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b14a_tea_3.1-3.3-2.pdf

Bus 14A Textbook Exercises and Answers

Finite Mathematics and Calculus with Applications – 10th edition

By Lial, Greenwell, and Ritchey

Complete:

3.1 24, 28, 30, 40, 42

3.2 2a, 6a, 10, 12

3.3 8, 14, 22

Check your answers:

3.1 24. Unbounded Graph

28. Bounded Graph

30. Bounded Graph

40. a) b)

0,0

206

8 2

1







yx

yx

yx

Glazed Unglazed Maximum

Number Made x y

Time of Wheel 1/2 1 8

Time in Kiln 1 6 20

40. c) Yes, 5 glazed and 2 unglazed planters can be made, since the point (5,2) lies within the

feasible region. No, 10 glazed and 2 unglazed planters cannot be made, since the point (10,2) lies

outside the feasible region.

42. a)

000,10

5000

3000



yx

y

x

b) Graph

3.2 2. a) The maximum value is 34 at (2,8). The minimum value is 8 at (4,1).

6. a) The minimum value is 10 at (0,10). No maximum. Feasible region is unbounded.

10. Maximize yxz 810  Subject to:

200

10

20045

10032







y

x

yx

yx

The maximum value is 400 when x = 200/7 and y = 100/7, as well as when x = 40 and y = 0 and

at all points in between.

12. Maximize yxz 54  Subject to:

0,0

12

1501020

100510









yx

yx

yx

yx

Since the region is unbounded, there is no maximum value, hence no solution.

3.3 8. Minimize yxz 1012  Subject to:

0,0

30024

80

75

100







yx

yx

y

x

yx

Ship 50 refrigerators to Warehouse A and 50 to Warehouse B for a minimum cost of $1100.

14. a) Maximize yxz 500350  Subject to:

0,0

260044

9002

360075









yx

yx

yx

yx

Maximum profit is $255,000 when 300 Flexscan sets and 300 Panoramic I sets are produced.

14. b) A maximum profit of $301,250 when 475 Flexscan and 175 Panoramic I sets are produced.

14. c) In the solution to part (a), 300 Flexscan and 300 Panoramic I sets are produced. There are

2600 – 2400 or 200 unused hours in testing and packing.

14. c) In the solution to part (b), 475 Flexscan and 175 Panoramic I sets are produced. There are

900 – 825 or 75 unused hours in the cabinet shop.

22. Minimize yxz 32  Subject to:

0,0

842

1035







yx

yx

yx

The minimum value is 46/7 ≈6.57 units of energy when 8/7 units of species I and 10/7 units of species II will meet the daily food requirements with the least expenditure of energy. However,

a predator probably can catch and digest only whole numbers of prey. This problem shows that

it is important to consider whether a model produces a realistic answer to a problem.