x,

Ji =2

ae and

des

=2

3

1

1

x, m= 4

x,

3 il -2

EXERCISE

4-6

165

ee te [3 aes : Ree : ta 3 i ris

1

ee

a

&

te

4

1(-2) + 4-1

Thus, x, = -8

eeeand,

x, = 2

adie : nee + ial ; °| Thus, x, = 0

2. saute

2°3

"(=1)2

Suegeand

3, = 4

s.[2 3] ]: . ibe 7 = iF 2,has BD)

i -1 |1 1 Mey oy (SLVR

Xx. an

inverse,

then

] = a1/5]. X, “i

°]: i -1 | i a - ig = oad Bae 4 i OF21 4-274 Gea |i 1

Se Rs

(-1)R,

0

>

R,

1 xea "7= 37 XS"

ee

R, + Rk, >

se

R,

ieee |

=2:

is. [} ace (8)

tad Opeewl:

17.

MMO Wjw

Seared

x,

The

equation

matrix

Ak

3

WooayP op

|x,

for

the

given

system

is:

k, -1

From

Exercise

Thus,

bk 7=

2S

166

1

Therefore,

CHAPTER

4

4-5,

Problem x

:

“1

SYSTEMS

21,

[

|

=

k | ]

14Lk,

OF

LINEAR

EQUATIONS;

MATRICES

:

=|

-2

mes, 1

xX.

eee a ee

=i]

atkSde

and

thUS7

2)

eco

Pa

ae

ore 2

[2PeoieIDis -| [3]3) ~“eeeee e-

cy The

2

| ee)

|

xX.

19.

JUS

a4)

%

Thus,

matrix

equation

rf 5) 2

7

for

the

given

system

is:

& ]

ILX,

k, -1

From

Exercise

4-5,

Problem

23,

2

|

=

me, [2]=[.2 2][8 (a)

xy

E

7

S (By |2| = x,

21.

The

d

ae i

25)

7

all *]: el

0

zy

-2>

matrix

xy

for

0 || m2

ky

0

all

Xx, | =

k,

4

x,

k,

=)

xX

and

Thus,

the

given

=

-2

xX, =

24

Geaee

a

system

is:

4 te Sr From

Exercise

4-5,

147

A THUS oyot ae =7 |) SAC

es)... 2,,-1

Thus,

Shoe

=2

ali-1

equation

a.

fa

ely |eal

5

X,

| a

idk =o)

X,

Ms

Al

Problem

eOOtae

25, | 0

1

i

2

=i

4

=)

25

eg24

3

2

-4

1

2

Sy

yea!

Thus,

My

bean

®t]

Ay

X, | =|

-2

aL:

k,

x,

2

oi (A)

-4

X>.) x

=

by

ab

Re)

219

+2

-4

2

k,

Beak

cM

1

Ome

bOe Tian

2

0

=

was ae

OF

x, =

0

EXERCISE

4-6

167

x (B)

a

=-5.

X5

=

x3

ve (C)

X5

=

The

5

+5,

9-12,

-2

5

get

2

See)

al

0

-7

dk

=

-2

=1

0

3

2

2

17

dl:

-2

al

equation

TP

-1

al

-4

2

matrix

3

-4

2

br 23.

-12

-2

for

vOnh

i X, | =

k,

2

x,

k,

=

5

1 the

"4

xy

cS

=)

U

xy

=

Ne,

}

Xp

=

—2F

Xo

=

3);

Exercise

X3

=

=

3

=f

-7 given

system

is:

4s

one? From

3

4-57"

Problem

277;

|>

ners)

i

0

62-b2em =[

2

ba

(-1)R,

+

1

AX

=

0 RH

3}

i

-8

1

0 2S

0 Bits

=o + R,

=

€C

elses,

i

tlhe | Gra l

(8,

ib

0

-

R,

ee

2R,

5)

has

infinitely

an

inverse.

real

number.

Solve

1

BY

hk

1

0

0

0

+ Rk; >

R,

An =

x

=

a

St.

x,

=

t,

t any

B33.) Ax Bx. = C

(A - B)X=C xa

al

ts:

(n1)R,

Ome=13

BX

3 i

system have

R, —> R,

0 0 Solutions: 31.

2

tee

ees

ae Dee | BO

the not

(A +

I)X

=

C,

B)**c

where

I is

order

n

the

identity

matrix

of

xK'e (A +t) we 35.

Ake

iE =

1D

—-

BX

AX + BX (A + B)X

No Lo)+ uu

x =_(al4.B)-1(c + D) Bilis

The

matrix

equation

fe ea ili

First

2

we

for

the

given

system

is:

a ]

X,

compute

k,

the

inverse

of 2

a

er

2

hi Py OOIIO 1

MAL ‘

"| -1)R,

greg

ste

1

2 001

i

50

Ge

0-001.

17-1

a

u

+

R,

a

DR

2

(-1000)Ro > 1R,

EXERCISE

4-6

169

by

1!

23007

Ol

ged

k

in 1000

0 | -2000

4 Pes

=

2

eae «

SS:

Es

Bras

(ay

39.

The

Nee aes

xX.

&

matrix

Fe sit 6°16.) Sons

=

2000

2

1000

-1000

2000

ere

1000

-1000/11

2000

ee

1000

-1000/Lo

2000 1000

apee 2 p -1000]L1

equation

rH | Waco 6 il) ated pe re

Pts,

&

-1

We

veeX.

2s ve

(-2.001)R,

ia

for

=

t

Ge 3 ae O76

equation

the

4

X,

1000

a

k| ||

-1000)LX,

0

pees dd?

eto

2008 F202, 0611, sx, = Lago00 -1000

system

135

ae XG and

is:

for

the

0.4

. 6-3 0.4

a

18:42

Xo

Pps}

given

2

145

xX

—Oe08m

HSSterimeoeod, 75 sOmioe

x, 3 = -15.2

250 ies

SYSTEMS

=

given

Bala SL 22.9

elle

CHAPTER

& 2000 |“ |

5%

1000

-1

Sage

system

is:

125

AD fF wo FP wu” ~ @omown7a)

170

4

0

=)'Sir2

matrix

“

135 ss 75

i

=

The

>R

i

: 2 ee

18.2

41.

+ R,

1-1000

+o

ce

-1000

oe

0 | Mi | 410607

er

0

OF

LINEAR

EQUATIONS;

MATRICES

—0.44) 1] 135

-0..68 | 0:84

[1155 75

3 eee

Gee 6. 4

bert) 7

Soh 1 250 3 195 1s

rst Pp

Cg ae ieeae Bee) AVERLT, BE 2

-0.25 shoo

145 125

.0.37) -0.4

0 0.5

0.28. 0.4

-0721 O12

|[i250 4.195

=0,16 +0.04 Ol Sib seat

. 0.28 Los

tei4s 125:

4

24

ore

24

big

=

5

and

a)

xX, =

15 43.

Let

The

-2

x,

= 15

x,

=

Number

of

$4

tickets

sold

x

=

Number

of

$8

tickets

sold

mathematical model wet x, = 10,000

aX, + The

as}

3

8x,

=.

,

corresponding

is:

k&k =

matrix

| alse

Compute

the

inverse

F ait MAG

tae) R, +

R, oad R,

aa

Concert

*] 7

2

ae 1:

the

iL ao

3

60,000,

coefficient

[1 Lad

7

Return

68,000

is:

matrix

4 é ad | OO oe

R,

(-1)R,

a eae

+

A.

_

R, >

R,

_t

ne

xi

Thus,

A? =

|

-

a

$56,000

2-3 eae

epee

equation

m > 1 crt ecko a

Mipotls2 0).

of

56,000,

: aa

-1 9 0FJL56,000)"* 14, 000

Thus,

6,000

Concert

2:

$4

tickets

Return

and

4,000

$8

tickets

must

be

sold.

$8

tickets

must

be

sold.

$8

tickets

must

be

sold.

$60,000

*] 4 2-7 Teo ee " bee Pet 2 tl60,o00) 15,000 Thus,

5,000

Concert

3:

x, | _ x,

a. -1

Thus,

3,000

$4

tickets

Return

2% fae ge 7 JL68,000 1

$4

and

5,000

$68,000

tickets

y eee 7,000 and

7,000

EXERCISE

4-6

171

(B)

$9,000

Return?

x,| 2

2s

x,

at

=,"

This

is

$3,000

a

*] x,

possible;

eS

1

te

is

Fix

9" 000

=7 7504

x, Cannot

not

negative.

k.

=

Thus,

gy

Since

x,

(L=anaes

x,

cannot

a:

20,000

22 (0),

negative.

k4

-

ae t

and

20,000

20,000 -10,000 +

=

-

-10,000

-10,000

k4°

+

-20,000

ks 22 (0)

|

tad taal od

AE 0 ae

X,

be

Then

alee] ohh k

is) =

|pate : ee

a lns000}

possible;

return

Since

be

Return?

x,

This

"aril

not

x,| 2

(C)

nina ||eeren E anes

80,000

+ Ae: 0

k => 42

10,000

k2

40,000

Thus, 40,000 < k § 80,000; any number between $40,000 and $80,000 is a possible return. 45.

