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CBSE Class 12 Math 2009 Solved Paper

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Question : 29 of 29
Marks: +1, -0
A diet is to contain at least 80 units of Vitamin A and 100 units of minerals. Two foods F1F_1 and F2F_2 are available. Food F1F_1 cost Rs. 4 per unit and F2F_2 costs Rs. 6 per unit. One unit of food F1F_1 contains 3 units of Vitamin A and 4 units of minerals. One unit of food F2F_2 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 two foods and also meets the minerals nutritional requirements.
Solution:  
Let x be the number of units of food F1F_1 and y be the number of units of food F2F_2.
LPP is,
Minimize Z = 4x + 6y such that,
3x + 6y ≥ 80
4x + 3y ≥ 100
x , y ≥ 0
Representing the LPP graphically
Corner points are [0,1003]\left[0, \frac{100}{3}\right] , [24,43]\left[24, \frac{4}{3}\right] , [803,0]\left[\frac{80}{3}, 0\right]
PointCost = 4x + 6y
(0,1003)\left(0, \frac{100}{3}\right) 4 × 0 + 6 × 1003\frac{100}{3} = 0 + 200 = 200
(24,43)\left(24, \frac{4}{3}\right) 4 × 24 + 6 × 43\frac{4}{3} = 96 + 8 = 104
(803,0)\left(\frac{80}{3}, 0\right) 4 × 803\frac{80}{3} + 6 × 0 = 3203\frac{320}{3} + 0 = 106.67
From the table it is clear that, minimum cost is 104 and occurs at the point (24,43)\left(24, \frac{4}{3}\right).
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