The objective function is $\mathrm{Z}=5 \mathrm{x}+10 \mathrm{y}$ contraints are $x+2 y \leq 120, x+y \geq 60, x-2 y \geq 0, \quad x, y \geq 0$


(i) The line $\mathrm{x}+2 \mathrm{y}=120$ passes through $\mathrm{A}(120,0)$ and $\mathrm{B}$ $(0,60)$ putting $x=0, y=0$ is $x+2 y \leq 120$ we get $0 \leq 12$ which is true.
$\Rightarrow \mathrm{x}+2 \mathrm{y} \leq 120$ lies on $\mathrm{AB}$ and below $\mathrm{AB}$.
(ii) The line $\mathrm{x}+\mathrm{y}=60$ passes through $\mathrm{P}(60,0)$ and $\mathrm{B}(0,60)$ putting $x=0$ and $y=0$ in $x+y \geq 60$, we get $0 \geq 60$ which is not true.
$\Rightarrow x+y \geq 60$ lies on $P B$.
(iii) The line $x-2 y=0$ passes through $O(0,0)$ and $Q(120,60)$ putting $x=60, y=0$ is $x-2 y \geq 0,60 \geq 0$ which is true.
$\Rightarrow x-2 y \geq 0$ region is on $O Q$
(iv) $x \geq 0$ lies on $y$-axis and on its right.
(v) $y \geq 0$ lies on $x$-axis $\Rightarrow$ feasible region is PARS which has been shaded.
(a) Sloving $\mathrm{OQ}: \mathrm{x}-2 \mathrm{y}=0$ and $\mathrm{AB}: \mathrm{x}+2 \mathrm{y}=120$ $\Rightarrow x=60, y=30 \Rightarrow R$ is $(60,30)$
(b) Solving $O Q: x-2 y=0$ and $P B: x+y=60, x=40$, $y=20 \Rightarrow S$ is $(40,20)$
At A $(120,0), \quad \mathrm{Z}=5 \mathrm{x}+10 \mathrm{y}=600 \quad \mathrm{max}^{\mathrm{m}}$
At R $(60,30), \quad Z=300+300=600 \quad m_x x^m$
At $\mathrm{S}(40,20), \quad \mathrm{Z}=200+200=400$
At $\mathrm{P}(60,0), \quad \mathrm{Z}=300+0=300 \quad \mathrm{~min}^{\mathrm{m}}$.
$\Rightarrow$ Minimum value of $Z$ at $P$ is 300 at $\mathrm{P}(60,0)$ and maximum value of $Z$ is 600 at all points joining $A(120,0)$ and $\mathrm{R}(60,30)$.