$A$ 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 $Rs.\,7$ 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?

(B) Let the manufacturer produce $x$ packages of nuts and $y$ packages of bolts.
Therefore,$x \geq 0$ and $y \geq 0$.
The given information can be compiled in a table as follows:

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

Corner point	$Z = 17.5x + 7y$
$O(0,0)$	$0$
$A(4,0)$	$70$
$B(3,3)$	$73.5$ (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 $Rs.\,73.50$.