Q6 of 118 Page 30

A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F1 and F2 are available. Food F2 costs ₹ 4 per unit F2 costs ₹ 6 per unit one unit of food F1 contains 3 units of vitamin A and 4 units of minerals. One unit of food F2 contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem and find graphically the minimum cost for diet that consists of mixture of these foods and also meets the mineral nutritional requirements.

The above information can be expressed in the following table:



Let the quantity of the foods F1 and F2 be ‘x’ and ‘y’ respectively.


Cost of food F1 = 4x


Cost of food F2 = 6y


Cost of Diet = 4x + 6y


Now,


⟹ 3x + 6y ≥ 80


i.e. the minimum requirement of Vitamins from the two foods is 80units, each of which contains 3units and 6units of Vitamins.


⟹ 4x + 3y ≥ 100


i.e. the minimum requirement of minerals firm the two foods is 100units, each of which contains 4unit and 3 units of vitamins.


Hence, mathematical formulation of the LPP is as follows:


Find ‘x’ and ‘y’ that:


Minimise Z = 4x + 6y


Subject to the following constraints:


(i) 3x + 6y ≥ 80


(ii) 4x + 3y ≥ 100


(iii) x,y ≥ 0 (∵ quantity cant be negative)



The feasible region is unbounded.


The corner points of the feasible region is as follows:



Z is smallest at


Let us consider 4x + 6y ≤ 104


As it has no intersection with the feasible region, the smallest value is the minimum value.


The minimum cost of diet is ₹104.


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A dietician mixes together two kinds of food in such a way that the mixture contains at least 6 units of vitamin A, 7 units of vitamin B, 11 units of vitamin C and 9 units of vitamin D. The vitamin contents of 1 kg of food X and 1 kg of food Y are given below :


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