Question

There are two types of fertilizers F₁ and F₂. F₁ consists of 10% nitrogen and 6% phosphoric acid and F₂ consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that she needs at least 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F₁ cost Rs 6/kg and F₂ costs Rs 5/kg, determine how much of each type of fertilizer should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?

Answer 1

Let the farmer buy x kg of fertilizer F₁ and y kg of fertilizer F₂. Therefore,
x ≥ 0 and y ≥ 0
The given information can be complied in a table as follows.

F₁ consists of 10% nitrogen and F₂ consists of 5% nitrogen. However, the farmer requires at least 14 kg of nitrogen. 10% of x + 5% of y ≥ 14

F₁ consists of 6% phosphoric acid and F₂ consists of 10% phosphoric acid. However, the farmer requires at least 14 kg of phosphoric acid.
6% of x + 10% of y ≥ 14

Total cost of fertilizers, Z = 6x + 5y
The mathematical formulation of the given problem is Minimize Z = 6x + 5y … (1) subject to the constraints, 2x + y ≥ 280 … (2) 3x + 5y ≥ 700 … (3) x, y ≥ 0 … (4)
The feasible region determined by the system of constraints is as follows.

It can be seen that the feasible region is unbounded.

The values of Z at these points are as follows.

As the feasible region is unbounded, therefore, 1000 may or may not be the minimum value of Z.
For this, we draw a graph of the inequality, 6x + 5y < 1000, and check whether the resulting half plane has points in common with the feasible region or not. It can be seen that the feasible region has no common point with 6x + 5y < 1000 Therefore, 100 kg of fertiliser F₁ and 80 kg of fertilizer F₂ should be used to minimize the cost. The minimum cost is Rs 1000.