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प्रश्न
A company manufactures two types of fertilizers F1 and F2. Each type of fertilizer requires two raw materials A and B. The number of units of A and B required to manufacture one unit of fertilizer F1 and F2 and availability of the raw materials A and B per day are given in the table below:
| Raw Material\Fertilizers | F1 | F2 | Availability |
| A | 2 | 3 | 40 |
| B | 1 | 4 | 70 |
By selling one unit of F1 and one unit of F2, company gets a profit of ₹ 500 and ₹ 750 respectively. Formulate the problem as L.P.P. to maximize the profit.
उत्तर
Let the company manufacture ‘x’ units of fertilizer F1 and ‘y’ units of F2.
The profit on one unit of F1 is ₹ 500 and on unit of F2 is ₹ 750.
∴ Total profit = ₹ (500x + 750y)
From the given table,
The raw material A required for x units of F1 and y units of F2 is (2x + 3y). The raw material B required is (x + 4y).
But maximum availability of raw materials A and B are 40 and 70 units respectively.
∴ The constraints are:
2x + 3y ≤ 40, x + 4y ≤ 70
Since x and y cannot be negative, we have x ≥ 0, y ≥ 0.
∴ Given problem can be formulated as follows:
Maximize Z = 500x + 750y
Subject to 2x + 3y ≤ 40, x + 4y ≤ 70, x ≥ 0, y ≥ 0.
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