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 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? Make it as an LPP and solve graphically. What is the maximum profit per week?

Answer 1

	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, $9x+12y\le 180\Rightarrow 3x+4y\le 60$
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 $=30.$

$x+3y\le 30$

Where x is the number of hours spent on finishing aid A while 3y on aid B.

So, the LPP becomes:
$Z\left(maximise\right)=80x+120y$

Subject to $3x+4y\le 60$
$x+3y\le 30$
$x\ge 0$
$y\ge 0$

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.