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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 the machine shop for finishing. The number of man-hours of labour required in each shop for production of A and B per unit and the number of man-hours available for the firm is as follows:
| Gadgets | Foundry | Machine shop |
| A | 10 | 5 |
| B | 6 | 4 |
| Time available (hour) | 60 | 35 |
Profit on the sale of A is ₹ 30 and B is ₹ 20 per units. Formulate the L.P.P. to have maximum profit.
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Solution
Let the number of gadgets A produced by the firm be x and the number of gadgets B produced by the firm be y.
The profit on the sale of A is ₹ 30 per unit and on the sale of B is ₹ 20 per unit.
∴ Total profit is z = 30x + 20y
This is a linear function that is to be maximized. Hence, it is the objective function. The constraints are as per the following table:
|
Gadgets A |
Gadgets B |
Total available Time (in hour) | |
| Foundry | 10 | 6 | 60 |
| Machine shop | 5 | 4 | 35 |
From the table, the total man-hours of labour required for x units of gadget A and y units of gadget B in the foundry is (10x + 6y) hours, and the total man-hours of labour required in a machine shop is (5x + 4y) hours.
Since the maximum time available in foundry and machine shops are 60 hours and 35 hours respectively.
Therefore, the constraints are 10x + 6y ≤ 60, 5x + 4y ≤ 35.
Since x and y cannot be negative, we have x ≥ 0, y ≥ 0. Hence, the given LPP can be formulated as:
Maximize z = 30x + 20y, subject to
10x + 6y ≤ 60,
5x + 4y ≤ 35,
x ≥ 0, y ≥ 0
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