Let x kg of food X and y kg of food Y are mixed together to make the mixture. Since one kg of food X contains one unit of vitamin A and one kg of food Y contains 2 units of vitamin A. Therefore, x kg of food X and ykg of food Y will contain x + 2y units of vitamin A. But the mixture should contain at least 10 units of vitamin A. Therefore,
x + 2y ≥ 10
Similarly, x kg of food X and y kg of food Y will produce 2x + 2y units of vitamin B and 3x + y units of vitamin C. But the minimum requirements of vitamins B and C are respectively of 12 and 8 units.
∴ 2x + 2y ≥ 12 and 3x + y ≥ 8
Since the quantity of food X and food Y cannot be negative.
∴ x ≥ 0, y ≥ 0
It is given that one kg of food X costs Rs 6 and one kg of food Y costs Rs 10. So, x kg of food X and y kg of food Y will cost Rs (6x = 10y).
Thus, the given linear programming problem is
Minimize Z = 6x + 10y
Subject to x + 2y ≥ 10
2x + 2y ≥ 12
3x + y ≥ 8
and, x ≥ 0, y ≥ 0
To solve this LPP, we draw the lines
x + 2y = 10, 2x + 2y = 12 and 3x + y = 8.
The feasible region of the LPP is shaded in fig.

The coordinates of the vertices (Corner-points) of shaded feasible regionA1P1P2B3 are A1 (10, 0), P1 (2, 4), P2 (1, 5) and B3 (0, 8). These points have been obtained by solving the equations of the corresponding intersecting lines, simultaneously.
The values of the objective function at these points are given in the following table.
Point (x, y)
|
Value of the objective function
Z = 6x + 10y
|
A1 (10, 0)
A2 (2, 4)
P2 (1, 5)
B3 (0, 8)
|
Z = 6 × 10 + 10 × 0 = 60
Z = 6 × 2 + 10 × 4 = 52
Z = 6 × 1 + 10 × 5 = 56
Z = 6 × 0 + 10 × 8 = 80
|
Clearly, Z is minimum at x = 2 and y = 4. The minimum value of Z is 52.
We observe that the open half-plane represented by 6x + 10y < 52 does not have points in common with the feasible region. So, Z has minimum value equal to 52.
Hence, the least cost of the mixture is Rs 52.