Let

Ay

=

number

of

hours

at

Plant

A

and

x

=

number

of

hours

at

Plant

B

Then,

The

the

model

frames)

+

8x,

=

k,

(number

of

car

5X,

+

8x,

=

k,

(number

of

truck

corresponding

matrix

equation

frames)

is:

ds % ie x,

k,

compute

the

84,

5 First

we

ie

8 |1

i; E

474

eRe

a

Ba

SV

is:

10x,

te

ae 10%

172

mathematical

(and

CHAPTER

om.

bo

71

4

SYSTEMS

inverse

OF

of lig

|

|i E On

mk

(-5)R,

+R,

LINEAR

EQUATIONS;

OR,

4 | i5

eee

a

ah ASa Z4%,

MATRICES

>R,

|

including)

10

Thus

for

For xy

For

order

|

2

ss.

x,

order

Let

Then, Rigo oie ds

3

=

X,

>

“

-5

k,

4@1L%,

Ea

1600

ale

25

x, =

280

hours

at

Plant

25

hours

at

Plant

B

Plant

A

=

1

-5

heen

peo

Bit

x,

-=SLGOmMHOUGS

at

xy

=

80

hours

at

Plant

A

and

te xX, =

225

hours

at

Plant

B

3: =

=

-5

ay

and

2: 5

xy

:

1:

2

1

a

ie

:

-= bites 7

a order

z

=

5.18

Now,

x)

oor

ee

x = Se 55 x, =

-
R, (-0.02)R, + R, 9 R, Ee 0.5 | 5 4 § ep he a °| Geers Mi Or an Oa & -2 20 20R, >

i i pee (A)

R,

(-0.5)R,

or 6 ie ON eed” | FV

Diet

1:

fie

Protein

-

60

Diet

2;

ounces

6

X,

oz,

20

Diet

3:

A,

R,

6 iad =a) 20

Fat

10)oz,

Z0

-

80

6 oz

ounces

Fate

4

ounces

of

Protein

-

|= fsa Thus,

mix

—

>

of

mix

B.

of

mix

B.

of

mix

B.

4) oz

22 |aa 4 (23

-2

Thus,

of

Protein

aig May

0 ounces

60 mix

10

of

A,

60

ounces

oz,

Fat

-

A,

100

ounces

faneediae

mix

6 oz

e]-(5 sl)-[2). sm a Protein

-

20

x.

x)

CHAPTER

oz,

Fat

-

is not

4

-

14

oz

?

_—

-2

This

174

hae

lee ee Thus,

(B)

20

+R,

A:

20J1L14

possible;

SYSTEMS

OF

LINEAR

240

x, and

x, must

EQUATIONS;

both

MATRICES

be

nonnegative.

Protein

-

20

1 -| 6 Ba l-2) x

This

EXERCISE

oz,

-

10z

?

tale ‘ asap se f20sl ad . L-20

=

is not

Fat

possible;

x, and

ae ee

x, must

BS

both

be

nonnegative.

4-7

Things 1.

to remember:

Given

two

industries

Cy

C, and

with

Cy

Ci\a 1

a

x.

Yi Uo ete, perp

where

as; is

worth

of

the

output

equation

d

=)

Technology Matrix

matrix

Cy,

dy

Output Matrix

input for

Final Demand Matrix

required

C;.

The

X = MX

from

C, to

solution

to

produce

the

a dollar's

input-output

+ Dis

x = (I - mb, where

1.

I is

the

identity

40¢ from A and output for A.

20¢

from

s.r-m=(? Real Got) 10.2) Converting

the

decimals

matrix,

E are

required

all “oa” to

fractions

SN

=

pa 5 Ry :

mets

:

+R,

=

(I

=

M)~*D,

2

0.4

0

016

-0.2

“|[8|

~H2

inverse.

a dollar's

calculate

the

worth

inverse,

5

re

Thus,

wlH wir

me oO

|

re sy ps

i3

4202‘ |ana r-n

0.9

x.

at

a

we

I

652,7,

OR,

thus, r-m=|

ae

to

al

7 Siok

produce

an

| : =3

0 >R,

to

I - M has

of

Bes ied |Loove One :

23,

assuming

Ea

9.2

and)

5

/

2

6

Sue

he w|e

Ro tk,

have:

OR,

£.8

*-Or4

i

Q4ur se

oc,

=

16.4,

Me Seok

ae

EXERCISE

4-7

175

7.

20¢

from

A,

Gollar's

1, 92.1.0 O°)

10¢

from

worth

DAioebvg “a Worl Oem

11..X =

of

B,

and

output

ousiel tacl-o0.20

10¢

for

one ; “Borer

from

E are

“O42 O.ic] Se (Ont

26s )}=|' 022" 0238.

90k4 1518s 0.22.

sn

oe

agriculture, ba1 lion.

$18

Ome 5 50 TSANG HIe( aeeoUias +

6

billion;

ise eee fs °|“ be? Crash) Ouaig) converting

the

decimals

i

‘ 1

0 4%,

4

18

On

Sheet

bot?

Oy)

4

|e

8

:

ee 7

®1 + Fo >

Thus

ee 1

ONS 1) Ole

OV

0.7 }-0el F°=0 53 =0'2) |a 206, 20k -Qx1 ,-0.47 Ong 10R,

> R,

10R,

> R,

$15.6

4

SYSTEMS

LINEAR

7

|

0

1

Ppurs Ble

‘

10oO

5

i

O 5 a8

85

Fy

Ro

BiH @lw

KS) I

P/O Uw

aak,

> R,

aa

146

Saeoll2s Pai Ome TH0sae

Se" 0 Ga Ousycd)

|S

46

0.8 2 Or 2) =.| hot Ww ee =

On 7 =O =0.2 059° —O81=0-1

S13 9, 22 Py to R, OR,

OF

energy,

1

| : 7

-10R,; > R,

CHAPTER

and

0 pile -||241

OL ies On) Or*

billion;

|

2

06>

= [5 O Ol) 1,

building,

tl | 16 = ig O76)

22>M=|'0')

5a 22.4

fractions.

mn AO — 70 +

R,

ie

to

3

7

(1.6)5 + (0.4)10 + “Ome 0222)5at) (1018)10 tnlG ey ie (0.38)5 + (0.22)10 + (ieee

ay -| 0.8 re] woe Oe Sten 07

2s 10

awow

-|

176

a

Omees0.2 =—0.2 insomee O.9°~ 0. L -0)52 SS)

2751.4 ve eee 1.9 + 2.2 | sPis.2

15.

to produce

M)~*D,

ey Therefore 7| 5 x;

Thus, $22.43

required

B.

EQUATIONS;

MATRICES

"206 -=00p .9

10 0 FO O1£10 £ 0 08 se Ove

+R,

>R,

©O _——— N\o

af |

ie

I ae

Posy (-7)R,

=

1 0

On

="

oO

al

1

-9

0

OF

Mahe

— 216

0

10

-20

oot dh @OPM

0

76

Cys ee il

>R, +R,

0 -10

R, > R,

AR,

=1 362

-8

OO

Veo

60

0

0

0

O92

10

8R, +R,

SAM

QO -10 e1:0

srt) PA

0

0

-2

OLeP

2R, +R,

ik

ree)

-10

ab

=LE820

0

Q

=7 518 -1.82

0

0)

45744

10

S200

70

0

>R,

45.44

a

oO”

G7.L8

0

-0.91

[he8)

ie

82

0

OS9L.

0

0

al

O22

"07.16

-8.18 1

82.

aa

WM AY

Oo

=0".92 O91

~ =8

08

eZ

je. CaS

eae

%3 73

i

0

0

1S

OBA

OAS

a0

al

0

One

doe

0.4

0

0

1

O22 a Oe Gn

ele?

1.82R, +R, > R, 7.18R,

+R,

>R,

ess!

Pe

02.247

10258

OO, PesAOZA62

1222

(2 = MO) =| O8e)

alan

DPSS}

ONAL

Beak = (2 - Mo *D=|0.4. OF22)

=

17.

(A)

The

40

D=

bp

.

0.4

0.16.

1.28

20

5 10

+

(OF: ZaySS

(Ons) 2

Osa

(e125) Sie elas) O

(GR22)210

+

HOES)

matrix

0.8,

M =

Smee

gas

:

input-output

COs Saeco

0825

ALP

:

The

woOhe sys:

1.2

(1758) 20

;

technology

love

Glee? 2) EO

and

the

‘

matrix

38.6 Seif)

equation

,

is

Li] 74 :

final

X

=

demand

MX

+

i

matrix

Dor

x

X=

he

Sale + halt where

The

solution

0.4

oe

0025

E 0

oe

ee

X =

40

is

X =

(I -

M) WD,

°| e-3 Pel ut

0.7 -0.1

0.1"5

provided

OG:

0925

-0.1

4-0.25),|

2

a Ah

ORS

0

fi

: ie -0.25 | 1 al

aah

53.

-7.5

O)h

I - M has

aa

an

inverse.

Now,

O75

ets:

Httate kK)

EXERCISE

4-7

177

a é

i

0.1

:

at (-0.7)R,

=O. ass] It

+ R, >

0

Ry

< ff -7.5 | 0 ee (0.2)R, > R, 07

Ss

1

~ kK ide e. S E 6h

atte

gaat

(7.5)R, + R, > R,

Te : iv a fees [tab ave 2) s ee ovo"fig 0.2 1.4JL40 64 for each million;

sector is: Manufacturing:

$64

million

If the agricultural output is increased by $20 million and the manufacturing output remains at $64 million, then the final demand Dis given by

RET

ap

are

The final demand final demand for

19.

Pater!

1 0.2 tis ool a.s si 2 | 6.2 olace

Thus, the output Agriculture: $80 (B)

7

Let

0.7 -0.1

Pee 0.75dL 64

4 i 38

for agriculture increases to $54 million manufacturing decreases to $38 million.

x

i=

total

output

of

energy

Hy

ie total

output

of

mining

demand

matrix

and

the

0.4x.

Then

the

final

equation

D =

C we and

is:

i

the

input-output

matrix

p2

* vs Gee fall | Hew X, This

0.4%

0.6x,

0.4x,

=

of

equations

0.3x,

total ouput of the the mining sector.

element

to

produce

be

a number

CHAPTER

3 4%,

x, =

Thus, the output of

178

+

0.4x, system

is equivalent to 0.4x, - 0.355) a0

or

Each

aexs,

the dependent 0.2x, + 0.3x,

which

74a

OFS

yields 0.6x, =

4

of $1

a

technology

dollar's

between

SYSTEMS

OF

worth

0 and

LINEAR

1,

energy

sector

matrix

represents

of

output

for

inclusive.

EQUATIONS;

MATRICES

should

C;.

the

be

75%

input

Hence,

of

the

needed

each

total from

element

Cc;

must

23.

The

technology

matrix

The

input-output

Ke

ie

The

solution

=|

matrix

OF

2 AORZ

Oz22

OA

equation

and

the

final

is X = MX

demand

matrix

D lia -

+ Dor

x. ih

gd,

ae.

