Question

A company manufactures three kinds of calculators : A, B and C in its two factories I and II, The company has got an order for manufacturing at least $6400$ calculators of kind A, $4000$ of kind B and $4800$ of kind C. The daily output of factory I is of $50$ calculators of kind A, $50$ calculators of kind B, and $30$ calculators of kind C. The daily output of factory II is of $40$ calculators of kind A, $20$ of kind B and $40$ of kind C. The cost per day to run factory I is Rs. $12,000$ and of factory II is Rs. $15,000$.
How many days do the two factories have to be in operation to produce the order with the minimum cost? Formulate this problem as an LPP and solve it graphically.

Answer 1

Let factories I and II should be operated for x and y number of days respectively. Then the problem can be formulated as in L.P.P. as:

Minimise $Z=12000x+15000y$
Subject to constraints
$50x+40y\ge 6400$ i.e., $5x+4y\ge 640$
$50x+20y\ge 4000$ i.e., $5x+2y\ge 400$
$30x+40y\ge 4800$ i.e., $3x+4y\ge 480$
$x\ge 0,y\ge 0$

We draw the lines $5x+4y\ge 640,5x+2y\ge 400,3x+4y\ge 480$ and obtain the feasible region (unbounded and convex) shown shaded in the adjoining figure. the corner points are $A(0,200),B(32,120),C(80,60)$ and $D(160,0)$.

The values of Z at these points are $3000000,2184000,1860000$ and $1920000$ respectively. As the feasible region is unbounded, we draw the graph of the half plane.
$12000x+15000y<1860000$
i.e.,

$12x+15y<1860$ and note that there is no point common with the feasible region, therefore Z has minimum and minimum value is Rs. $1860000$.

It occurs at the point $(80,60)$ i.e., Factory I should be operated for $80$ days and factory II should be operated for $60$ days to minimise the cost.

Running Cost per day (in Rs )		$A$	$B$	$C$
$12000$	$I$	$50$	$50$	$30$
$15000$	$II$	$40$	$20$	$40$