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Question
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.
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Solution
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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