A manufacturing company makes two types of teaching aids A and B of Mathematics for Class XII. Each type of A requires 9 labour hours of fabricating and 1 labour hour for finishing. Each type of B requires 12 labour hours for fabricating and 3 labour hours for finishing. The maximum labour hours available per week are 180 and 30 respectively. The company makes a profit of Rs.80 on each piece of type A and Rs.120 on each piece of type B. How many pieces of type B should be manufactured per week to get a maximum profit?

Fabricating Hours Finishing Hours
A 9 1
B 12 3
Let pieces of type A manufactured per week = x
Let pieces of type B manufactured per week = y
Companies profit function which is to be maximized: Z = 80x + 120y as shown in the tabular column:
Constraints:
Maximum number of fabricating hours = 180
Therefore, 9x + 12y ≤ 180 ⇒ 3x + 4y ≤ 60
Where 9x is the fabricating hours spent by type A teacher aids, and 12y hours spent on type B and maximum number of finishing hours = 30.
x + 3y ≤ 30
Where x is the number of hours spent on finishing aid A while 3y on aid B.
So, the LPP becomes:
Z(maximise) = 80x + 120y
Subject to
3x + 4y ≤ 60
x + 3y ≤ 30
x ≥ 0
y ≥ 0
Solving it graphically:
Z = 80x + 120y
at (0, 15) = 1800
Z = 1200 at (0, 10)
Z = 1600 at (20, 0)
Z = 960 + 720 at (12, 6) = 1680
Maximum profit is at (0, 15).
Therefore, Teacher aid A = 0
Teacher aid B = 15 should be made.