Let the company manufacture x and y numbers of toys A and B.
∴According to the question,
5X + 8y ≤ 180, 10x + 8y ≤ 240, x ≥ 0, y ≥ 0
Maximize Z = 50x + 60y
The feasible region determined 5X + 8y ≤ 180, 10x + 8y ≤ 240, x ≥ 0, y ≥ 0 is given by
The corner points of feasible region are A(0,0) , B(0,22.5) , C(12,15) , D(24,0).
The value of Z at corner point is
Corner Point |
Z = 50x + 60y |
|
A(0, 0) |
0 |
|
B(0, 22.5) |
1350 |
|
C(12, 15) |
1500 |
Maximum |
D(24, 0) |
1200 |
|
The maximum value of Z is1500 and occurs at point (12,15).
The company should manufacture 12 A toys and 15 B toys to earn profit of rupees 1500.