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प्रश्न
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 and 20 minutes available for cutting and 4 hours available for assembling. The profit is Rs. 50 each for type A and Rs. 60 each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximize profit? Formulate the above LPP and solve it graphically and also find the maximum profit.
उत्तर १
Let no.of souvenirs of type A to be manufactured = x
Type B = y.
Type |
Quant. | Profit | Cutting time | A.T |
A | x | 50 | 5min | 10min |
B | y | 60 | 8min | 8min |
Total time for cutting = 3h 20min
= 200min
Total time for Assembling = 240 min
Profit = 50x + 60y (this is obj function)
5x + 8y ≤ 200 .....(i)
10x + 8y ≤ 240 .....(ii)
Also x, y ≥ 0
Corner points | Value of objective |
O(0,0) | 0 |
B(8 , 20) | 400 + 1200 = 1600 |
A(0 , 25) |
1500 |
C (24 , 0) | 1200 |
So the maximum profit is obtained by producing 8 units of A and 20 units of B And max. profit is Rs. 1600
उत्तर २
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 x 60 + 20 = 200 |
Assembling (min) | 10 | 8 | 4 x 60 = 240 |
The profit on type A souvenirs is Rs 50 and on type, B souvenirs is Rs 60. Therefore, the constraints are i.e.,
5x + 8y ≤ 200.
10x + 8y ≤ 240 i.e., 5x + 4y ≤ 120
Total profit, Z = 50x + 60y
The mathematical formulation of the given problem is
Maximize Z = 50x + 60y ..… (1)
subject to the constraints,
5x + 8y ≤ 200 ..… (2)
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 = 50x + 60y | |
A (24, 0) | 1200 | |
B (8, 20) | 1600 | → Maximum |
C (0, 25) | 1500 |
The maximum value of Z is 1600 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 1600.
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