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 the she needs atleast 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F₁ costs ₹6/kg and F₂ costs ₹5/kg, determine how much of each type of fertilizer should be used so that the nutrient requirements are met at minimum cost. What is the minimum cost? 

Answer 1

Suppose x kg of fertilizer F₁ and and y kg of fertilizer F₂ is used to meet the nutrient requirements.

F₁ consists of 10% nitrogen and F₂ consists of 5% nitrogen. But, the farmer needs atleast 14 kg of nitorgen for the crops.

∴ 10% of x kg + 5% of y kg ≥ 14 kg

\[\Rightarrow \frac{x}{10} + \frac{y}{20} \geq 14\]
\[ \Rightarrow 2x + y \geq 280\] 

Similarly, F₁ consists of 6% phosphoric acid and F₂ consists of 10% phosphoric acid. But, the farmer needs atleast 14 kg of phosphoric acid for the crops.

∴ 6% of x kg + 10% of y kg ≥ 14 kg

\[\Rightarrow \frac{6x}{100} + \frac{10y}{100} \geq 14\]
\[ \Rightarrow 3x + 5y \geq 700\]
The cost of fertilizer F₁ is ₹6/kg and fertilizer F₂ is ₹5/kg, therefore, total cost of x kg of fertilizer F₁ and and y kg of fertilizer F₂ is ₹(6x + 5y).

Thus, the given linear programming problem is

Minimise Z = 6x + 5y

subject to the constraints

2x + y ≥ 280

3x + 5y ≥ 700

x, y ≥ 0

The feasible region determined by the given constraints can be diagrammatically represented as,

The coordinates of the corner points of the feasible region are \[A\left( \frac{700}{3}, 0 \right), B ( 100,80 )   \text{ and C }( 0,280 )\] .

The value of the objective function at these points are given in the following table.

Corner Point	Z = 6x + 5y
\[\left( \frac{700}{3}, 0 \right)\]	\[6 \times \frac{700}{3} + 5 \times 0 = 1400\]
(100, 80)	6 × 100 + 5 × 80 = 1000 → Minimum
(0, 280)	6 × 0 + 5 × 280 = 1400

The smallest value of Z is 1000 which is obtained at x = 100, y = 80.

It can be seen that the open half-plane represented by 6x + 5y < 1000 has no common points with the feasible region.

So, the minimum value of Z is 1000.

Hence, 100 kg of fertilizer F₁ and and 80 kg of fertilizer F₂ should be used so that the nutrient requirements are met at minimum cost. The minimum cost is ₹1,000.