Question

A manufacturing company makes two types of teaching aids A and B of mathematics for class XII. Each type of A requires 9 labor hours of fabricating and 1 labor hour for finishing. Each type of B requires 12 labors hour for fabricating and 3 labor hour for finishing. For fabricating and finishing, the maximum labor hours available per week are 180 and 30 respectively. The company makes a profit of ₹80 on each piece of type A and ₹120 on each piece of type B. how many pieces of type A and 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 LLP and solve graphically. What is the maximum profit per week?

Answer 1

Let the company make x no of 1^st type of teaching aid and y no of 2^nd type of teaching aid.

∴According to the question,

9x + 12y ≤ 180, x + 3y ≤ 30, x ≥ 0, y ≥ 0

Maximize Z = 80x + 120y

The feasible region determined by 9x + 12y ≤ 180, x + 3y ≤ 30, x ≥ 0, y ≥ 0 is given by

The corner points of feasible region are A(0,0) , B(0,10) , C(12,6) , D(20,0).

The value of Z at corner point is

Corner Point	Z = 80x + 120y
A(0, 0)	0
B(0, 10)	1200
C(12, 6)	1680	Maximum
D(20, 0)	1600

The maximum value of Z is 1680 and occurs at point (12,6).

The company should make 12 of 1^st type and 6 of 2^nd type of teaching aid. Maximum profit is Rs.1680.