Let x and y be number of kilograms of fertilizer I and II
∴According to the question,
0.10x + 0.05y ≥ 14, 0.06x + 0.10y ≥ 14, x ≥ 0, y ≥ 0
Minimize Z = 0.60x + 0.40y
The feasible region determined by 0.10x + 0.05y ≥ 14, 0.06x + 0.10y ≥ 14, x ≥ 0, y ≥ 0 is given by
The feasible region is unbounded. The corner points of feasible region are A(0,280) , B(100,80) , C(700/3,0).
The value of Z at corner points are
Corner Point |
Z = 0.60x + 0.40y |
|
A(0. 280) |
112 |
|
B(100, 80) |
92 |
Maximum |
C(700/3, 0) |
140 |
|
The minimum value of Z is 92 at point (100,80).
Hence, the gardener should by 100 kilograms o fertilizer I and 80 kg of fertilizer II to minimize the cost which is Rs.92.