Let x and y be number of units of products of A and B.
∴According to the question,
36x + 6y ≥ 108, 3x + 12y ≥ 36, 20x + 10y ≥ 100, x ≥ 0, y ≥ 0
Minimize Z = 20x + 40y
The feasible region determined 36x + 6y ≥ 108, 3x + 12y ≥ 36, 20x + 10y ≥ 100, x ≥ 0, y ≥ 0 is given by
The feasible region is unbounded. The corner points of feasible region are A(0,18) , B(2,6) , C(4,2) , D(12,0).The value of Z at corner points are
Corner Point |
Z = 20x + 40y |
|
A(0, 18) |
720 |
|
B(2, 6) |
280 |
|
C(4, 2) |
160 |
Minimum |
D(12, 0) |
240 |
|
The minimum value of Z is 160 at point (4,2).
Hence, the firm should buy 4 units of fertilizer A and 2 units of fertilizer B to achieve minimum expense of Rs.160.