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Question
A company manufactures two types of toys A and B. A toy of type A requires 5 minutes for cutting and 10 minutes for assembling. A toy of type B requires 8 minutes for cutting and 8 minutes for assembling. There are 3 hours available for cutting and 4 hours available for assembling the toys in a day. The profit is ₹ 50 each on a toy of type A and ₹ 60 each on a toy of type B. How many toys of each type should the company manufacture in a day to maximize the profit? Use linear programming to find the solution.
Solution
| Toy A | Toy B | Time in a day | |
| Cutting time | 5 min | 8 min | 180 min |
| Assembling time | 10 min | 8 min | 240 min |
| Profit | 50 | 60 | |
| Assumed quantity | x | y |
Profit function z = 50x + 60y
x ≥ 0, y ≥ 0
5x + 8y ≤ 180
10x + 8y ≤ 240 or 5x + 4y ≤ 120
5x + 8y = 180
| A | B | |
| x | 0 | 36 |
| y | 22.5 | 8 |
5x + 4y ≤ 120
| C | D | |
| x | 0 | 24 |
| y | 30 | 0 |
| Corner point | z = 50x + 60y |
| At O (0, 0) | 0 |
| At D (24, 0) | 1200 |
| At E (12, 15) | 1500 |
| At A (0, 22.5) | 1350 |
Hence, the maximum profit is Rs 1500 at E (12, 15).
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