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?
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
Total profit, Z = 5x + 6y
The mathematical formulation of the given problem is
Maximize Z = 5x + 6y … (1)
subject to the constraints,
`5x+4y<=120` … (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|
|B(8, 20)||160||→ Maximum|
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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