There two types of fertilisers 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 atleast 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 farmer use x kg of F₁ and y kg of F₂.
Let Z be minimum cost.
Table
 

We are to minimise
Z = 6x + 5y
subject to constraints
                  
   
                       
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of 2 x + y = 280
For x = 0, y = 280
For y = 0, 2 x = 280 or x = 140
∴  line meets OX in A(140, 0) and OY in L(0, 280).
Again we draw the graph of 3x + 5y = 700
For x = 0,  5y = 700   or   y = 140
For y = 0,  3x = 700  or  
  line meets OX in 

Since feasible region satisfies all the constraints.
∴ shaded region is the feasible region which is unbounded and has corner points are 

∴ smallest value = 1000 at (100, 80)
Since feasible region is unbounded.
∴ we are to check whether this value is minimum.
For this we draw the graph of
6x + 5y < 1000    ...(1)
Since (1) has no common point with feasible region.
∴ minimum value = Rs. 1000 at (100, 80)
∴ minimum cost is Rs. 1000 when 100 kg. of fertilizer F₁ and 80 kg. of fertilizer F₂ are used.