The objective function is $\mathrm{Z}=5 \mathrm{x}+3 \mathrm{y}$ constraints are
$$
3 x+5 y \leq 15,5 x+2 y \leq 10, x \geq 0, y \geq 0
$$


(i) Consider the line $3 x+5 y=15$ which passes through $\mathrm{A}(5,0)$ and $\mathrm{B}(0,3)$ putting $\mathrm{x}=0, \mathrm{y}=0$ in $3 \mathrm{x}+5 \mathrm{y} \leq 15$ $\Rightarrow 0 \leq 15$ which is true.
$\therefore \quad$ Region $3 \mathrm{x}+5 \mathrm{y} \leq 15$ lies on and below $\mathrm{AB}$.
(ii) The line $5 x+2 y=10$ passes through $P(2,0)$ and $Q(0,5)$, Put $x=0, y=0$ in $5 x+2 y \leq 10$
$\therefore \quad 0 \leq 10$ which is true.
$\therefore$ Region $5 \mathrm{x}+2 \mathrm{y} \leq 10$ lies on and below $\mathrm{PQ}$.
(iii) $\mathrm{x} \geq 0$ Region lies on and to the right of $\mathrm{y}$-axis.
(iv) $y \geq 0$ lies on the above $x$-axis.
(v) The feasible region is the shaded area OPRB. Solving the equation $3 x+5 y=15$ and $5 x+2 y=10$
$\Rightarrow \mathrm{x}=\frac{20}{19}$ and $\mathrm{y}=\frac{45}{19}$
$\Rightarrow \mathrm{AB}$ and $\mathrm{PQ}$ intersect at $\mathrm{R}\left(\frac{20}{19}, \frac{45}{19}\right)$
At $\mathrm{P}(2,0) \mathrm{Z}=5 \mathrm{x}+3 \mathrm{y}=10+0=10$,
At $\mathrm{R}\left(\frac{20}{19}, \frac{45}{19}\right)$,
$\mathrm{Z}=\frac{100}{19}+\frac{135}{19}=\frac{235}{19}=12 \frac{7}{19}$
At B $(0,3) \mathrm{Z}=9$, AT $0(0,0) \mathrm{Z}=0$,
$\therefore$ Maximum value of $Z=\frac{235}{19}$ at $\left(\frac{20}{19}, \frac{45}{19}\right)$