+ ee where

so%

10

is X =

ee 1 3 =) a 9

~

Ge)

}

1.

provided

0.4

i5, 6

+R,

&

0

1

me

+S

Thus,

(ah =

ear

Be

=

;

2

oes:

10

10

2g

>,

2

j=

:

2 2

0

:

a

Ro +R,

OR,

and

1n8

0.4

we

20

1.

Sip

26]

the

output

for

technology

matrix

M =

D = ial The

mj

60

xX +

e.

each

sector

ei

0.30

0715

input-output

where

X

is:

coal,

$28

billion;

steel,

$26

i

0.20

matrix

eit

=

w

10

provided

--

Now,

Spalted o 22]

Therefore, billion.

1 5 x=!/, 20

Wet

inverse.

= Sao toad ay 0 a¢.4 Cinehiy wer le

Ee 5k,

m)~+

an

aes

2

>R,

has

a

|sal ea 5 bar Gard) 6

Bil:

- M)

0.6

2

0

Matrix

(I

ied chy

~0.2

2

1

The

M)~1D,

a] ab #3 > iL 3 at ar hes l. 0.

R,7R,

ge (e

25.

(I -

anh

x,

be i E he ed -| 0.9 eae

10 9

X =

+,

The

=|,

uo

a

=

| and

the

final

demand

Je ean) oO

equation

solution

is X = MX

is ‘X=

+ Dor

(I - m)~*p,

es

(I - M) 1 0

has an i

inverse.

Now nt,

Ope

a

* 2

°|‘4 LE 20

|a

3

3

#20 7

io

0

PR aR AGsAR

met. 3

at

0

10 5

ee - 30

: :

20

i

RR OR, R,+R,

ae 5

3)

8

to foe

|

°

ce

0

al

TOM ate

7

3

bei? Al

0

2

z fy

| of

Dns 5 Rp > yarRp

4

8

Ry R 07 eR Bea

EXERCISE

4-7

LHS.

Thu

lacked ®) babes

s (r'-m7

Therefore,

Ue Wiun|r co |

the

output

Agriculture—$148

for

each

million;

sector

The

technology

matrix

M=]0.2

10

148

80

146

is:

Tourism—$146 0.2

27.

60

0.2.

POR4)

ORs

0.1

0.1]

ORI

OR

million and

the

final

demand

matrix

Dae 20

The

input-output O25

x = be

MOR'4:

0.8."

ot

20

al

0

0

(ER

1

Of)

0

0

al

"0,0 W=022,

(I - mM)1D, O22" i

provided

I - M has

OR 4ar 023

0.8

-0.3 -0.1

—0 29> =0'.1

0.9

0).12.

0.32%

OPEL

it

0

0

8

er omleemes meas -2

inverse.

0.9

Ore

(0.25

an

-0.4

ORR2 a Oe

S051) 0h 1009) 1, 0!

10R,

+ Dor

es

Ore

is X =

-04.4%+=0%3

-0527 “0.2

is X = MX

10

Ode

solution

wa a

equation

LO.3

oe eS

OZ e Oras

The

matrix

-4

-3

10

oe -1

>t 9

0

0

Oo Sor Ores OF

70 10

£

> R,

& 2)Ry > Ry

10R, >R 10R,

> R,

[110 0 0 Alar eg ea 00, Sao eC ee

84

6-46

“et —2".

i eer 9 sO =5 cummed Gu 3 24°10; 190]

9 —3)

1)

S)

0

OF

ato

-2

=

R, OR,

9

0

SG 0

Of

0

(8) RK, +R, — Rp

2R iF By R3

1) 0 OR

9 = 5) 5)1 32 10" b 10

0 OF

9 ok 1Fo

lO)

9QR,

CHAPTER

+R,

4

+R,

Oa aa)

sOmees 0

2 BRS

R

1 al 0 820.7 Te (-32)R,

180

R, =>

Onion9

taoaeee

aaeo 40 Oe).

So

iy

don

On

1t

t.8.

-5,)

0

m7) all

10). 0

40 a

0 ak

R, &

geeeate

Line

=4 eh FOR

ep Ose.

0! Om

1 Zama

eller 0

Jey Ose

pl 95

7R,

0 O

237,

OR,

SYSTEMS

OF LINEAR

EQUATIONS;

MATRICES

R,

Gesi

> 95

Le) 1 Sas

Now,

mee) ain) On

X44

||

R,

(ft -/M)- =)

ag

IOs

Omer

Tee

i

te

0 See

=. 28

0.28.11

15

O68

2h21232

81

(iG) a0) +)

(0'. 78)al-

KOmA BO! +s

(1.. 32)

(O24) 10

(0432)05R

$40.1

technology

matrix

The

input-output

is

matrix

M =

Set 5

(Ome 220 (eZ

ert aS

equation

0naeHa

29.4 34.4

manufacturing,

910 Sivan

0.07 0225) Onda

SZ

40.1

(le28)i2Z0

billion;

O05 The

F0832. WO s

BORA r 032s

10

4.32...

+

Aly

S102

(0 162

Ona)

Therefore, agriculture, energy, $34.4 billion.

29.

Pel a OF W Oke OS

0.26 qr- and

Ora

Oeee Ola 7:8

(OP

1.32

Mie AT-M) D ="'9.4

{OMe

OD

Gea?

ies6geO.Ome O62

Sale

On

$29.4

billion;

and

aOR 23°" 07.09

0.12 0.15 0.19 JRO Siam OO Sammons “OLS Cr2:Same Ores6

is

XK =

MX

+ D

A

where

X =

7" and

D is

the

final

demand

matrix.

Thus,

X =

(I

- m)~?D,

M.

0,95

Phare

How,

tT - uM =|

0-07 -0.25 =O

—0 alvin

0.88 -0.08 2 0) 1.9

S022SeenOF109

-0.15 0297 0.28

Lev SameO eStpem

Onsde? 4.0%. 4.0

OPA Nessie

S95 Wn»

x =| 0-26

1.83,

Oe SiO O.Ah

0, Adie 0.48

1:

65

D =|

41] 18 Sail

ana

Agriculture:

$65

billion;

Manufacturing:

$88

(I

-

m)~1pD ~ | 83 qal 88

Energy:

D =|

48]

$83

billion;

Labor:

$71

billion;

83 99 $97 billion;

Labor:

$83

billion;

billion

32

Year 2:

|p

a0) 66 aussi

PAS Year

-0.19 9082 0.84

81

ana

(1 - mtv =|

Dat 33 Agriculture: $81 billion; Energy: Manufacturing: $99 billion

97

EXERCISE

4-7

181

55

nae

35 Agriculture: $117 billion; Energy: Manufacturing: $120 billion

120 $124

CHAPTER

y=)

and

-

(J=*P' =

4 RS REVIEW

eR

ne

127

(I - M)"!pD =

Year

PT

ES

I

EEE

ES

billion;

a

EE

tT

Labor:

ILE AEE

2

point

of

intersection adie!

a

2

+

4x

256 pe

zis

ae ‘aie

Gh

-4= 3

is the

fFThis

solution.

(2m 1

into,

(1)

(2)

equation

Substitute 2x

The

billion;

AES ELL NEE

2.

(1)

26

ee: Y.= 5x+

$106

Substitute

x = 4

(is)

anton

yro2-4-42=4

(aaa)

Solution: x = 4,'Y = @ataaa

0) (A)

-

is not

3

second row is not to the right [condition 1 in the first row. R, o

(B)

form;

ek 0

in reduced

the

left-most

1 in the

of the (d)]

left-most

R,

1

©

2

is not

‘9

3

3

element

in reduced in

row

form;

2 is

not

left-most

the 1.

[condition

nonzero (b)]

3%, gies) hi 0 0

Fi

3

Rw

y:

is not

1

1

3

second row is not the only non-zero its column. [condition (c)]

(SL)

5

182

Ome?

sta)

a-|5

Ba

ei

aus

(C)

AB

is

BA

is defined.

4

defined;

SYSTEMS

Rat Rie

form;

the

left-most

1 in the

element

in

(4-3)

Ry

4

|.e-| -10

A> 46-2" x" 5: MBs 3EK+2 ao, = Shy, ais = 2% b3, =/—1, not

in reduced

hap

(A) (B)

CHAPTER

form.

is in reduced

a

10 (D)

|2

ceecemme

the the

7

by» =a

number number

of of

columns rows of

of B.

A #

(4-2,

OF LINEAR

EQUATIONS;

MATRICES

4-4)

Ei -2 1

‘|. E -2 | ‘|i re

-3

Z

(-1) 8, +R, ~

is 0

‘ 1

calculate

as

|? 1 B=

-1

-2

the

(2)R, +R, x,

inverse

:

z

0

($1) RoR,

‘| Therefore, 2

or

A+

0 R,

a

=

of

Sr

the

3] ona [1]-[

Sree 2

ri

ifpe 2

Be +21:

ist

4

ye)&

coefficient

ig ala

1 +2

>R,

=

8 7; 2]

matrix

xX,

i

=

Sy

A:

2

Ae

4

(4-4)

(4-2,

S62

ella The matrices B and D cannot be added because their dimensions different.

of > af > af! 1

AC

is not

defined

Ofecris 1

gyall 200 40713" 200 eal 23-400

ie

1,400,

6

200)

Solution:

Bon

EAA

Bs,

4 5

1

7

are

| 10)

Z

Ss)

O44

1002 1

0

0

edit oeOle

SAN

cll

5

+

0

Of

DON

meri

4ii0) ih

2

ee

3%

wand

40 30

2

0

5

3

7

2

1

4

5

$0),

0

1 ay

4

3

1

4

we dl gO)

0

tac

OR

p0ebed

~

1

5

49 430

ras

4

O°

0

CHAPTER

4

0]

ty

20

R,

a

5

0

0

0

> ®

3

Olbe

ete

tM)"

ip

it

1

3

az

4

0

0."

70

iien

a ae

aes

45%;

7

ee)

5

0

a

0

0

13 2 #2

Se

O°

yO kee ew

Ory

J).

