Let x and y be number of kilograms of fertilizer I and II
∴According to the question,
0.10x + 0.05y , 0.06x + 0.10y
Minimize Z = 0.60x + 0.40y
The feasible region determined by 0.10x + 0.05y , 0.06x + 0.10y 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
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.
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