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प्रश्न
A company manufactures two types of toys A and B. Type A requires 5 minutes each for cutting and 10 minutes each for assembling. Type B requires 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours available for cutting and 4 hours available for assembling in a day. The profit is Rs 50 each on type A and Rs 60 each on type B. How many toys of each type should the company manufacture in a day to maximize the profit?
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उत्तर
Let x toys of type A and y toys of type B were manufactured.
The given information can be tabulated as follows:
| Cutting time (minutes) | Assembling time (minutes) | |
| Toy A(x) | 5 | 10 |
| Toy B(y) | 8 | 8 |
| Availability | 180 | 240 |
The constraints are
\[5x + 8y \leq 180\]
\[10x + 8y \leq 240\]
The profit is Rs 50 each on type A and Rs 60 each on type B. Therefore, profit gained on x toys of type A and y toys of type B is Rs 50x and Rs 60 y respectively.
Total profit = Z = \[50x + 60y\] The mathematical formulation of the given problem is
Max Z = \[50x + 60y\] subject to
\[5x + 8y \leq 180\]
\[10x + 8y \leq 240\]
First we will convert inequations into equations as follows:
5x + 8y = 180, 10x + 8y = 240, x = 0 and y = 0
Region represented by 5x + 8y ≤ 180:
The line 5x + 8y = 180 meets the coordinate axes at A1(36, 0) and \[B_1 \left( 0, \frac{45}{2} \right)\] respectively. By joining these points we obtain the line 5x + 8y = 180. Clearly, (0,0) satisfies the 5x + 8y = 180. So, the region which contains the origin represents the solution set of the inequation 5x + 8y ≤ 180.
Region represented by 10x + 8y ≤ 240:
The line 10x + 8y = 240 meets the coordinate axes at C1(24, 0) and \[D_1 \left( 0, 30 \right)\] 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 ≤ 180, 10x + 8y ≤ 240, x ≥ 0 and y ≥ 0 are as follows.
The feasible region is shown in the figure
The corner points are B1 \[\left( 0, \frac{45}{2} \right)\] E1(12, 15) and C1(24, 0).
The values of Z at the corner points are
| Corner points | Z = \[50x + 60y\] |
| O | 0 |
| B1 | 1350 |
| E1 | 1500 |
| C1 | 1200 |
Thus, for maximum profit, 12 units of toy A and 15 units of toy B should be manufactured.
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