0

Rieke

OL

Ae

eee

(SES

él, 2R3

SYSTEMS

OF

+R

beer

OR,

LINEAR

EQUATIONS;

O eIX nin

3

_i

2R,+R,>R,,

190

0

Oe

27S

53

2

Pe

cart

Lano-so ll er

0

wy

0

tine

5

0

(-75 }R2+ Bs > Rs

~

9

4

4

a

12h

0

>

2 (-3

-5

Gl

1 4

2

R, OR, 0

=5

40510

- Ry

“ae

(4-3)

2,400

OE

0

Oo

-20

io

a

Won

tes

Ey

5

ou | 2 5

0°

eo

= 1,400 = 3,200 x, = 2,400

BR 200)

0.3

e) 7

1

aa

7;

0-2

xX,

O1/=|5

Ss

fo

Bee

x,

4

2

0

To.

2 4ee

WhO

1

34.M=

aren

(-9)R, +R, > R, :

MATRICES

Sle b

ie

5|~ uly o

13

7

b}

TT Thus

(I

-

yt

M

eee. eR

35.

0.4

ve | =|

One

LG

5

GlL*YR0Rs

5

i273 =10.2 Oi

fr = mp

1.3

I SL5,

10

Xe=

eC)

0.7 || 40 6. 023. 2ZOun 0.8) «=~ 4. AO

(A) The system has a unique solution. (B) The system either has no solutions

81 1.49 62

es

or

(4-7)

infinitely

many

solutions.

(4-6) 36.

37.

(A) The

has

a unique

solution.

(B) The

system

has

no

(C)

system

has

infinitely

The

The

third X

Each

38.

system

Let

step x =

Sete

=)

in

the

(A) C(x) R(x) (B)

step

—SMX

in (IT —

(B)

(A) M)X

is

number

6,480

solutions.

incorrect: not X(T >—

mM)

(4-6)

of

machines

produced.

+ 22.45x

sarc):

59.95x = 243,000 37.5x = 243,000 x = 6,480 If

many

is correct.

= 243,000 = 59.95x (x)s

solutions.

+ 22.45x

machines are produced, C = point (6,480, 388,476).

R =

$388,476;

break-even

(C) ASDrOfeoccurs) Yy

1f

x >16,/4807a

loss

occurssitt

x
R, (-0.02)R, +R, >R, 100R, > R, , 2 2 |) 7:00 °|4 B 0 | 500 cial 0 1] -200 100 0 1] -200 100 i}

°| ey

(-2)R, +R, OR, Thus,

the

xy

Hence,

a fe ‘ x, Now

192

CHAPTER

the

4

matrix

-200

solution

500 -200)

500

re

=200

“ 260)"

100

is:

hee bs |

=

-900

10

1000

+

=

250

tons

of

ore

A.

oS

100

tons

of

ore

B.

-

be 100

2aghl 2a 7 ie - ae r ee 2OCdLeS -460 + 500 40

solution

SYSTEMS

is

rallead

500

=

X,

the

Again

inverse

OF

is:

xX,

150

tons

of

ore

A.

5)

40

tons

of

ore

B.

LINEAR

EQUATIONS;

MATRICES

(4-6)

41.

Let

“Sige

number

of

A

trucks

trucks

xX,

=

number

of

model

B

xX,

=

number

of

model

C trucks

Then

X,

+

and

18, 000x,

+

or

xX, +

xX, +

9x,

The

model

+

X=

We

22,000x,

X,

+

11x,

+

augmented

+

es

15x,

30,000x,

=

793007000

12

=

matrix

5y(0)

corresponding

to

this

system

is is

:

1 |ofl

EL? Now

ie rhs |ie é e Saee

15

(250

HOR

0

6 | 42

0

iaR, > Ry

(=9)R, + R5 > 1R, The

a - 2

2

corresponding

system

of

qo.

0 ex . E

0 -2 Fe

ul

cl

1

eal

Now,

i. Sa Lg pee 5) + 3x, = 21 since

persis

-/ Of

For

5: 6: 7:

sts= ie i] t =

42.

X,,

(A)

Xj,

equations

t =U5066)

1 medel

A A A

3 model 5 model

x, are lor

600

6,000

1,400

Supplier

A

-(§ 1620.) total

SOOM

The

These

Zia.

Kwe= Oa 21 x,=¢

integers,

rs

we

9 SE must

have.

7).

6 model: 3 model 0 model

%Bstrucks, B trucks, B trucks,

sC"trucks C trucks C trucks

(4-3)

of materials for each alloy from also defined, but does not have an this problem.

300

ORS

0.70

att

6.50

6.70

0, 40)

820450

Supplier

5S model 6 model 7 model

B

$13,930]

costs

of

Alloy2

materials

from

Supplier

A is:

from

Supplier

B is:

from

the

DES ESAe Oe asics, 5010

total

Stipes

2k

53 Waa

$13,880 The

are

nonnegative

truck,’ trucks, trucks,

bie

Sb

is

luti solutzons

The elements of MN give the cost each supplier. The product NM is interpretation in the context of

at.

(C)

ana ath the

and

0

(1), FR, 7? Fy Pe

a

A Sa 50

costs

of

+4 515,930

values

7 Supplier

4

can

be

| 7,620 13-880) B will

materials 521, 460

obtained

matrix

product

dah 13,930

provide

the

materials

at

lower

cost.

CHAPTER

(4-4)

4 REVIEW

193

15

0. 20

£0: 25

0.05]

6835

2 4

The labor LSS Omoor

cost

for

one

model

B calculator

at

the

California

plant

(B) The elements

of MN give the total labor costs for each calculator each plant. The product NM is also defined, but does not have an interpretation in the context of this problem.

|

(C) MN

44.

Let

x

and

X,

all

0.5

Ui. 20

0325

0.20

invested

at

5%

at

10%.

Go

1

|

1 Od

Hence,

The

xy

matrix

|

0.1x,

1

0.05

Dice

we

+R,

=

$2000

Lagi

0

dl

CHAPTER

+R,

5%,

$6.35

S5!,20

given

above

‘i

il

it

5000

0

1

3000

si se

%2 7

X, =

: inverse

|

Model

A

Model

B

(4-4)

is:

(-1)R, +R,

*®,

$3000

at to

Fe

1 0

2000 i

3000

|

the

|

>R,

10%.

(253)

system

||

in

Problem

35

is:

5000

20

SYSTEMS

400

OF

il

of

|1

Oin'O OSA Ube 0. 05

of

i matrix

Oop)

0.05

aes

>R,

inverse

210)

4

et 0.050

1 0.05 0.

the

| EL I I 194

system

corresponding

the

0.05

mrs

oo 200

400

0

Si

the

|iy; | ak

the

Texas

SB (5S

5000

al

al

Thus,

at

equation

i

(-0.05)R,

T° 0)

>R,

compute

|

for

|q

40 0

Callus.

5000 400

2

matrix

x

1

Now

x

| 5000

(-0.05)R,

45.

+

augmented

0.05

4

invested

Por

1

10

amount

0.05x,

;

12

amount

Then,

The

15 OF 05 et 12 0 ; 4

at

E-00 05 ®2

1

> Ry

coefficient

LOVO0G.; =), DOOR

LINEAR

0

I

34 1 °|-[2 i 2 a4 Pca ZO LD, e (-1)R, +R, R, 1 0

matrix

is

| 8,000 I 3000

23; 000

EQUATIONS;

2000

MATRICES

aoe au and

-1 SO

20

Boe 1 Rk

ae |: 2 -1°°'20 oo)

is:

matrix

°| “1

A.

° 2 a PY 420

Rk,

-| :

equations return)

the

Ot

PR, >} R,

of

5000

2

=-1>

20

k = $200? x.if

..

2

Xx,

-1

This

is

k =

$600?

tut

X, is

|

20

not

-1000

x. >» Cannot

alee

=

x

=

600

be

Al

negative.

x, Aees2,000,

x, © s7fo00

7000

possible,

k.

007 x,0= $6,000) ix; 2 $11,000

200

x, Cannot

be

negative.

Then

2 Belle z pas ong : a =i | 20)tbek -5,000 + 20k

xX, =

Since

Ge :

possible,

a return

x] x, SO

not

-1

This

a 20

2

xX.

Fix

sf

10,000

x, 2 0,

we

-

20k,

have.

=

3

-5,000

10,000

-

+

20k2

20k

0

20k Ss! £0, 900 ks 500

Since

x, 2 0,

we

have

-5,000

+ 20k2

0

20k 2 5,000 k28250 The

possible

annual

yields

must

satisfy

250

< k < 500.

CHAPTER

(4-6)

4 REVIEW

195

47.

Let

x,

number

of

$8

tickets

X,

=

number

of

$12

tickets

x,

=

number

of

$20

tickets

$8

tickets

Since

the

number

of

must

equal

the

number

of

$20

tickets,

have yo =. Xq0 OF

Xe

Wirz ian

Also, since all seats are x, + xX, + x, = 25,000

Finally, the return is 8x, + 12x, + 20x, =

Thus,

the

system

x,

LY 1s 6

Soy

xy

+

X5

+

8x,

+

12x,

+

First,

we

(-8)R,

=

20x,

the

eo) 25)

000

or

al

8

inverse

of

the

06

20

cme

1

R,

Concert

Oe

sche ctevis

Hab

x,

+

2

Thus | % | =|-3 =

CHAPTER

25,000

eA

coefficient

R

matrix

%d kl) Se

DORE Si Dien

2 Ftarns7 at iim

100s

0 3.aaar

DR,

2

1

Thus,

the inverse

Lean

is|-3

Yoyo eis

+

=e

196

30

=

7

REE

3

As

1:

=I x,

0

xX,

(-2) +R, R, OR,

sae

0.:

ail

12

R,+R,

70-1.

fio

a

le 0-2} ft | 6.) ING. Sees tao i ricwese | eo |@) (-12)R, +R, R,

O27 SY Py, 2 igheieon Peer a)femeewa, Omen

Oca

required) .

