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Asked by Ananya Last Modified
Answer 
Vishaka Bhutia
Home Tutor
Let the farmer mix x bags of brand P and y bags of brand Q.
The given information can be compiled in a table as follows.
| Vitamin A (units/bag) | Vitamin B (units/bag) | Vitamin C (units/bag) | Cost (Rs/bag) | |
| Food P | 3 | 2.5 | 2 | 250 |
| Food Q | 1.5 | 11.25 | 3 | 200 |
| Requirement (units/bag) | 18 | 45 | 24 |
The given problem can be formulated as follows.
Minimize z = 250x + 200y … (1)
subject to the constraints,
The feasible region determined by the system of constraints is as follows.
The corner points of the feasible region are A (18, 0), B (9, 2), C (3, 6), and D (0, 12).
The values of z at these corner points are as follows.
| Corner point | z = 250x + 200y | |
| A (18, 0) | 4500 | |
| B (9, 2) | 2650 | |
| C (3, 6) | 1950 | → Minimum |
| D (0, 12) | 2400 |
As the feasible region is unbounded, therefore, 1950 may or may not be the minimum value of z.
For this, we draw a graph of the inequality, 250x + 200y < 1950 or 5x + 4y < 39, 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 5x + 4y < 39
Therefore, the minimum value of z is 1950 at (3, 6).
Thus, 3 bags of brand P and 6 bags of brand Q should be used in the mixture to minimize the cost to Rs 1950.
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