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Question
A manufacturing firm produces two types of gadgets A and B, which are first processed in the foundry and then sent to machine shop for finishing. The number of man hours of labour required in each shop for production of A and B and the number of man hours available for the firm are as follows:
| Gadgets | Foundry | Machine Shop |
| A | 10 | 5 |
| B | 6 | 4 |
| Time available (hours) | 60 | 35 |
Profit on the sale of A is ₹ 30 and B is ₹ 20 per unit. Formulate the L.P.P. to have maximum profit.
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Solution
Let the manufacturing firm produces x units of gadget A and y units of gadget B.
The profit on 1 unit of A is ₹ 30 and on 1 unit of B is ₹ 20.
∴ Total profit on selling x units of A and y units of B is ₹ 30x + 20y.
Thus the profit function Z = 30x + 20y
A and B are the products while the time required in the foundry and machine shop are constraints, we construct the given table with the products written column wise and the constraints row-wise.
|
Constraints/Gadgets |
A (x) |
B (x) |
Time available in hours |
| Foundry | 10 | 6 | 60 |
| Machine shop | 5 | 4 | 35 |
1 unit of A requires 10 hours in the foundry and 1 unit of B requires 6 hours.
∴ x units of A requires 10x hours and y units of B requires 6y hours in the foundry. But the maximum time available in the foundry is 60 hours.
∴ The 1st constraint is 10x + 6y ≤ 60.
The constraint for the machine shop is 5x + 4y ≤ 35.
Since number of gadgets cannot be negative, we have x ≥ 0, y ≥ 0.
∴ Given problem can be formulated as follows:
Maximize Z = 30x + 20y
Subject to 10x + 6y ≤ 60, 5x + 4y ≤ 35, x ≥ 0, y ≥ 0.
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