OR,

1 23

Two

return

is:

=e

the

(Qi Steet, ta

~“i0° CM

R is

Uae

Ta tovtess [co we ge Ay Oot Di i seni OR,

+R,

(where

equations

x,

compute

(OF 20h Aromat (19. SOR (-1)R, +R,

1.)

of

R

sold

1

4

SYSTEMS

ee) xX.

=

257,000

Eb

S351

g

7

--$]|)

-3

OF

Or

0 25,000 | =]

2]L320,000

LINEAR

EQUATIONS;

Dy

Earns | Ie

1

1

5,000

15,000 5,000

MATRICES

alt

X5

0 =

25,000

we

and

=

5,000

$8

tickets

x, =

15,000

$12

tickets

x, =

5,000

$20

tickets

Concert

2:

x,

eae

x,

ae

Xo

+

O

x,

=

25,000

or

fo

he

A,

4

Swe aL

en X,

0 =

25,000

=

25,000

4

an

Bin

Thus | x, | =|

4

0

-3.._7..-¢.||.

25,000

x; and

13023

=

7,500

$8

tickets

=

10,000

$12

tickets

x,

=

7,500

$20

tickets

Xx,

°

x3

4

2-3

x,

a

=

25000

or

7

4

0

7 -3]]

25,000]

M3

=

10,000

$8

tickets

5,000

$12

tickets

x, =

10,000

$20

tickets

Problem 47, if it is not tickets and $12 tickets, then x, + x, + x, = 25,000

8x, + 12x,

+ 20x,

k

augmented

matrix

is:

a

ibe Xo

0

10,000

=|.

5,000 10,000

required to have an equal number the new mathematical model is:

(return

of

$8

requested)

i

k

a |ge

puei2 . 20

k

1

oe

1 (-1)R,

: E

1

QO)

4

k

(-8) Ry ce

:

-L

od

2. 20)

5

Om A,

(4-6)

From

ie aaa

ee al.

11L340,000

x, =

ey

7500

Sorry: 20 >

Thus |*|=]-3

The

|-10,000

3:

x,

x,

7,500

|=

330,000

x

x,

48.

7

x,

Concert

and

bere

1 | 25,000

m

+

= 50,000 R, >

25,000

12.1

ee

4 Ro =

R,

2

tears R,

| 3.)

|

| 2

5

Or

2008000

ood

|-¢ +

953

pe

= _ 50,000

R,

CHAPTER

4 REVIEW

197

Concert

1 QO

©

0 FT.

eke=5S8207000-

-2] 237

-5,000)| s0rEnG

= =

x, = 2 - 5,000 x, = 30,000 - 3t

,

t

ae Since

X11

X, 2 0,

Concernta2a,

1 eee Since

x,,

Fa Lo

3:

k=

The

x,,

satisfy ==

t must

technology

‘

we

lis _

-;

x

=

Ze

=

35,000

-

xX,

=

¢

tickets,

3

2)

gefou

0

li

of

integer

t $ 11,666.

(4-3)

ack

Pet

5

2

AR

Pe

(896 |

24°

*

+R,OR

Be 0.60

- mM

aiyeeo

1

AO

I

wees

odemebaer

oe

oe

oc (h 2h

a

7

huss

Dd:

for

the

output

xX =

4

and

SYSTEMS

ie i 2

output

agriculture

peg

4

x, =

CHAPTER

inverse

Gar

2

198


R,

Peel P

(A)

5

R,

X,

11.

4-0

Qyez

Grea

7

oro 1

0

0

il

form:

is:

a = 5

a

(minimum)

ete

0

i pivot column

EXERCISE

5-4

235

(C)

His

$s,

8

-

s,/1(@

Py

1b

oO

CY t0

S,

3

0

i:

OF)

1

Pils

Oi;

L0F

| te

ih

21. Fo) © 0.8 a)fot

5

8

Oy

s

0

1

O

{10

3

ees,5 a. |

se

(-1)R, + 2,OR, 30R, on,

X,

acu ~

eats

$5]

0

The

0

simplex

Faas

So

pt

60

15:

tableau

S,

2

S,

pps

pivot _,.

1g)

Oy

|e

bial MOR

80"

for

1

5 So

a. |sp

this

Enter X, Ss,

x,

al

i)

S,

S3

0

0

0

lame’

oye

Pag

3

SA riOl

aL

©) OMPOUMBIE

oi.

gm fost or

eee

Thus,

is:

|10

deea

|10

gk

[Note: The have been

CHAPTER

5

LINEAR

+ R, => R,,

AND

and

40R,

+

ha

m=

mmOn

eh

|= =) 3. faim inant)

eyo ul

és |a1

copes

INEQUALITIES

pivot elements circled. ]

= = 6 (minimum)

af

236

row

P = 150

5; »,0= 0) s, = 0; smemer

19) 6 0) O20} 1 |240 Peet

max

ee

(-1)R,

weLaaton

in the last

nonnegative.

ta?

a

al

the elements

are

P.

ere

ee

All

ati,

problem

Os

i

Paver >

S,

2 eels

soe

Po 13.

X,

—? R,

LINEAR

PROGRAMMING

=

R, => R,

-10

0

2. (-3). +

15.

%

SF

Sy

0

0

i

38

x,

il

0

0

2

|,

9

1

The

simplex ;

Sp

B20

a (-3

112240

>

+

R,

>

R3,

and

10R,

+

R, =>

R,

83

1g

le

0

ot

ail

0

2

ee:

1

0

°

oo

ter

1 to6t |) A es ey = Psy Ss OP ts = 0.

for

this

tableau

Optimal

problem

solution:

max

P =

260 0 at

is:

Enter

Exit —

OF

R, oad R,,

X,

Pees

0

x,

+s;

X,

-2

8,

Mlle,

Cosy

OVP

@

al

0

0

0

FL

8G

AT

So

CO | 5

0

0

0

Ee

eee

a

2

2

2

(minimum)

2-5

a

PLi-2

-3

0

+

Divot

t-a1PRoace-Re-—>

A

¥

ay

il

ils

tabi pivot

~

row

**

2

R.,

(1)

2

0

0

0

ee lee

eo

oats

3

fs ee and

3k)

3

1

44k

RFR

4

4

2= 3 A

DRS 0

pies

2

. =e)

“)

ML 3

0

re, 0

il:

a0= 2

(minimum)

6

7

i ea -2

il

i

1

ig ia

@)

0

-8

0

0

qi 8; 0

0

ey 2

Ss Cogito Bee

Seyamo 3}

2R, + R,+>R,, and

1 203

0

257

0/2

0

ilk

6

whew

0

Ss

Ss

Iz

0

0

Al

0

ee

3

6

0 OG Be Bp Se | tN 4

1

Ogee

g0e

ls

sein

2

0

QO

0

4

ilies

(2,

(-1)R, +R, OR,

-1

4 , pivo a

8R, +R,>R,

Since there are no positive elements dashed line), we conclude that there

s Ge

in is

the pivot column no solution.

(above

EXERCISE

the

5-4

237

17.

The

simplex

tableau Mae

this

Sin

1Sop

eee

0

none

2 = 2 (minimum)

a

toe

OR

hee

a 24

Side

0

See

(101s

Pe

pee

Xo,

pve ahs: |oe Soh

for

Ce

si"

3"

"0

S3

NEY cad

oOo

P

Te

870)

e2) T

eek 1

@)

OG

oa)

-3 Pein

0 sae

ep

row

gt

alt

4

+ Ro

0

0

0

Le

SOR

OMe

£62?

04

is:

Ry

4R, t F)—>

2

112 ne

pivo

3 < pivot

| [Note: We positive

coals

Pivot

a

®

0

“3

,itebOe

Mada”

RSsse eee AR outta | Oi

=1'

50:0)"

25)

Boe

io ->

2 oO

$

19.

The

9R,,

R, +R,

simplex

0

pe 8%a

ea

s

3R, +R,

2) 2 Us.

1

this S,

So

1 -2

1

0

ee

OO

pivot

gt

Be 0

ut 5%,

238

Ley

CHAPTER

3 >

5

Ry,

X33

242)"

fa

0

et -4

lo

( 2) Ry +

1

5

the above

in the

S3

P

Bh

R, >

PROM: we el 5

150

0

ee Oe (aei +12 ; aaa a)

oe

Optimal

solution:

xX, =

10,

5,

Clee

Ory

max

Ss, =

3

eae P = 7 at Xx, = 3,

0,

Ss, =

21.

is:

i

PO

R,,

SR,

+

R, >

R,

il

1 2

-1

1 2

0 1

0

3

-4

5

0

150

is

110

|PA Dea tie|er 0 Ro

AND

0

0410 |22 = 10 (minimum) OP Ores 15

0 | 10

INEQUALITIES

0

P

0

R,

LINEAR

S%

Nee

SB

problem

NCL 0

S,

CeO

for

Xp,

Prow 281]@

wv

line

column.]

X,

il

ambeeds

> R,

X,

i

or

tableau

P

ey

omer, ce 6 8 Oe

0

Syaie cite ee re eo cae

and

+R,

row

1

-1

R, +R,

2R,

2 Re -, Ro X,

A

and

only use elements

the dashed

A

column

&,,

2

0 1 0 ce ik” pas omen

wwlrose

si N

MUON

(-3)R,

-1

pivot

problem

LINEAR

+

Rk,

>

R,,

PROGRAMMING

4R,

+

R,

>

R,

.

3

1

1 3) 0

£

a

un 0

7

cr . =

=

19)

3

go

FP

Al

=p 2

1i-


R,

X,

x;

S;

S»

al

0

ay

1

0

0

0

1

A.

0

al

0

0

0

al

2

3

at a alee

simplex

ivot he

—)

~1

I

0

@

(3)

b

3

me -4

-3

R,

ts | 4

270

£1

50

0

2

2

Lo

R,

and

3

Optimal xX, =

this

problem

solution:

3,

x,

=

13 4

R,

+

S,

S>

S3

Ss,

3}

Py

5

ak

0

0

OM23

Sy

@)

att

aL

0

tk

0

0

-2

0

0

t pivot column

iail 2 ae i

Ry

&

Ce

>

DO

P

7

R,

0,

8

5

Lire

=

4

.

