Let x kg of fertilizer A and y kg of fertilizer B be used. As per the question :

Here cost is minimum at (30,210) and is ₹1,980. Since the region is unbounded, we have to draw

10x+8y <1980

:. L : 10x+8y = 1980

Clearly open half plane has no common point with the feasible region is minimum value of ₹1,980.

Farmer has to use 30 kg of fertilizer A and 210 kg of fertilizer B so that nutrient requirements are met at a minimum cost.