1

Answers to Chapter 6

Exercise 6A 1 Number of boxes of gold assortment � x Number of boxes of silver assortment � y

Objective: maximise P � 80x � 60yConstraints• time to make chocolate, 30x � 20y � 300 � 60 which simplifi es to 3x � 2y � 1800 All units now in minutes.• time to wrap and pack 12x � 15y � 200 � 60 which simplifi es to 4x � 5y � 4000• ‘At least twice as many silver as gold’ 2x � y• non-negativity x, y � 0In summary: maximise P � 80x � 60ysubject to 3x � 2y � 1800 4x � 5y � 4000 2x � y

x, y � 0 2 number of type A � x number of type B � y

Objective: minimise c � 6x � 10yConstraints• Display must be at least 30 m long x � 1.5y � 30 which simplifi es to 2x � 3y � 60• ‘At least twice as many x as y’ 2y � x• At least six type B y � 6• non-negativity x, y � 0In summary: minimise c � 6x � 10ysubject to:

2x � 3y � 60 2y � x y � 6 x, y � 0 3 Number of games of Cludopoly � x Number of games of Trivscrab � y

Objective: maximise P � 1.5x � 2.5yConstraints• First machine: 5x � 8y � 10 � 60 which simplifi es to 5x � 8y � 600

All units now in minutes.• Second machine: 8x � 4y � 10 � 60 which simplifi es to 2x � y � 150• At most 3 times as many x as y 3y � x• non-negativity x, y � 0In summary: maximise P � 1.5x � 2.5ysubject to:

5x � 8y � 600 2x � y � 150 3y � x x, y � 0 4 Number of type 1 bookcases � x Number of type 2 bookcases � y

Objective: maximise S � 40x � 60yConstraints• budget 150x � 250y � 3000 which simplifi es to 3x � 5y � 60• fl oor space 15x � 12y � 240 which simplifi es to 5x � 4y � 80

• ‘At most 1 _ 3 of all bookcases to be type 2’ y � 1 _ 3 (x � y) which simplifi es to 2y � x• At least 8 type 1 x � 8• non-negativity x, y � 0In summary: maximise S � 40x � 60ysubject to: 3x � 5y � 60 5x � 4y � 80 2y � x x � 8 x, y � 0

Answers to Chapter 6

2

ANSWERS

5 Let x � number of kg of indoor feed and y � number of kg of outdoor feedObjective: maximise P � 7x � 6yConstraints• Amount of A 10x � 20y � 5 � 1000 which simplifi es to x � 2y � 500

All units now in grams.• Amount of B 20x � 10y � 5 � 1000 which simplifi es to 2x � y � 500• Amount of C 20x � 20y � 6 � 100 which simplifi es to x � y � 300• At most 3 times as much y as x y � 3x• At least 50 kg of x x � 50• non-negativity y � 0 (x � 0 is unnecessary because of the previous constraint)In summary: maximise P � 7x � 6ysubject to x � 2y � 500 2x � y � 500 x � y � 300 y � 3x x � 50 y � 0

6 number of A smoothies � x number of B smoothies � y number of C smoothies � zObjective: maximise P � 60x � 65y � 55zConstraints• oranges x � 1 _ 2 y � 2z � 50 which simplifi es to 2x � y � 4z � 100• raspberries 10x � 40y � 15z � 1000 which simplifi es to 2x � 8y � 3z � 200• kiwi fruit 2x � 3y � 2 � 100• apples 2x � 1 _ 2 y � 2z � 60 which simplifi es to 4x � y � 4z � 120• non-negativity x, y, z � 0In summary: maximise P � 60x � 65y � 55zsubject to: 2x � y � 4z � 100 2x � 8y � 3z � 200 2x � 3y � z � 100 4x � y � 4z � 120 x, y, z � 0

7 Let number of hours of work for factory R � x Let number of hours of work for factory S � yObjective: minimise C � 300x � 400yConstraints• milk 1000x � 800y � 20 000 which simplifi es to 5x � 4y � 100• yoghurt 200x � 300y � 6000 which simplifi es to 2x � 3y � 60

