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,
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,
Therefore, A/Q, constraints are,
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,
The region represented by 6x + 10y ≥ 1400: line 6x + 10y = 1400 passes through A(\(\frac{700}{3}\) ,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\((\frac{700}{3},0)\)
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.
