There are two types of fertilizers F_{1 }and F_{2}. F_{1 }consists of 10% nitrogen and 6% phosphoric acid and F_{2 }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 F_{1 }costs ₹6/kg and F_{2 }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?
Solution
Suppose x kg of fertilizer F_{1 }and and y kg of fertilizer F_{2} is used to meet the nutrient requirements.
F_{1 }consists of 10% nitrogen and F_{2 }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, F_{1 }consists of 6% phosphoric acid and F_{2 }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 F_{1 }is ₹6/kg and fertilizer F_{2 }is ₹5/kg, therefore, total cost of x kg of fertilizer F_{1 }and and y kg of fertilizer F_{2 }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 
It can be seen that the open halfplane 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 F_{1 }and and 80 kg of fertilizer F_{2} should be used so that the nutrient requirements are met at minimum cost. The minimum cost is ₹1,000.