Let x kg of Type I fertilizer and y kg of Type II fertilizers are supplied.

The quantity of fertilizers can not be negative.

So, x,y 0

A gardener has a supply of fertilizer of type I which consists of 10% nitrogen and Type II consists of 5% nitrogen, and he needs at least 14 kg of nitrogen for his crop.

So,

(10100) + (5100) 14

Or, 10x + 5y 1400

A gardener has a supply of fertilizer of type I which consists of 6% phosphoric acid and Type II consists of 10% phosphoric acid, and he needs at least 14 kg of phosphoric acid for his crop.

So,

(6100) + (10100) 14

Or, 6x + 10y 1400

Therefore, A/Q, constraints are,

10x + 5y 1400

6x + 10y 1400

If the Type I fertilizer costs 60 paise per kg and Type II fertilizer costs 40 paise per kg. Therefore, the cost of x kg of Type I fertilizer and y kg of Type II fertilizer is Rs0.60x and Rs 0.40y respectively.

Total cost = Z(let) = 0.6x + 0.4y is to be minimized.

Thus the mathematical formulation of the given LPP is,

Min Z = 0.6x + 0.4y

Subject to the constraints,

10x + 5y 1400

6x + 10y 1400

x,y 0

The region represented by 6x + 10y 1400: line 6x + 10y = 1400 passes through A(,0) and B(0,140). The region which doesn’t contain the origin represents the solution of the inequation 6x + 10y 1400

As (0,0) doesn’t satisfy the inequation 6x + 10y 1400

Region represented by 10x + 5y 1400: line 10x + 5y = 1400 passes through C(140,0) and D(0,280). The region which doesn’t contain the origin represents the solution of the inequation 10x + 5y 1400

As (0,0) doesn’t satisfy the inequation 10x + 5y 1400

The region, x,y 0: represents the first quadrant.

The corner points are D(0,280), E(100,80), A(,0)

The values of Z at these points are as follows:

The minimum value of Z is Rs 92 which is attained at E(100,80)

Thus, the minimum cost is Rs92 obtained when 100 kg of Type I fertilizer and 80 kg of TypeII fertilizer is supplied.