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Home » A housewife wishes to mix together two kinds of food, and , in such a way that the mixture contains at least 10 units of vitamin units of vitamin and 8 units of vitamin The vitamin contents of of each food are given below.
A housewife wishes to mix together two kinds of food, X and Y, in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C . The vitamin contents of 1 \mathrm{~kg} of each food are given below.
CBSE | Class 12 | Exercise 33B | Linear Programming | Maths | RS Aggarwal
A housewife wishes to mix together two kinds of food, X and Y, in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C . The vitamin contents of 1 \mathrm{~kg} of each food are given below.

If 1 \mathrm{~kg} of food \mathrm{X} cost \pm 6 and 1 \mathrm{~kg} of food \mathrm{Y} costs \pm 10, find the minimum cost of the mixture which will produce the diet.

Solution:

Let x and y be number of kilograms of food X and Y.
\therefore According to the question,
x+2 y \geq 10,2 x+2 y \geq 12,3 x+y \geq 8, x \geq 0, y \geq 0
Minimize Z=6 x+10 y
The feasible region determined x+2 y \geq 10,2 x+2 y \geq 12,3 x+y \geq 8, x \geq 0, y \geq 0 is given by

The feasible region is unbounded. The corner points of feasible region are A(0,8), B(1,5), C(2,4), D(10,0). The value of Z at corner points are

    \[\begin{tabular}{|l|l|l|} \hline Corner Point & $\mathrm{Z}=6 \mathrm{x}+10 \mathrm{y}$ & \\ \hline $\mathrm{A}(0,8)$ & 80 & \\ \hline $\mathrm{B}(1,5)$ & 56 & \\ \hline $\mathrm{C}(2,4)$ & 52 & \\ \hline $\mathrm{D}(10,0)$ & 60 & \\ \hline \end{tabular}\]

The minimum value of Z is 52 at point (2,4).
Hence, the diet should contain 2 kgs of food X and 4 kgs of food Y for the least cost of Rs. 52 .

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