Question -
Answer -
Let the diet contain x units of food F1 andy units of food F2.Hence,
x ≥ 0 and y ≥ 0
The given information can be compiled in a table is given below
| 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 isRs 6 per unit
Hence, 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 can be writtenas
Minimise Z = 4x + 6y …………… (i)
Subject to the constraints,
3x + 6y ≥ 80 ………… (ii)
4x + 3y ≥ 100 ………. (iii)
x, y ≥ 0 …………. (iv)
The feasible region determined by the constraints is given below
We can see that the feasible region is unbounded.
A (80 / 3, 0), B (24, 4 / 3), and C (0, 100 / 3) are the cornerpoints
The values of Z at these corner points are given below
| Corner point | Z = 4x + 6y | |
| A (80 / 3, 0) | 320 / 3 = 106.67 | |
| B (24, 4 / 3) | 104 | Minimum |
| C (0, 100 / 3) | 200 | |
Here, the feasible region is unbounded, so 104 may or not be theminimum value of Z.
For this purpose, we draw a graph of the inequality, 4x + 6y< 104 or 2x + 3y < 52, and check whether the resulting half plane haspoints in common with the feasible region or not
Here, it can be seen that the feasible region has no commonpoint with 2x + 3y < 52
Hence, the minimum cost of the mixture will be Rs 104