Let x and y be number of kilograms of food X and Y.
∴According to the question,
x + 2y ≥ 10, 2x + 2y ≥ 12, 3x + y ≥ 8, x ≥ 0, y ≥ 0
Minimize Z = 6x + 10y
The feasible region determined x + 2y ≥ 10, 2x + 2y ≥ 12, 3x + y ≥ 8, x ≥ 0, y ≥ 0 is given by
The feasible region is unbounded. The corner points of feasible region are A(0,8) , B(1,5) , C(2,4) , D(10,0).
The value of Z at corner points are
Corner Point |
Z = 6x + 10y |
|
A(0, 8) |
80 |
|
B(1, 5) |
56 |
|
C(2, 4) |
52 |
Minimum |
D(10, 0) |
60 |
|
The minimum value of Z is 52 at point (2,4).
Hence, the diet should contain 2 kgs of food X and 4 kgs of food Y for the least cost of Rs. 52.