Question

A manufacturer produces nuts and bolts. It takes $1$ hour of work on machine $A$ and $3$ hours on machine $B$ to produce a package of nuts. It takes $3$ hours on machine $A$ and $1$ hour on machine $B$ to produce a package of bolts. He earns a profit, of Rs.$17.50$ per package on nuts and Rs.$7.00$ per package on bolts. How many packages of each should be produced each day so as to maximise his profit, if he operates his machines for at the most $12$ hours a day?

Answer 1

Let the manufacturer produce $x$ package of nuts and $y$ packages of bolts.

Therefore, $x\ge 0$ and $y\ge 0$
The given information can be compiled in a table as follows:

	Nuts	Bolts	Availability
Machine A (h)	$1$	$3$	$12$
Machine B (h)	$3$	$1$	$12$

The profit on a package of nuts is Rs. $17.50$ and on a package of bolts is Rs. $7$.

Therefore, the constraints are
$x+3y\le 12$
$3x+y\le 12$
Total profit, $Z=17.5x+7y$
The mathematical formulation of the given problem is
Maximise $Z=17.5x+7y$ .........(1)
subject to the constraints
$x+3y\le 12$ .......(2)
$3x+y\le 12$ ......(3)
$x,y\ge 0$ .....(4)
The feasible region determined by the system of constraints is as shown.
The corner points are $A(4,0),B(3,3)$ and $C(0,4)$.
The values of $Z$ at these corner points are as follows:

Corner point	$Z=17.5x+7y$
$O(0,0)$	$0$
$A(4,0)$	$70$
$B(3,3)$	$73.5$	$\to$ Maximum
$C(0,4)$	$28$

The maximum value of $Z$ is Rs.$73.50$ at $(3,3)$.
Thus, $3$ packages of nuts and $3$ packages of bolts should be produced each day to get the maximum profit of Rs.$73.50$.