• At least 1 _ 3 of total time for R x � 1 _ 3 (x � y) which simplifi es to 2x � y

• At least 1 _ 3 of total time for S y � 1 _ 3 (x � y) which simplifi es to 2y � x• non-negativity x, y � 0In summary: minimise C � 300x � 400ysubject to: 5x � 4y � 100 2x � 3y � 60 2x � y 2y � x x, y � 0

ANSWERS

3

Exercise 6B 1

f

c

0

f � c � 28

2f � 3c � 60

R(feasible region)

f � 2c � 36

2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 4002468

1012141618202224262830

4

x

y

0 1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

6

7

8

y � 1

x � 3

3x � 4y � 24

2x � 3y

R

2

x

y

0 80 160 240 320 400 4800

80

160

240

320

400

480

y � 80

x � 200y � 2xR

x � y � 400

5

x

y

0 2 4 6 8 10 12 14 16 18 200

2

4

6

8

10

12

14

16

18

20

y � 10

x � y � 20

5x � 6y � 60

2x � y

R

3

x

y

0 1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

6y � x

y � 5

2x � 3y � 18

R

6

x

y

0 1 2 3 4 5 6 7 80

1

2

3

4

5

6

7

8

y � 3

y � 6 � 2x

y � 2x � 3

R

Exercise 6C 1 a Need intersection of 4x � y � 1400 and 3x � 2y � 1200

(320, 120) m � 760 Objective line passes through (200, 0) and (0, 400).

b (0, 400) N � 1600 Objective line passes through (400, 0) and (100, 0).

c Need intersection of x � 3y � 1200 and 3x � 2y � 1200 Objective line passes through (200, 0) and (0, 200). ( 171 3 _ 7 , 342 6 _ 7 ) P � 514 2 _ 7

d (350, 0) Q � 2100 Objective line passes through (100, 0) and (0, 600).

4

ANSWERS

2 a (0, 90) E � 90b Need intersection of 6y � x and 3x � 7y � 420

(100.8, 16.8) F � 168c Need intersection of 9x � 10y � 900 3x � 7y � 420 Objective line passes through (80, 0) and (0, 60).

( 63 7 __ 11 , 32 8 __ 11 ) G � 321 9 __ 11

d Same intersection as in b (100.8, 16.8) H � 201.6 Objective line passes through (120, 0) and (0, 20).

3 a Need intersection of 3x � y � 60 and 5y � 3x ( 16 2 _ 3 , 10 ) J � 56 2 _ 3

b Need intersection of y � 4x and 9x � 5y � 450

( 15 15 __ 29 , 62 2 __ 29 ) K � 77 17

__ 29

c Need intersection of 3x � y � 60 and y � 4x Objective line passes through (10, 0) and (0, 60).

( 8 4 _ 7 , 34 2 _ 7 ) L � 85 5 _ 7

d Need intersection of 9x � 5y � 450 and 5y � 3x Objective line passes through (40, 0) and (0, 80).(37.5, 22.5) m � 97.5

4 a C b A c B d D e C f A g B h D i C j D

5 a

x

y

20 40 60 80 100 120 140 160 180 200O

20

40

60

80

100

120

140

160

2x � y � 160

x � 3y � 180

x � y � 120

C1 � 2x � 3y(objective for (c))

C1 � x � y(objective for (c))

C � 3x � 2y(objective for (b))

A (40, 80)R

B (90, 30)

bVertices C � 3x � 2y

(0, 160) 320

(40, 80) 280

(90, 30) 330

(180, 0) 540

so minimum is (40, 80) value of C � 280c (90, 30) C1 � 270d C2 is parallel to x + y � 120 so all points from A to B

are optimal points.

