Question -
Answer -
Let the mixture contain x kg of food P and y kg of food Q.
Hence, x ≥ 0 and y ≥0
The given information can be compiled in a table as given
| 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 11units of vitamin B. Hence, the constraints are
3x + 4y ≥ 8
5x + 2y ≥ 11
Total cost of purchasing food is, Z = 60x + 80y
So, the mathematical formulation of the given problem can bewritten as
Minimise Z = 60x + 80y (i)
Now, 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 isgiven below
Clearly, we can see that the feasible region is unbounded
A (8 / 3, 0), B (2, 1 / 2) and C (0, 11 / 2)
The values of Z at these corner points are given below
| Corner point | Z = 60x + 80 y | |
| A (8 / 3, 0) | 160 | Minimum |
| B (2, 1 / 2) | 160 | Minimum |
| C (0, 11 / 2) | 440 | |
Here the feasible region is unbounded, therefore, 160 may or maynot be the minimum value of Z.
For this purpose, we graph the inequality, 60x +80y <160 or 3x +4y <8, and check whether the resulting half plane has points in common with thefeasible region or not
Here, it can be seen that the feasible region has no commonpoint with 3x + 4y < 8
Hence, at the line segment joining the points (8 / 3, 0) and (2,1 / 2), the minimum cost of the mixture will be Rs 160