Let $\mathrm{x}$ nuts and $\mathrm{y}$ bolts are produced.


Machine $A$ is used for $\mathrm{x} \times 1+\mathrm{y} \times 3$ hours, Maximum time avaliable $=12$ hours,
$$
\therefore \quad \mathrm{x}+3 \mathrm{y} \leq 12
$$
Machine $B$ is used for $3 \times x+1 \times y$ hours.
Maximum time avaliable $=12$ hours
$$
\Rightarrow 3 \mathrm{x}+\mathrm{y} \leq 12
$$
Profit function $\mathrm{Z}=(17.50) \times \mathrm{x}+(7.00) \mathrm{y}$. Thus the objective function $Z=17 \cdot 5 x+7 y$, subject to constraints are $x+3 y \leq 12$, $3 x+y \leq 12$ and $x, y \leq 0$
(i) The line $\mathrm{x}+3 \mathrm{y}=12$ passes though $\mathrm{A}(12,0), \mathrm{B}(0,4)$
Putting $x=0, y=0$ in $x+3 y \leq 12$, we get $0 \leq 12$ which is true. $\Rightarrow x+3 y \leq 12$ lies on and below $A B$.

(ii) $\mathrm{x}+3 \mathrm{y}=12$ passes thorugh $\mathrm{C}(4,0)$ and $\mathrm{D}(0,12)$ putting $x=0, y=0$ in $3 x+y \leq 12$, we get $0 \leq 12$ which is true. $3 \mathrm{x}+\mathrm{y} \leq 12$ lies on and below $C D$.
(iii) $x \geq 0$ is the region which lies on and to the right of $y$-axis.
(iv) $y \geq 0$ is the region which lies on and above the $x$-axis.
(v) The shaded area $\mathrm{BPCO}$ is the feasible region.
The point $\mathrm{P}$ is the intersection of the lines.
The point $\mathrm{P}$ is the intersection of the lines.
$\mathrm{AB}: \mathrm{x}+3 \mathrm{y}=12$ ....(i)
$\mathrm{CD}: 3 x+y=12$ ....(ii)
Multiplying equ. (i) by 3 and subtracting (ii)
from (i) $8 y=36-12=24, \therefore y=3$
from (i) $x=12-3 y=12-9=3$
$\therefore \quad$ The point $P$ is $(3,3)$ and $z=17 \cdot 5 x+7 y$

Maximum profit is $₹ 73-50$ when 3 nuts and 3 bolts packages are produced.