Question

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 hour for finishing. For fabricating and 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 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?

Answer 1

$\text{Let x units of type A and y units of type B}\text{teaching aids be produced.}$
$\text{Then given L.P.P. is}$
$\text{Maximize z= 80x + 120y, subject to}$
$9x+12y\le 180x+3y\le 30,x\ge 0,y\ge 0$
$\text{1st constraint}$
$9x+12y\le 180$
$Whenx=0,y\le 15,$ $wheny=0,x\le 20$
$\text{Two point on it are A(0,15), B(20,0)}$
$\text{2nd constraint}$
$x+3y\le 30$
$\text{when x = 0, y}\le \text{10, when y = 0, x}\le \text{30}$
$\text{Two points on it are C(0,10) and D(30,0)}$
$\text{Plotting these constraint, we obtain the feasible region (shaded portion)}$
$\text{with corner points O(0,0), C(0,10), M(12,6) and B(20,0)}$



$AtC(0,10),Z=80\times 0+120\times 10=1200$
$AtM(12,6),Z=80\times 12+120\times 6=1680$
$AtB(20,0),Z=80\times 20+120\times 0=1600$
$\therefore MaxZ=1680atx=12,y=6$
$\text{Hence, 12 pieces of type A and 6 pieces of type B should}$
$\text{be prepared to get maximum profit of Rs. 1680 per week.}$