6 a

x

y

5 10 15 20 25 30 35 40 45O

10

20

30

40

50

60

70

14x � 9y � 630

R

P � x � 3y

2y � x

y � 5x

Q � 6x � y4x � y � 60

(10 , 53 )4059

2359

(34 , 17 )237

137

(6 , 33 )23

13

(13 , 6 )13

23

b i ( 13 1 _ 3 , 6 2 _ 3 ) P � 33 1 _ 3

ii ( 34 2 __ 37 , 17 1 __ 37 ) Q � 221 13 __ 37

ANSWERS

5

7 a

x

y

20 40 60 80 100 120 140 160 180 200O

50

100

150

200

250

300

350

400

450

2y � x � 100

2x � y � 400

z � 5x � y

x � 2y

R

y � 10x x � 120

(5 , 52 )519

1219

(33 , 333 )

(120 , 160)

(120 , 110)

13

13

b i (120, 160) z � 760

ii ( 5 5 __ 19 , 52 12 __ 19 ) z � 78 18

__ 19

c Optimal point ( 33 1 _ 3 , 333 1 _ 3 ) optimal value of x � 2y � 700

8

x

y

50 100 150 200 250 300 350 400 450 500O

50

100

150

200

250

300

350

400

450

500

x � 2y � 500

x � y � 300

2x � y � 500

R

y � 3xx � 50

7x � 6y � 2100

9

x

y

5 10 15 20 25 30 35 40O

5

10

15

20

25

30

2y � x

y � 2x

2x � 3y � 60

5x � 4y � 100

R

Objective line passes through (0, 15) and (20, 0).Intersection of 2x � 3y � 60 5x � 4y � 100 ( 8 4 _ 7 , 14 2 _ 7 ) value � 8285 5 _ 7

Objective line passes through (0, 350) and (300, 0). Maximum point is (200, 100). Pmax � 2000.

Exercise 6D 1

x

y

22222 44444 66666 88888 00110O

222

444

666

888

10100

3x � 2y22

R

4xx �� 33yyy33 �� 242

33xxx ��� 444yyy444 ��� 24242

x � 5y5 � 101

11

44

22

Maximum integer value (5, 1) 3x � 2y � 17

6

ANSWERS

2

x

y

2 4 6 8 10 12 14 16 18O

2

4

6

8

10

12

14

16

18

20

22

24

26

28

30

5x � 6y � 60

4x � y � 28

Objective

2x � y

R

3

x

y

20 40 60 80 100 120 140 160 180 200O

20

40

60

80

100

120

140

160

180

200 y � 4x

5x � 4y � 800

2y � x

Objective

R

5x � 2y

Solving 2y � x and 5x � 4y � 800 simultaneously gives

( 114 2 _ 7 , 57 1 _ 7 ) Test integer values nearby.

Point 2y � x 5x � 4y � 800 5x � 2y

(114, 57) � � 684

(114, 58) � � —

(115, 57) � � —

(115, 58) � � —Minimum integer value (6, 5) 2x � y � 17

so optimal point is (114, 57) value 684

4

x

y

100 200 300 400 500 600 700 800O

50

100

150

200

250

300y � x

Objective2x � y

3x � 16y � 2400

3x � 5y � 1500R

Solving 3x � 16y � 2400 3x � 5y � 1500

simultaneously gives ( 363 7 __ 11 , 81 9 __ 11 ) Taking integer point

Point 3x � 16y � 2400 3x � 5y � 1500

(363, 81) � �

(363, 82) � �

(364, 81) � �

(364, 82) � �

So optimal integer point is (363, 82)Value 808

5

O

y

x200 400 600 800 1000

100

200

300

400

500

600

700

800

900y � 2x

4x � 5y � 4000

3x � 2y � 1800

Objective

4x � 3y

R

Intersection of 4x � 5y � 4000 and 3x � 2y � 1800

gives ( 142 6 _ 7 , 685 5 _ 7 ) Testing nearby integer points

Point 4x � 5y � 4000 3x � 2y � 1800 80x � 60y

(142, 685) � � 52 460

(142, 686) � � 52 520

(143, 685) � � 52 540

(143, 686) � � —

so maximum integer solution is 52 540 pennies at (143, 685).