2 ches

“At

Msi

,

2

fae

a

og

0

1

a!

3

Oo

0

[aa

1

te

ho

va

2

O13

ds

0

47

0

Q) +

Qo -&

-2

0

0

0

al

0

QO

-1

0

>R,, and

*

(minimum)

2)

+R,

0.

:

8

(-1)R,

17 at x, = 4,

Ss, =

0

0

|

max P= =

ees 8

ab

123

s,

=

0 med oe7

0

i a4el]

0,

2

(-3)R, +R, >R,, 4R, + R, od R,

R;

is:

x;

RF.

1

>

X,

ete eet

+

:

eee

ak il 2 0 ft Sean aectieee teenth azL Ac SeeG Se

mraz)

45,

4

for

-3

(minimum)

OF a meanest

x,

-4

3

P

eee

tableau

eR

1;

R,

x,

See

The

y-1 Pp of

2 =

2R,

0

>

2

0

ah ||) aus

R,

EXERCISE

5-4

239

Re Ke Se 1 1 1 S35 92 CUCU) aon

ae L 0,0 FO, Oke ee

OAPs

OF

->:

| 08

Oe

S08) A Ona

De

Te

~5= JR, +R, 7, R,—R

R OR,

3+ Rk,

X,

0 Bewig pay

07 i] 0k

07 2° 0--15 1. 3,

P

OO

an

FE

83

SQ

S$,

X3;

X,

as

F

3

‘

2

1

15 -14 -2ps 0% 10 OS-le Pa

3

25a

1

OO

OUR eee sone

d

R, +R, > R, Optimal Sa

25.

=2

3

solution:

22

ge

at

iL, anus

6,

Bo

the

third by 10 problem is: PEC al

first

to

clear Ss

2

problem

1

LT Goi ke aia OF) 328 rece 00. (320305 09

the

constraint

fractions.

Ss

Ss

P

0

te” Or

eOn Near ooO On er cog Lomia aoe Tee t1 0

2

3

OCISee “Oe FOm

by a

Then,

the

the

i600...

i,

2. i080 =

1,350

5 +800 - 900

2

1

1

| z alt

Cs

oO

| LW)

©»

O'=1,

0

0

0

800

he

(a)

oO

ws jo)cS

"0.

"Lior

too

800

_ iT]

parl00

1,600

400 Ce

al

CHAPTER

5

LINEAR

INEQUALITIES

AND

second

simplex

au

240

01)

s, =

Si

ope

OF

08

Multiply

4 S, s ye B

P =

max

LINEAR

PROGRAMMING

by

100,

tableau

and the

for

this

xy

xX

Sy

8

se

aoe

0

2M

me

0

A)

Pee

Pe

So

P

10

600 Cl

400

0

oo &

Optimal

S3

300

6 © “1/26, 000

solution:

max

P =

26,000

at

Kina

e001

X=

600,

Ss =

0,

Be =

Sb

S3

1g

0,

s, = 300. 27.

The

simplex X,

tableau

X,

apa mee

2 3

ee

meme

2

for

this

X3;

S$,

Sy

Pe

Paso O Sd. 50

problem

Sz

is:

12

="0=| 600 -20r' |600

oe oe

75 300

Se Oh. G1 A080 | AO Rea 3

Cte

0

aE > R,

‘

oe, 2) & Mees 62 90 meee at oO meee 623 §0 (-2)R,

+ R, >

pape)

06 71 90 60

Ry,

-;

oye 75 =04] 600 208} 400 21 0

lr LPR

he

$ G) 0

DO o® sa 60

2

+ R, >

GO)

0)

0

0

ge

R;,

and

TEP }450

3R,

+ RnR,

306

|422 ~ ig9

ed

La

) Ce RS ae od br 2p, > R, X,

X,

X3;

S,

pee 0 1¢1)4 2 o~d 0 Stormo 7s

xf

=

0

1

+ @

>|

+

1

O-=

0-4

ime te 4-0 2 len 2 + Rk, 7 eau,

k, + R, >

7

0

0}180

OO 0

eG [szouy 011/225

R,, (-rae + R, > R,

|

s-%

0

2

0 0 {180

ees) we“Oro sese Feo 1 ELGG eon creer Oe R3,

Optimal ai ae

0 | 30

20 | io in| 450

solution: max P = 450 30; A ea se a

EXERCISE

5-4

241

29.

The

simplex x,

$,fa2.

tableau

X>

Sy

82

22h

for

S

this

S3

S, | totes

(a0.

glia

Sleliey.4:

GO

Od

problem

is:

P

ne

S,

$0) h 0a

a)

ao

Cum

O-

fous

48 =

OLE

GicE

60 4

16

ol WO ooteb@ |Sry 9 3°. onh oiiehae | algae OlneOe | One),

Pulte

(-2)R, + R, > R,, and

SR,

+

240.

>

(-3)R, + 2, OR,

fei.

OF)

-2

Teo

90),

4.0

balay

ORE23,-aeulanG

2

iH}

Oo’

@

0

0

0

ape-4

4

i

ney

O

Atul

Oe"

OO

(-1)R, + R,

Shoe

eon

3R

2

OR,

he

O01)

a

OO.

40t gD

Ay

0)

+

R, a

R,

aia

On ie oaakoo nS 20) eGo 2 ener Rs Acta togy

0 TAO (Ome R,

Cig

DR, (-1)R, + R, OR,

0920) 19 Oi) 300 Rad [i MLE

and

R3,

12

a ate

tay

+

R, =

R, => R,

di0

(-2)R,

+

oe

4R,

eo, lad

and

2R,

2 2

ra

14

4

ii

+ R, OR

5

322)

et

Ones

wedi Stan ade ge eae

Oo 1 0 -1

1

0

ea

OmiES aie

0/12 [4 = 22 1

R,

242

+

CHAPTER

R,

>

R,,

5

LINEAR

3R,

oo R; >

INEQUALITIES

R,

AND

(-1)R,

LINEAR

PROGRAMMING

5

LS

a OL ON

S,

0

0

1

S,

0

0

Ou

Petes x,

(A)

Veer

0

Solution

1

0

0

4

ad

0

1

0

6

aus vind)

0...

0.

-S0ulod

iL

0

0

0

the

first

aed

using

=2

ag

2

3

8

column

as

the

pivot

column

P

iS Ls al

MT

>

10 MV 206

EXERCISE

5-4

243

Of

Ggeley

1)

OF

Sos OU

6%

23

ayjota

68

oll

0 40 =ie Cyd 0.70 “LT Sia0.

olla

Opes oteiaia

$= 3

iaR, > R,

AGO) ;,

(nde!)

20 ey

FO

ee

eh oeae: tahoe 2R,

+

x5

R, =)

Ras

+

R, >

X,

S$,

Sp

$3

0

a

0

0

1

:

»

-

la

ae

se

ukQileesieleners

ae

So

(-1)R,

X,

i! ~

aOe ul

JA

P

Ord

0-0)

0

: Md

So

S3

S,

2.4)

1

1

BO

ee

Ome

s,

1°@

Oty

Oi

Pte

0am

Comune

(-1) R, +

tao Ry

>

G)i(0 ots MR pee wie a

CHAPTER

Os

5

Ry,

Poe Mele Ve

Okan

LINEAR

INEQUALITIES

solution:

at™x,

33° x,s=

er

i) a2

16

10

Daley fae = 40

Oe R,

R,

Optimal

P

32 ]a_2 0% @ 0 MO, te

244

fe

Sy

Meh

R, >

{10

X,

3)

R, +

P

x,

P

Rom

ae +

R, ae R,

Ga vom e voles || uc

6

t= 3 2 - 6

eee

AND

LINEAR

PROGRAMMING

=

oS

107

max

P = 13

sia

‘

1

0

=1

0

al

0

0

a,

Gg. 0

0

(-1)R,

3

0

0

6

1.

oO

Mane

if

i |)Au

+

R,

0 =>

R,,

R, +

X,

X,

S,

Sy

x,

1

0

=i

0

5,

Gago

“>

at :

ot

;

Ne

@

ewes

P

Oh

Choosing the same

The

0

ee

-1

33.

-i:

simplex

R, >

S83

i5

R,

P

0

3

=

oon

ee)

“

alae

Optimal

solution:

at

Sir f=

Rens as oocie sewecnet wee

aocii)

-

ii

yeae

a

Se

=

ays,

=

max P = 13 Oy

$,

=

0,

0

dope

either solution produces optimal solution.

tableau

for

this

X,

problem

is:

iz)

opus. (i) 2

(A)

Solution

using

the

first

column 20

vt

Mem

Bee

4

the

pivot

column

20

O81) Poll se | 22 ote

i2F,

7

1

1

2

al

0

OF

eee

65%

Gs

1s)

Pot ae

a

as

Ry

1

PAO)

EXERCISE

5-4

245

Ale

6

o

to3

2

eeeeewe

wee

ee

vo eee

0-2 4 2k, > R, al

0

Ges

ew ee

2

-1

got

Oe 2

hae

(-3)" * R 2 7? oR Ror oh Xy

Xx.

s

0

ab

0

2

x,

Siete

SEY

(B)

Solution

using

X,

Si]

X,

P

-1

0

LO

8

a eee

Optimal

8

the

second

S,

25)

cdpOe

c=

column

So

solution:

max

P = 60

at x, = 12, x, = 8, x, = 0,

Oneal

a).

X3;

1G)

+ sk ® ee 3 S>

SQ) eed

OO.

Pil.

AS/

8

3

x,

(ecm

ew ws

0

eS 3

ew ole

1448

3

Z

eS

oO ie

1

ew ewe

0

ee. ec

52

oS

eee

3

0

ae: -7

of

P

as

the

OF

pivot

oo

0

colum

Le

SON) 208)

eet 20

$2] 2.2.4 0 20 132 |22 = 32 P

|-3

seile2

(-1)R, xy

tookg

+ Rot xX,

x,

ods

te

Rey

3R,

Sy

So

PER

Re,

P

od Mate ei Saat eS MPSot jcgde (AE

=

Bye)

Pil

LO

9)

Od)

Duae ds

ree

ON

eon

Oe

at NcO

The maximum value of P is 60. at two corner points, (12, 8, line segment connecting these 35.

