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 costs Rs 6/kg and F₂ costs Rs 5/kg, determine how much of each type of fertiliser should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?

Let the quantity of fertiliser F₁ be x kg and F₂ be ‘y’ kg.

Minimize  Z = 6x + 5y

s.t.    10/100 x + 5/100 y ≥ 14

6/100x + 10/100 y ≥ 14

x ≥ 0, y ≥ 0      

On simplifying the above constraints.     

2x + y ≥ 280 

3x + 5y ≥ 700

x ≥ 0, y ≥ 0   

The feasible region is an unbounded region with corner points A, B, C 

Corner points Z = 6x + 5y

A(700/3,0)   1400

B(100,80)      1000

C(0,280)       1400          

1000 is the minimum value from the table. As the region is unbounded, we have to check if 1000 is the minimum value of objective function or not.

6x + 5y > 1000

The region represented by the above inequality doesn’t have any points common with the feasible region.

Hence 1000 is the minimum value

Quantity of fertilizer F₁ = 100kg,         F₂ = 80kg