Let x and y be number of units of X and Y.
∴According to the question,
2x + y ≥ 8, x + 2y ≥ 10, x ≥ 0, y ≥ 0
Minimize Z = 5x + 7y
The feasible region determined 2x + y ≥ 8, x + 2y ≥ 10, x ≥ 0, y ≥ 0 is given by
The feasible region is unbounded. The corner points of feasible region are A(0,8) , B(2,4) , C(10,0).
The value of Z at corner points are
Corner Point |
Z = 5x + 7y |
|
A(0, 8) |
56 |
|
B(2, 4) |
38 |
Minimum |
C(10, 0) |
50 |
|
The minimum value of Z is 160 at point (4,2).
Hence, the dietician should mix 2 units of X and 4 units of Y to meet the requirements at minimum cost of Rs.38.