A dietician wishes to mix two types of foods in such a way that vitamin contents of the mixture contain atleast 8 units of vitamin A and 10 units of vitamin C. Food ‘I’ contains 2 units/kg of vitamin A and 1 unit/kg of vitamin C. Food ‘II’ contains 1 unit/kg of vitamin A and 2 units/kg of vitamin C. It costs Rs. 50 per kg to purchase Food ‘I’ and Rs 70 per kg to purchase Food ‘II’. Formulate this problem as a linear programming problem to minimise the cost of such a mixture.

Let the mixture contain x kg. of Food ‘I’ and y kg. of Food ‘II’.
Clearly x ≥ 0, y ≥ 0. We make the following table from the given data:
Table

Since the mixture must contain at least 8 units of vitamin A and 10 units of vitamin C.
∴  2x + y ≥ 8
and    x + 2y ≥ 10
Total cost Z of purchasing x kg. of food ‘I’ and y kg. of Pood ‘II’ is
Z = 50x + 70y
∴   mathematical formulation of the problem is:
Minimise    Z = 50x + 70y
subject to the constraints:
2x + y ≥ 8
x + 2y ≥ 10
x, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Let us draw the graph of 2 x + y = 8.
For x = 0, y = 8
For y = 0, 2 x = 8 or x = 4
∴ line meets OX in A(4, 0) and OY in L(0, 8).
Again, we draw the graph of x + 2y = 10
For x = 0, 2 y = 10 or y = 5
For y = 0, x = 10
∴ line meets OX in B(10, 0) and OY in M(0, 5)
Since feasible region satisfies all the constraints.
∴ shaded region is the feasible region and it is unbounded.

Corner points are B(10, 0), C(2, 4), L(0, 8).
At B(10, 0), Z = 500 + 0 = 500
At C(2, 4), Z = 100 280 = 380
At L(0, 8), Z = 0 + 560 = 560
∴ smallest value of Z is 380 at (2, 4).
Since feasible region is unbounded.
∴ we are to check whether this value is minimum.
For this, we draw the graph of
50x + 70y < 380 or 5x + 7y < 38    ...(1)
Since (1) has no common point with the feasible region.
∴  minimum value = 380 at (2, 4).
∴ cost is minimum when the dietician mixed 2 kg. of food I with 4 kg. of food II. Minimum cost is Rs. 380.