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A diet of a sick personmust contains at least 48 units of vitamin A and 64 units of vitamin B. Two foods F_(1) and F_(2) are available. Food F_(1)costs Rs. 6 per unit andFood F_(2)costs Rs. 10 per unit. Oneunit of food F_(1)contains 6 units of vitamin A and 7 units of vitamin B. Oneunit of food F_(2) contains 8 units of vitamin A and12 unitsofvitamin B. Find the minimum cost for thediet that consists ofmixture of these two foodsand also meeting theminimum nutritionalrequirements. |
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Answer» Solution :Let x units of food `F_(1)` andy units of food `F_(2)` be included in the diet of the sick person. Then their total costis Z Rs. `(6x + 10y)` We have to minimize the above cost function . Theconstraints are as per given the following table. `{:(,"Food "F_(1),"Food "F_(2),"Minimum"),(,(x),(y),"requirement"),("Vitamin A",6,8," "48),("Vitamin C",7,12," "64):}` Hence the CONSTRAINTS are ` 6x + 8y ge 48` ` 7x + 12y ge 64` Also the no. of units of food ` F_(1) and F_(2)` cannot be negative. ` x ge 0 and y ge 0` Hence themathematical formulation of given LPP is Minimize `Z = 6x + 10y` Subject to ` 6x + 8y ge 48` ` 7x + 12y ge 64` ` rArrx/(64//7) + y/(16/3) = 1` Plot these equations on GRAPH paper we get feasible region shaded on graph paper. ![]() The vertices of the feasible region are ` C(64/7, 0), P ` and B (0, 6). P is the point of intersection of these lines ` 6x + 8y = 48` and ` 7x + 12y = 64` On solving these equations we get `P-=(4,3)` . Values of OBJECTIVE function at these vertices are ` Z(C) = 6 (64/7) + 10 (0) = (384)/7` ` = 54*85` ` Z(P) = 6(4) + 10(3)24 + 30 = 54` ` Z(B) = 6(0)+10(6) = 60` Minimum value of Z is 54 at point P(4,3) hence, 4units of food `F_(1) and 3` units of food `F_(2)` should be included in the diet of sick person to meet theminimal nutritional requirements, in order to have the minimum cost of Rs. 54. |
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