246

Optimal

solution:

ae

ae

See the number of A components rad9 ul the number of B components x, = the number of C components The mathematical model for this problem Maximize P = 7x, + 8x, + 10x Subject to 2X, + 3X, + 2x, < 1000 BS e or 2x,

3)

2

dt,

0

this

obtain

=

1000

=

1S010

P=

0

Sy

10x,

tableau

s, 2 to

ns

S; +

xX,

oie

s

problem

the

equivalent

form:

is:

P

OY

1000

1ee

500

$2} [email protected].__9 | 800. | 202. = ao0 2

iayar-se-f0*

[0

.90)»

1

0

2%, 7

Peet

29 irDiOO

Zz

1

mens

=

G0),

0°

eet

60280 © 70>

(-2)R,

+

R, _

; @

(ee

os

Ak

a

tO)

[1000

4

3.0 Oba Ry,

1400 t

10R,

= 0

m

26)

905

0 +

R, ~

R,

R200

|200

_

0

A400

_

400

PERUVE bocce sausccs Ce

heePA

Boeeeaeeg

t}-4000

TD

ee 800

ZL

2% 7

OOM, at

ee

E

LZ

i

aeara

1 * oO.

Bade1)\p

=

a

eae a 1

sOnges oO

De

1

-0

1.400

4S © 7214000

R © R oF? R. Bon 2937 Atty thy

Q

1

n0

it 1 bees ae 2

0:

[p-LOOy

Races ones enn e hahaa

7

0

eg

a

0

mete

et

0

0

oh

1

174300

2R,

>

1

=

350

PyR 100 =—— teas = 200

|320

1/4

-

1400

R,

EXERCISE

5-4

247

i

12

(-a)R.

+B

>

x,

X,

x3

S,

L

2

0

fille

x,

Re

F,

+

S>

R,

>

R,

P

al

0

200

the maximum profit is $4400 when Optimal solution: components and 300 C components are manufactured.

37.

Let

x, =

the

amount

invested

in

2

the

amount

invested

in mutual

and

=

the

amount

invested

The

mathematical Maximize

model

for

.08x,

+

P =

Subject

to:

xigt

X, 41

We

introduce 2S

slack

es

res

- 08x,

The

simplex

Ss,

S.|

X51

.15x,

+

xX, +

+

X,

-

-15x,

s, it and

0

+

Pa=

0

problem

is:

Ie

il

Al

1

il

0

0

1SQOP

Sy

100,000

|100,000

ein

Stone

0

Log)

0

et

the

equivalent

form:

0) 1010/0

=

So

sD) ja Fone

obtain

See +

this

s, 2 to

S;

S,

S130,

funds.

is:

2e0

xX,

-1

100,000

+ Gje-punimand fo(opm

©).

Ge

elaine

-1

f°

A464.

G

23) 1

2,02)"

L0um

10) ian eee

aR,

CHAPTER

X3

xX,

for

market

X, < x,

X,

(-1)R,

248

+

x,

tableau

bonds,

as < 100,000

variables

.13x,

+

0 B

funds,

problem

x,

P | +08

~

-

in money this

.13x,

25.

government

A components,

200

>

5

noon 00 1

fag

aay

0

= 23.02 Ra

INEQUALITIES

AND

LINEAR

-+

Oyo

Cae

R,

LINEAR

Oia

1.

Rak,

PROGRAMMING

0 ]50,000

8

-18,000

Ry,

24,000R,

0

0

+ R =

@ -2000

fo) =

elu ale

0

0

I

e|"°So i) tv oir LZ

ap oO

(2) [o)

0

1 | 240,000

« 22 Ra + RB, 5 'R,, . 2000R, + R,

oO oO

R,

EXERCISE

5-4

249

xX,

X,

30

fred

xf ett

Xx

+

al x o

P

0

S)

0

S»

ws

-1

(0

5

+ -o

on

s2

10

1000

10

4000

het and

(A)

x, X, x3

=

the the the

The

number number number

of of of

mathematical

Maximize

P =

:

Subject

1 | 260,000

number of potential customers is 260,000 when 5 prime-time ads, and bP 0 late-night ads

Optimal solution: maximum x, = 10 daytime ads, x = are placed. 41.

Ip

colonial houses, split-level houses, ranch-style houses.

model

20,000x,

for -

this

problem

18, 000x,

+

is:

24,000x,

pi 2x,

+

BS2X

+

60, 000x,

+

60, 000x,

+

80,000x,

4,000x,

+

3,000x,

+

4,000x,

to:

x,

S30 < 3,200,000 < 180,000

BE Lypee Deey OX. 3 > 0 We simplify the inequalities obtain the initial form:

es 2*1

+

Ay 2 X2

+

6x,

+

6x,

+

4x,

+

-20,000x,

[Note:

3X,

~

This

The

simplex

S; +

+

change

tableau

for

this

Xo

problem

X3

Sy

0

0} 320 |222

olan

40s)apn LS 0sup eee mame wee Aaa

0

oO

1

R3,

24,000R,

0

1

3

ABE= ow 0.

mn

4

ee

met)

he ee rae ele oem meen

1:-20;,000

5

0.

y218;0007%-24,,000°°0 R, ->

Ro,

(-4) R, +

il

uf

3

en

pe

2

ay

@)

ieee

R, —>

nn

30

=—@

He

10

0

80

Rie

dC

lg

40

60

0 ~ 0.

AND

LINEAR

..1.1720; 000

PROGRAMMING

30 a

vag

4o

0

..0

aus INEQUALITIES

P

100-00

-6000),00924,000.

LINEAR

of

0 |p 30

g

CHAPTER

interpretation

cy)

6

3

the

0

OCs

6

=8000 ut

80

S3

5,

(-8) R, +

=a

P=

So

$1

P

9320

variables

is:

$1

www ow

9°30

= S,

+

Ss,

S;

a= Sy

24,000x,

will

slack

variables. ]

xy

250

xX, +

introduce

4x,

-

simplification

slack

then

8x,

+

18, 000x,

and

+

R, —

R,

the

to

~

-8000 -6000 0 24,000 0 0 1 1720,000 ple & 2 Rs +iR, > Ry, (-2)R, + Ry > Ry, 8000R, 4 R, > R,

0

=; 1

2

©

aaa

15

0

ipso

1-4

+.

See

20

eee

iee...

2

el

+

0

30

0 -2000 0 8000 0 400011 960,000 (-4). +R, X,

STA xX,

X,

0

Plo are (B)

R,

3

GS TOP

Recta Optimal houses,

>

(-2)% + Bouse |Kee, S3

2000R,

+R,

Sy

S5

hla

-5

0

0

10

sal

0

20

i

OQ”

-4

a

jo.

0...0

-+

—?

R,

Je

1i06

20

0 0 0 2000 2000 11 1,000,000 solution: maximum profit is $1,000,000 when x, = 20 colonial » Mos 20 split-level houses, and bp 10 ranch-style houses

built.

The mathematical model for this problem is: Maximize P = 17,000x, + 18,000x, + 24,000x, Subject

to:

Bae + 60,000x,

+

FX, + 60, 000x,

x, < 30

+

80, 000x,

S 37200,000

4,000x, + 3,000x, + 4,000x,

Roe

-24,000

30]

*-0") 380

= 30 ane ay

0

(= 4) R

EXERCISE

5-4

251

:

=1

iL

=.

ig

i:

To

Swe

2

@)

0

- g.

21,

Boi,

(0

2

fi.

36

=4

90)

ga

40

30 [75 = 60 so | —80= = 4p

30

60 | 60 _

ee

-5000

-6000

0

24,000

a

0

0

1

1720,000

ais Rio ;

ie

1

1.

£20

Seo

oo

30

1 2

Cis ine

0

A =a)

oar oF

| 0 ar” o

40 60

0

0

-5000

-6000

(-i2 )R, + Rea X,

X,

ay

Oa

Omer

Xp

Ld

sas

built.

0

24,000

Ri,

Ua

X;

pO4n

40

Ree

Ze

1 | 720,000

eee,

3

oh

60, 000R, + Rive,

Sy

S5

S3

P

AL:

as

Om!

AO

10

2

NO!

yo

40

eye

de

0

20

eae

Tin

Optimal houses,

(C)

ee

fono

solution: maximum profit is $960,000 when x, es 40 split level houses and x, = 10 ranch

In

this

case,

Ss, = 20

(thousand)

labor

hours

= 0 colonial houses are

are

not

used.

the

simplex

The mathematical model for this problem is: Maximize P = 25,000x, + 18,000x, + 24,000x,

; Subject

af 2%,

+

4 2%

+

60, 000x,

a

60,000x,

+

80, 000x,

4,000x,

+

3,000x,

+

4,000x,

to:

Following

the

solutions

xX, = 30

in parts

(A)

and

S 13/7200, 000 < 180,000 (B),

we

obtain

tableau: x,

X,

x3

S)

S5

te

G4

1

1

=

2

=

:

i

poe

3

6

6

S780.

s

- 257000). ne

43

252

CHAPTER

5

LINEAR

S187 000mme 245000".

0"

AND

LINEAR

30

0/1320

30 eee

ep

te

328 = 53.33

preset keertate fie OC 0°

£0."

aR,

INEQUALITIES

P

HOH

1.40.

re| She ©)don-nereBenn toons acta Pel.

S3

PROGRAMMING

ff

0

ib

z

6 ©

:

al

Z

i

6 a

oy 1p

Paes Omeee

0

ei kire>

Ki,

-6R,

X3

+ Koeao

x,

X,

S,

Sp

71

20

5

s

A. | OOP

S|

0

$

o,

0

he,

0

ChE

| 370 Ole | 45

0

1] 0

|

-25,000 -18,000) -24,000. 0.0 (-3)

0

25,000R,

S3

+ R, >

R,

1

eG

Fo

30

+

7.5

0

50

er | ert {e-*- Gran” 9 + 0 45 p |circcirrtereciesrecceseesesceresteseeeposeeesscseees O50

1000

0

0

6250

1 Ud, i225 £009

Optimal solution: maximum profit is $1,125,000 when X= 45 colonial houses, xX, = 0 split level houses and xX, = 0 ranch houses are built. In this case, a 7.5 acres of land, and Sy = 50(10,000) = $500,000 of capital are not used.

