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

(12, 6)

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 + 3y = 30

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)