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 number of package of nuts to be made is $X$ and number of package of bolts to be made is $Y$

Since, nuts package requires $1$ hours and bolts package requires $3$ hours of machine A work time. Also, there is maximum $12$ hours of machine A time available.

$\therefore X+3Y\le 12...\left(1\right)$

Since, nuts package requires $3$ hours and bolts package requires $1$ hours of machine B work time. Also, there is maximum $12$ hours of machine B time available.

$\therefore 3X+Y\le 12...\left(1\right)$

Since, count of package can't be negative.

$\therefore X\ge 0,Y\ge 0...\left(3\right)$

We have to maximize profit of the manufacturer.

Here, profit on nuts package is $17.5Rs$ and on bolts package is $7Rs$

So, objective function is $Z=17.5X+7Y$

Plotting all the constraints given by equation (1), (2) and (3), we got the feasible region as shown in the image.

Corner points	Value of $Z=17.5X+7Y$
A $(0,4)$	$28$
B $(3,3)$	$73.5$ (maximum)
C $(4,0)$	$70$

Hence, manufacturer should produce 3 packages of each to maximize his profit.