43.

Let

x 1 Ul the

number

of

boxes

of

Assortment

I,

Kygh=

the

number

of

boxes

of

Assortment

II,

the

number

of

boxes

of

Assortment

III.

and

x,

(A)

The

=

profit

S2408—

The

per

box

[4K0.20)

profit

per

of +

box

Assortment

4(0.25))

of

+

I is:

L2(0530)a-

Assortment

II

=

S400

is:

eeo Om 1 2500S?O) Re =4 CO).25)) 4 4 0OR 30) = = ts3-00 The profit per box of Assortment III is: if OO—wbSiiOaZ0) +ae(On25) + S(O). 30) —=S 5200 The mathematical model for this problem is: Maximize P = Ax, + 3X, + 5x

Subject

to:

4x,

+

12x,

+

a

Ax,

+

Ax,

+

8x,

< 4000

12x,

+

4x,

+

8x,

< 5600

X11

We

introduce

slack

4x,

+

12x,

+

8x,

4x,

+

4x,

+

8x,

12x,

+

Ax,

+

8x,

~4x,

=

3X,

-

5X3

Xp,

X,

variables +

< 4800

2

0

to obtain

Sy) + +

S, +. P=

the

=

4800

=

4000

=

5600

initial

form:

0

EXERCISE

5-4

253

wetea | fs.

Sif

4

ae

S5|

4

A

Ss

Fi

goa

8.

83

10)

0

Oma

OL...)

uo

aoe = 600

SOMmano?

see - 500

3] 12489040

P

=6

23)

510

0)

5600 |$600 _ a9

OCmmre

0

abo

Oe

2

Sere

sects

ai) 120.

“64

(1

ep)

Ok

ce6

(Ol

"4

Wh

Met

=4)

ete

=34455

MO * GoM

W0) , Oke 0

Cee

2040.0)

(-8)R,

+ R; >

R,,

SR,

+ Roop.

Ay

OO." ioe

x

mus

Gyo 3

=o)

a

eo)

1)

a0

* 0%

oO

ae

Ee

£0

RO.

442

© &. O04

-pF8] 5

yg

Se

eCo0 ee,

3

0

500

10

193200

ie

0 f)

OLiS 1 0 0 +

SA" 0 0

ChOn igbwreaeyO (400 0 -— R,

aa A BA

40)

5-5.

(aka

ay

eer

eceeeewwrereereere

ener

and

(A)

in parts

solutions

is:

< 6000 < 6000 < 5600

S,

=4.°

*:

8x, 8x, 8x,

X,

ewsewwee

P

+ + +

12x, 4x, Ax,

X,

Aritq math

+ + +

the

Following tableau:

Ss]

4x, 4x, 12x,

to:

Subject

problem

for this + 5X3

model + 3X,

The mathematical Maximize P = 4x,

PROGRAMMING

a?

R,

the

simplex

xX,

X,

-1

5 X;

a ~

X3

S,

Sy,

1

0

2

0

-8

0

0

2

0

1

0 1 ois]

ao Optimal

45.

Let and

The

0

solution:

assortment

I,

assortment

III

candies

are

> ois the x= the x, = the

Subject

are

0

50

10 3 0 @igaee

maximum 50

12

-%a

UC |CO

Xo. =

not

400 675

e525 profit

boxes

of

produced.

is

In

this

of of of

grams grams grams

of of of

food food food

A, B, C.

model for this problem 3x, + 4x, + 5x,

to:

$3,525

assortment

when II,

case,

used.

number number number

mathematical Maximize P =

r

$3

x,

and

s, 2 =

= x,

400

0 boxes =

675

of

boxes

of

fruit-filled

is:

x + 3X, + 2x, < 30 2x, ne

i 2x, S$ 24 X11 Xp, Xy a)

We

The

introduce X,

+

2x,

tan

-3X,

-

simplex x,

Sy

s)

ail

slack

3X,

+

kot 4x,

-

variables

2x,

+

S,

2x,

+

S,

5x,

tableau

for

XX,

SS,

So

3

2

1

0

@ 9

S, to

=

SiO

="

24

+2 P=

X,

2 1

s, ot and

this

obtain

the

initial

form:

£10

problem

is:

P

O./

1

30 303la5

Gf

15

24

0}

|p = 2

is aR, > R,

Tepe My rety ~-}2 7 @ 9 Bored) 5-0. Rae

Rt

Ry

eR,

BH Py 7 o}12) 0 at.| 0 SR,

+

R;

>

R;

“1° @)‘o 1 a o fs] S- 3 1 1 | 2 2 2 0 2 of42 [ae og Bier Oy 80 teem d 60 i2%,

7

EXERCISE

5-4

257

wR

Orie 2 oy aie

aT

2

ae

Cone ea aR ) ia oo emia Rte Nigar Roe

Let

ary =

and

x x3

58S)

109)

2 t-2) a Ne

rie

8s bn mk

f 0- 7 ~ =< 3

oe

ae

5

SS, 1 #

a

a

Pee oeigc ae 129

units when 10.5 grams of

of protein is 64.5 of food B and Aye

undergraduate students, graduate students, faculty members.

of of of

number number number

the the the

Il

100)

3

the maximum amount Optimal solution: 0 grams of food A, x, = 3 grams Pa food C are used.

47.

5 Se

eRe ile” «ae 1 1 60 a,

rg

of

ee

Sd eek ie Ce ee 2

Ou

4

1S as ST) ase 3 2 2 aa

au)

el

Xa

Bee

1

The mathematical model for this problem is: Maximize P = 18x, + 25x, + 30x, < 20 x, Si et Subject to: 100x,

+

150x,

200x,

=

a hte 5 second

the

Divide

introduce

slack

x,

6+ xX, + 3X, + + - 25x, =

2X, -18x,

The

X,

Ss,

X,

ul

x, 4x, 30x,

3

af

oe

S,

to

Sy

= =

Sy

+

+

P

So

iP

pe

0

0

PAO, 64

=0

is:

problem

this

S,

simplify the arithmetic. Then obtain the initial form.

to

50

Ss, and

+

for

tableau

simplex

ees by

inequality

variables

< 3200

2 0

20 EBoks

20

20

=| 2.3 © o 1 0}64 [eb ie P | -18

-25*>

-30

0

1

ab ma al

0

al

0

A 1

ee

-18 -25

-30

ee

0

R,,

54 R, >

(=1)

0

0

1

30R,

+

aew

Ss

0

R, >

Ge ee

-2

(-3)": A

258

CHAPTER

0

0

pore Rex,

5

LINEAR

1] 3R,

pa

ee

480 3 i

INEQUALITIES

1

Q

NY

i

AND

LINEAR

oe

1/2 6

480

871/2 =16

o 2-4 of Ei

Pos

erie OEeee

2R,

8

114 Dal. 26 | ee

1]

#

—ale

Sy

By der ¢ R,

0

0

2R, >

R,

eee

deh tO.) eat Oe

2

3

z

ne

i1

-2

-3

(Qy 72° 00 2) ee ore i

4 0 6%

3 ati 1 PSD

4

0

1-4

0

2

Q

20

0

2 MO 0. Lia

~ leneoe ieee

086) >

Ry

PROGRAMMING

ine

21

[Pecos

ie = 24

~

goals ati e4 Pec 1

(-3):

+

—>

XX,

5,

4

2

1

0

ymalgor

ia

pete,

jugt

@

wdidatéww

R,,

3 |. soa |) Aetna Ry

+

R,

—>

Over

yee

0

16

1:00)

|4

=2

x,

tones eeteozC

R,

Optimal solution: the maximum number of interviews undergraduates, X, = 16 graduate students, and i= are hired.

EXERCISE

18

16

es. R,

X,

0

= ii 1

Olas

1 1

2

X,

is 520 when cil 0 4 faculty members

5-5

Things

to

remember:

1.

Given a matrix A. The transpose of A, denoted A’, is the matrix formed by interchanging the rows and corresponding columns of A (first row with first columnn, second row with second column, and so on.)

2.

FORMATION

Given

3.

OF

THE

DUAL

a minimization

PROBLEM

problem

with

2 problem

constraints:

Step

1. Use the coefficients and constants in the problem constraints and the objective function to form a matrix A with the coefficients of the objective function in the last row.

Step

2.

Step

3. Use the rows of A’ to form < problem constraints.

THE

Interchange the rows and columns of the matrix A’, the transpose of A.

FUNDAMENTAL

PRINCIPLE

OF

matrix

a maximization

A to

form

problem

with

DUALITY

A minimization problem has a solution if and only if its dual problem has a solution. If a solution exists, then the optimal value of the minimization problem is the same as the optimal value of the dual problem. 4.

SOLUTION

OF

A

MINIMIZATION

PROBLEM

Given a minimization problem the objective function: (i)

with

nonnegative

coefficients

Write all problem constraints as 2 inequalities. may introduce negative numbers on the right side problem constraints. )

in

(This of the

EXERCISE

5-5

259

(ii)

Form

(iii)

(iv)

(v)

the

dual

Write the initial system of the variables from the minimization variables.

dual problem, using the problem as the slack

Use

the

the

simplex

method

Read the solution bottom row of the (Note:

If

the

minimization

1, 0A =

problem.

E59 oes

megeteys

to

solve

dual

problem.

of the minimization problem from the final simplex tableau in Step (iv).

dual

problem

problem

has

has no

no

solution,

then

the

solution.]

-5 0 ie al 8

aoe

4

3 cae

a Ses

8G. 4]

4 uh ie

+2 [2a BH atl a Oo Ayn

:

pee

PONIgy Kyte wy ; A’ = | PO epee dies