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Asked by Apoorva Last Modified
Answer 
Vishaka Bhutia
Home Tutor
Let the diet contain x units of food F1 and y units of food F2. Therefore,
x ≥ 0 and y ≥ 0
The given information can be complied in a table as follows.
| Vitamin A (units) | Mineral (units) | Cost per unit (Rs) | |
| Food F1 (x) | 3 | 4 | 4 |
| Food F2 (y) | 6 | 3 | 6 |
| Requirement | 80 | 100 |
The cost of food F1 is Rs 4 per unit and of Food F2 is Rs 6 per unit. Therefore, the constraints are
3x + 6y ≥ 80
4x + 3y ≥ 100
x, y ≥ 0
Total cost of the diet, Z = 4x + 6y
The mathematical formulation of the given problem is
Minimise Z = 4x + 6y … (1)
subject to the constraints,
3x + 6y ≥ 80 … (2)
4x + 3y ≥ 100 … (3)
x, y ≥ 0 … (4)
The feasible region determined by the constraints is as follows.
It can be seen that the feasible region is unbounded.
The corner points of the feasible region are 
.
The corner points are
.
The values of Z at these corner points are as follows.
| Corner point | Z = 4x + 6y | |
|
|
| |
|
| 104 | → Minimum |
|
| 200 |
As the feasible region is unbounded, therefore, 104 may or may not be the minimum value of Z.
For this, we draw a graph of the inequality, 4x + 6y < 104 or 2x + 3y < 52, 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 2x + 3y < 52
Therefore, the minimum cost of the mixture will be Rs 104.
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