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Question
A factory manufactures two types of screws, A and B. Each type of screw requires the use of two machines, an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on hand operated machines to manufacture a package of screws A, while it takes 6 minutes on automatic and 3 minutes on the hand operated machines to manufacture a package of screws B. Each machine is available for at the most 4 hours on any day. The manufacturer can sell a package of screws A at a profit of Rs 7 and screws B at a profit of Rs10. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximize his profit? Determine the maximum profit.
Solution
Let the factory manufacture x screws of type A and y screws of type B on each day. Therefore,
x ≥ 0 and y ≥ 0
The given information can be compiled in a table as follows.
| Screw A | Screw B | Availability | |
| Automatic Machine (min) | 4 | 6 | 4 × 60 =240 |
| Hand Operated Machine (min) | 6 | 3 | 4 × 60 =240 |
The profit on a package of screws A is Rs 7 and on the package of screws B is Rs 10. Therefore, the constraints are
`4x+6y <= 240`
`6x + 3y < 240`
Total profit, Z = 7x + 10y
The mathematical formulation of the given problem is
Maximize Z = 7x + 10y … (1)
subject to the constraints,
`4x+6y <=240`… (2)
`6x +3y <= 240` … (3)
x, y ≥ 0 … (4)
The feasible region determined by the system of constraints is
The corner points are A (40, 0), B (30, 20), and C (0, 40).
The values of Z at these corner points are as follows.
| Corner point | Z = 7x + 10y | |
| A(40, 0) | 280 | |
| B(30, 20) | 410 | → Maximum |
| C(0, 40) | 400 |
The maximum value of Z is 410 at (30, 20).
Thus, the factory should produce 30 packages of screws A and 20 packages of screws B to get the maximum profit of Rs 410.
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