Q. 245.0( 1 Vote )

A gardener has a supply of fertilizers of the type 1 which consist of 10% nitrogen and 6% phosphoric acid, and of the type II which consist of 5% nitrogen and 10% phosphoric acid. After testing the soil condition, he finds that he needs at least 14kg of nitrogen and 14 kg of phosphoric acid for his crop. If the type - I fertilizer costs 60 paise per kg and the type - II fertilizer costs 40 paise per kg, determine how many kilograms of each type of fertilizer should be used so that the nutrient requirement are met at a minimum cost. What is the minimum cost?

Answer :


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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