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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:
| Fertilizers→ | F1 | F2 | Availability |
| Raw Material ↓ | |||
| A | 2 | 3 | 40 |
| B | 1 | 4 | 70 |
By selling one unit of F1 and one unit of F2, the company gets a profit of ₹ 500 and ₹ 750 respectively. Formulate the problem as LPP to maximize the profit.
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Solution
Let the company manufactures x units of fertilizers F1 and y units of fertilizers F2. Then the total profit to the company is z = ₹(500x + 750y).
This is a linear function that is to be maximized. Hence, it is an objective function.
| Fertilizers→ | F1 | F2 | Availability |
| Raw Material ↓ | |||
| A | 2 | 3 | 40 |
| B | 1 | 4 | 70 |
The raw material A required for x units of Fertilizers F1 and y units of Fertilizers F2 is 2x + 3y. Since the maximum availability of A is 40, we have the first constraint as 2x + 3y ≤ 40.
Similarly, considering the raw material B, we have x + 4y ≤ 70.
Since, x and y cannot be negative, we have, x ≥ 0, y ≥ 0.
Hence, the given LPP can be formulated as:
Maximize z = 500x + 750y, subject to
2x + 3y ≤ 40, x + 4y ≤ 70, x ≥ 0, y ≥ 0
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