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.

Answer 1

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.