Reshma wishes to mix two types of food P and Q in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 11 units of vitamin B. Food P costs Rs 60/kg and Food Q costs Rs 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?
Solution
Let the mixture contain x kg of food P and y kg of food Q. Therefore,
x ≥ 0 and y ≥ 0
The given information can be compiled in a table as follows.
Vitamin A (units/kg) | Vitamin B (units/kg) | Cost (Rs/kg) | |
Food P | 3 | 5 | 60 |
Food Q | 4 | 2 | 80 |
Requirement (units/kg) | 8 | 11 |
The mixture must contain at least 8 units of vitamin A and 11 units of vitamin B. Therefore, the constraints are
3x + 4y ≥ 8
5x + 2y ≥ 11
Total cost, Z, of purchasing food is, Z = 60x + 80y
The mathematical formulation of the given problem is
Minimise Z = 60x + 80y … (1)
subject to the constraints,
3x + 4y ≥ 8 … (2)
5x + 2y ≥ 11 … (3)
x, y ≥ 0 … (4)
The feasible region determined by the system of constraints is as follows.
It can be seen that the feasible region is unbounded.
The corner points of the feasible region are .`A(8/3,0),B(2,1/2) and C (0, 11/2)`
As the feasible region is unbounded, therefore, 160 may or may not be the minimum value of Z.
For this, we graph the inequality, 60x + 80y < 160 or 3x + 4y < 8, and check whether the resulting half plane has points in common with the feasible region or not.
It can be seen that the feasible region has no common point with 3x + 4y < 8
Therefore, the minimum cost of the mixture will be Rs 160 at the line segment joining the points `(8/3,0) and (2,1/2)`.