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There are two types of fertilizers 'A' and 'B'. 'A' consists of 12% nitrogen and 5% Phosphoric acid. 'B' consists of 4% nitrogen and 5% Phosphoric acid. After testing the soil conditions, farmer finds that he needs at least 12 kg of nitrogen and 12 kg of Phosphoric acid for his crops. If  'A' costs ₹10 per kg and 'B' costs ₹ 8 per kg, then graphically determine how much of each type of fertilizer should be used so that nutrient requirements are at a minimum cost.

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Let x kg of fertilizer A and y kg of fertilizer B be used. As per the question :

Here cost is minimum at (30,210) and is ₹1,980. Since the region is unbounded, we have to draw

10x+8y <1980

:.  L : 10x+8y = 1980

Clearly open half plane has no common point with the feasible region is minimum value of  ₹1,980.

Farmer has to use 30 kg of fertilizer A and 210 kg of fertilizer B so that nutrient requirements are met at  a minimum cost.

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