Question

A manufacturing company makes two types of teaching aids A and B of Mathematics for Class X. Each type of A requires $9$ labour hours for 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 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? Formulate this as Linear Programming Problem and solve it. Identify the feasible region from the rough sketch.

Answer 1

Let the pieces of type $A$ manufactured per week be $x$.

Let the pieces of type $B$ manufactured per week be $y$.

$\therefore$ maximum profit $z=80x+120y$

Fabricating hours for $A$ is $9$ and finishing hours is $1$

Fabricating hours for $B$ is $12$ and finishing hours is $3$

Maximum number of fabricating hours is $180$.

$\therefore 9x+12y\le 180$

$\Rightarrow 3x+4y\le 60$

Maximum number of finishing hours is $30$.

$\therefore x+3y\le 30$

Thus we can formulate the Linear Programming Problem as follows:

$maxz=80x+120y$ subjected to the constraints, $3x+4y\le 60,x+3y\le 30,x\ge 0,y\ge 0$

Solving the constraints we get $x=12,y=6$

Now the area of feasible region is as shown in the figure.

We can see that the points bounded by the feasible region are $A(0,10),B(12,6),C(20,0)$

Now let us calculate the maximum profit.

At $A(0,10),z=80\left(0\right)+120\left(10\right)=1200$

At $B(12,6),z=80\left(12\right)+120\left(6\right)=1680$

At $C(20,0),z=80\left(20\right)+120\left(0\right)=1600$

Hence the maximum profit is at $(12,6)$

Therefore $12$ pieces of type $A$ and $6$ pieces of type $B$ should be manufactured per week to get a maximum profit of $Rs.1680$