A manufacturing company makes two models $\mathrm{M}_1$ and $\mathrm{M}_2$ of a product. Each piece of $\mathrm{M}_1$ requires 9 labour hours for fabricating and one labour hour for finishing. Each piece of $\mathrm{M}_2$ require 12 labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available are 180 and 30 respectively. The company makes a profit of Rs. 800 on each piece of $\mathrm{M}_1$ and Rs. 1200 on each piece of $\mathrm{M}_2$
The maximum profit will be at the point
Answer & explanation
Correct answer: option 3
Z = 800x + 1200y
Constraints
$3 x+4 y \leq 60$
$x+3 y \leq 30$
$x_1 y \geq 0$ → solution in first quadrant
first plotting
3x + 4y =60
| x | 20 | 0 |
| y | 0 | 15 |
x + 3y = 30
| x | 30 | 0 |
| y | 0 | 15 |
for 3x + 4y ≤ 60
checking for O(0, 0)
⇒ 0 ≤ 60
⇒ solution lies to side of 3x + 4y = 60 containing (0, 0)
for x + 3y ≤ 30
checking for O(0, 0)
⇒ 0 ≤ 30
⇒ solution lies to side of x + 3y = 30 containing (0, 0)
Corner points obtained checking
A(0, 10) Z(x, y) = 800x + 1200y for points
B(12, 6) Z(10, 0) = 12000
C(20, 0) Z(12, 6) = 16800
D(0, 0) Z(20, 0) = 16000
Z(0, 0) = 0
Maximum profit is at point = (12, 6)