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Question
Reshma wishes to mix two types of food P and Q in such a way that the vitamin contents of the mixture contains at least 8 units of vitamin A and 11 units of vitamin B. Food P costs ₹60/kg and food Q costs ₹80/kg. Food P contains 3 units/kg of vitamin A and 5 units/kg of vitamin B while food Q contains 4 units/kg of vitamin A and 2 units/kg of vitamin B. Determine the minimum cost of the mixture.
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Solution
Let x units of food P and y units of food Q are mixed together to make the mixture.
The cost of food P is ₹60/kg and that of Q is ₹80/kg. So, x kg of food P and y kg of food Q will cost ₹(60x + 80y).
Since one kg of food P contains 3 units of vitamin A and one kg of food Q contains 4 units of vitamin A, therefore, x kg of food P and y kg of food Q will contain (3x + 4y) units of vitamin A. But, the mixture should contain atleast 8 units of vitamin A.
∴ 3x + 4y ≥ 8
Similarly, x kg of food P and y kg of food Q will contain (5x + 2y) units of vitamin B. But, the mixture should contain atleast 11 units of vitamin B.
∴ 5x + 2y ≥ 11
Thus, the given linear programming problem is
Minimise Z = 60x + 80y
subject to the constraints
3x + 4y ≥ 8
5x + 2y ≥ 11
x, y ≥ 0
The feasible region determined by the given constraints can be diagrammatically represented as,
The coordinates of the corner points of the feasible region are A \[\left( \frac{8}{3}, 0 \right)\] , B \[\left( 2, \frac{1}{2} \right)\] and C \[\left( 0, \frac{11}{2} \right)\]
The value of the objective function at these points are given in the following table.
| Corner Point | Z = 60x + 80y |
|
\[\left( \frac{8}{3}, 0 \right)\]
|
\[60 \times \frac{8}{3} + 80 \times 0 = 160\] → Minimum
|
|
\[\left( 2, \frac{1}{2} \right)\]
|
\[60 \times 2 + 80 \times \frac{1}{2} = 160\] → Minimum
|
|
\[\left( 0, \frac{11}{2} \right)\]
|
\[60 \times 0 + 80 \times \frac{11}{2} = 440\]
|
The smallest value of Z is 160 which is obtained at the points \[\left( \frac{8}{3}, 0 \right)\] and \[\left( 2, \frac{1}{2} \right)\]
It can be verified that the open half-plane represented by 60x + 80y < 160 has no common points with the feasible region.
So, the minimum value of Z is 160. Hence, the minimum cost of the mixture is ₹160.
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