Upload
vuduong
View
226
Download
0
Embed Size (px)
Citation preview
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.