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. For fabricating and finishing, the maximum labour hours available per week are 180 and 30 respectively. The company makes a profit of Rs80 on each piece of type A, and Rs120 on each piece of type B. How many pieces of type A and type B should be manufactured per week to get a maximum profit? Make it as an LPP and solve graphically. What is the maximum profit per week?

Let pieces of type A are manufactured per week and b pieces per week for type B

Profit on type A is 80rs per piece hence weekly profit is 80a. Similarly, profit on type B is 120rs per piece hence weekly profit is 120b

Total profit be z = 80a + 120b

Now we have to maximize profit which means we have to maximize z based on some constraints

Let us identify the constraints now

Constraint 1: labour hours for fabricating per week

Each piece of type A requires 9 labour hours of fabricating hence weekly there would be 9a hours of fabricating for type A

Now for type B 12 labour hours for each piece hence weekly for b pieces 12b labour hours of fabricating

The maximum labour hours for fabricating is given as 180hours weekly

Which means the total labour hours for fabricating of both type A and type B should not exceed 180

⇒ 9a + 12b ≤ 180

⇒ 3a + 4b ≤ 60 …(i)

Constraint 2: labour hours for finishing per week

Each piece of type A requires 1 labour hours of finishing hence weekly there would be 1a hours of finishing for type A

Now for type B 3 labour hours for each piece hence weekly for b pieces 3b labour hours of finishing

The maximum labour hours for finishing is given as a 30hours weekly

Which means the total labour hours for fabricating of both type A and type B should not exceed 30

⇒ a + 3b ≤ 30

⇒ a + 3b ≤ 30 …(ii)

Also, a, b represents the number of pieces manufactured per week hence it cannot be negative hence a ≥ 0 and b ≥ 0

Plot equations (i) and (ii)

Scale

On X-axis

1 cm = 2 pieces

On Y-axis

1 cm = 2 pieces

Now the corner points are O, A, B and C

Let us find values of z at these points

Hence the maximum value of z is 1680 at B(12, 6)

Hence for maximum profit, a = 12 pieces should be manufactured for type A and b = 6 pieces for type B per week

And the maximum profit is 1680rs