CBSE Class 12 Math 2009 Solved Paper

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

F_1

and

F_2

are available. Food

F_1

cost Rs. 4 per unit and

F_2

costs Rs. 6 per unit. One unit of food

F_1

contains 3 units of Vitamin A and 4 units of minerals. One unit of food

F_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

F_1

and y be the number of units of food

F_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,{100}/3]

[24,4/3]

[{80}/3,0]

Point	Cost = 4x + 6y
$(0,{100}/3)$	4 × 0 + 6 × ${100}/3$ = 0 + 200 = 200
$(24,4/3)$	4 × 24 + 6 × $4/3$ = 96 + 8 = 104
$({80}/3,0)$	4 × ${80}/3$ + 6 × 0 = ${320}/3$ + 0 = 106.67

From the table it is clear that, minimum cost is 104 and occurs at the point

(24,4/3)

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