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Question

A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours of assembling. The profit is Rs 5 each for type A and Rs 6 each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximize the profit?

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Solution

Let the company manufacture x souvenirs of type A and y souvenirs of type B. Therefore,

x ≥ 0 and y ≥ 0

The given information can be complied in a table as follows.

Type A

Type B

Availability

Cutting (min)

5

8

3 × 60 + 20 =200

Assembling (min)

10

8

4 × 60 = 240

The profit on type A souvenirs is Rs 5 and on type B souvenirs is Rs 6. Therefore, the constraints are

i.e.,

Total profit, Z = 5x + 6y

The mathematical formulation of the given problem is

Maximize Z = 5x + 6y … (1)

subject to the constraints,

… (2)

… (3)

x, y ≥ 0 … (4)

The feasible region determined by the system of constraints is as follows.

The corner points are A (24, 0), B (8, 20), and C (0, 25).

The values of Z at these corner points are as follows.

Corner point

Z = 5x + 6y

A(24, 0)

120

B(8, 20)

160

→ Maximum

C(0, 25)

150

The maximum value of Z is 200 at (8, 20).

Thus, 8 souvenirs of type A and 20 souvenirs of type B should be produced each day to get the maximum profit of Rs 160.


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