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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 20 minutes available for cutting and 4 hours available for assembling. The profit is 50 paise each for type A and 60 paise 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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उत्तर
Let the company manufacture x souvenirs of type A and y souvenirs of type B.
Number of items cannot be negative.
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 |
3x60+20=200
|
| Assembling (min) | 10 | 8 | 4 x 60 = 240 |
Therefore, the constraints are
5x + 8y ≤ 200
10x + 8y ≤ 240
The profit on type A souvenirs is 50 paise and on type B souvenirs is 60 paise.Therefore, profit gained on x souvenirs of type A and y souvenirs of type B is Rs 0.50x and Rs 0.60y respectively.
Total profit, Z = 0.5x + 0.6y
The mathematical formulation of the given problem is
Maximize Z = 0.5x + 0.6y
subject to the constraints,
5x + 8y ≤ 200
10x + 8y ≤ 240
x ≥ 0 and y ≥ 0
First we will convert inequations into equations as follows:
5x + 8y = 200, 10x + 8y = 240, x = 0 and y = 0
Region represented by 5x + 8y ≤ 200:
The line 5x + 8y = 200 meets the coordinate axes at A1(40, 0) and B1(0, 25) respectively. By joining these points we obtain the line 5x + 8y = 200 . Clearly (0,0) satisfies the 5x + 8y = 200. So,
the region which contains the origin represents the solution set of the inequation 5x + 8y ≤ 200.
Region represented by 10x + 8y ≤ 240:
The line 10x + 8y = 240 meets the coordinate axes at C1(24, 0) and D1(0, 30) respectively. By joining these points we obtain the line 10x + 8y = 240. Clearly (0,0) satisfies the inequation 10x + 8y ≤ 240. So,
the region which contains the origin represents the solution set of the inequation 10x + 8y ≤ 240.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 5x + 8y ≤ 200, 10x + 8y ≤ 240, x ≥ 0 and y ≥ 0 are as follows.
The corner points of the feasible region are O(0, 0), B1(0, 25), E1(8, 20), C1(24, 0).
The values of Z at these corner points are as follows.
| Corner point | Z = 0.5x + 0.6y |
| O(0, 0) | 0 |
| B1(0, 25) | 15 |
| E1(8, 20) | 16 |
| C1(24, 0) | 12 |
The maximum value of Z is 160 at E1(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 16.
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