ANSWERS

7

6

Objective6x � 10y

y

xO 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32

2468

10121416182022

y � 6

x � 2y

2x � 3y � 60

R

Intersection of 2x � 3y � 60 and y � 6(21, 6)

cost c � 6x � 10y so minimum cost � 6 � 21 � 60 � £186

7

x

y

20 40 60 80 100 120 140O

20

40

60

80

100

120

140

160

5x � 8y � 600Objective1.5x � 2y

2x � y � 150

x � 3yR

(54 , 40 )611

1011

Intersection of 5x � 8y � 600

2x � y � 150 giving ( 54 6 __ 11 , 40 10 __ 11 )

Points 5x � 8y � 600 2x � y � 150 1.5x � 2y

(54, 40) � � 161

(54, 41) � � 163

(55, 40) � � 162.5

(55, 41) � � —

so maximum value is 163 at (54, 41).

8

x

y

4 8 12 16 20O

4

8

12

16

20

3x � 5y � 60

5x � 4y � 80Objective line40x � 60y

x � 8

2y � x

R

Using Method 1 from Example 13 shows you that the optimal integer solution is (11, 5) giving 740 m of shelving.

Using Method 2 gives you the following solution: Intersection of 3x � 5y � 60

5x � 4y � 80 giving ( 12 4 __ 13 , 4 8 __ 13 )

Points 3x � 5y � 60 5x � 4y � 80 40x � 60y

(12, 4) � � 720

(12, 5) � � —

(13, 4) � � —

(13, 5) � � —

Maximum value is 720 at (12, 4).

In this instance, the solution produced by Method 2 is actually incorrect, but it requires a very particular set of circumstances to create this discrepancy. It is generally safe to assume that a solution found using Method 2 will be correct, but do check your graph to see whether there could be an alternative optimal integer solution.

Mixed exercise 6E 1 a fl our: 200x � 200y � 2800 so x � y � 14

fruit: 125x � 50y � 1000 so 5x � 2y � 40b Cooking time 50x � 30y � 480 so 5x � 3y � 48

8

ANSWERS

c

x

y

4 8 12 16 202 6 10 14 18O

4

8

12

16

20

2

6

10

14

18

x � y � 14

5x � 2y � 40

5x � 3y � 48Objective lin

e

3.5x � 1.5y

R

d P � 3.5x + 1.5ye Integer solution required (6, 5)f Pmax � £28.50

2 b cost: 6x � 4.8y � 420 5x � 4y � 350c Display 30x � 2 � 20y 3x � 4y d

x

y

10 20 30 40 50 60 70 80 90O

10

20

30

40

50

60

70

80

90 5x � 4y � 350

x � y � 80

3x � 4y

R

Objective3x � 2y

e Integer solution required (43, 33).

3 a i Total number of people b

x

y

2 4 6 8 10 12O

2

4

6

8

10

12

14

16

18

x � y � 12

9x � 4y � 72

Objective84(4x � 3y)

x � 7

R

54x � 24y � 432 so 9x � 4y � 72ii number of adults x � y � 12iii number of large coaches x � 7

c minimise C � 336x � 252y� 84(4x � 3y)

d Objective line passes through (0, 4) (3, 0)e Integer coordinates needed (7, 3)

cost � £3108

ANSWERS

9

4 a 4x � 5y � 47 y � 2x � 8 4y � y � 8 � 0 x, y � 0

b Solving simultaneous equations y � 2x � 8 4x � 5y � 47 6 3 __ 14 , 4 3 _ 7

c i For example where x and y ii (6, 4)• types of car to be hired• number of people etc

5 a 2 1 _ 2 x + 3y � 300 [5x + 6y � 600] 5x + 2y � 400 2y � 150 [ y � 75]b Maximise P � 2x + 4yc

x

y

20 40 60 80 100 120 140O

20

40

60

80

100

120

140

160

180

200

5x � 2y � 400

2 x � 3y � 300Objective

y � 75

R

12

(30, 75) P � 360

d The optimal point is at the intersection of y � 75 and 2 1 _ 2 x + 3y �300.So the constraint 5x + 2y � 400 is not at its limit.At (30, 75) 5x + 2y � 300 so 100 minutes are unused.