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Question
A company manufactures two types of chemicals A and B. Each chemical requires two types of raw material P and Q. The table below shows number of units of P and Q required to manufacture one unit of A and one unit of B.
| Raw Material \Chemical | A | B | Availability |
| p | 3 | 2 | 120 |
| Q | 2 | 5 | 160 |
The company gets profits of ₹ 350 and ₹ 400 by selling one unit of A and one unit of B respectively. Formulate the problem as L.P.P. to maximize the profit.
Solution
Let x units of chemical A and y units of chemical B are manufactured by the company.
Here, (3x + 2y) units of material P and (2x + 5y) units of material Q is required and 120 units of material P and 160 units of material Q are available.
∴ The constraints are :
3x + 2y ≤ 120,
2x + 5y ≤ 160
Since x and y cannot be negative, we have x ≥ 0, y ≥ 0
Now, Profit on one unit of chemical A is ₹ 350.
∴ Profit on x units of chemical A is 350x.
Profit on one unit of chemical B is ₹ 400.
∴ Profit on y units of chemical B is 400y.
∴ Total Profit, Z = 350x + 400y
This is the objective function to be maximized.
∴ Given problem can be formulated as,
Maximize Z = 350x + 400y
Subject to 3x + 2y ≤ 120, 2x + 5y ≤ 160, x ≥ 0, y ≥ 0.
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