Solve the following LPP graphically:
Minimize $\mathrm{Z}=5\mathrm{x}+10\mathrm{y}$ subject to the constraints $\mathrm{x}+2\mathrm{y}\le 120,\mathrm{x}+\mathrm{y}\ge 60,\mathrm{x}-2\mathrm{y}>0$ and $\mathrm{x},\mathrm{y}\ge 0$

Our problem is to minimise
$\mathrm{z}=5\mathrm{x}+10\mathrm{y}\dots .. (\mathrm{i})$
$\mathrm{Subject} \mathrm{to} \mathrm{constraints}$
$\begin{array}{l}\mathrm{x}+2\mathrm{y}\le 120\dots \text{ (ii) }\\ \mathrm{x}+\mathrm{y}\ge 60\dots \text{ (iii) }\\ \mathrm{x}-2\mathrm{y}\ge 0\dots \text{ (iv) }\end{array}$
$\text{ and }\mathrm{x}\ge 0,\mathrm{y}\ge 0$
Table for line $\mathrm{x}+2\mathrm{y}=120$is 

$\mathrm{Put} (0,0) \mathrm{in} \mathrm{the} \mathrm{inequality} \mathrm{x}+2\mathrm{y}\le 120,\mathrm{we} \mathrm{get}$
$\begin{array}{l}0+2\left(0\right)\le 120\\ \Rightarrow 0\le 120\text{ (which is true) }\end{array}$
So, the half plane is towards the origin. Secondly, draw the graph of the line $\mathrm{x}+\mathrm{y}=60$
 

On putting (5,0) in the inequality x-2 y $\ge$0, we get
$5-2\left(0\right)\ge 0\Rightarrow 5\ge 0 (\mathrm{which} \mathrm{is} \mathrm{true})$
$\text{ Thus, the half plane is towards the }\mathrm{X}\text{-axis. Since, }\mathrm{x},\mathrm{y}\ge 0$
$\therefore$ The feasible region lies in the first quadrant.

Clearly, feasible region is ABCDA.
On solving equations x-2y=0 and x+y=60,
we get D(40,20) and on solving equations
$\mathrm{x}-2\mathrm{y}=0\text{ and }\mathrm{x}+2\mathrm{y}=120\text{, we get }\mathrm{C}(60,30)$ The corner points of the feasible region are A(60,0), B(120,0), C(60,30) and D(40,20). The values of Z at these points are as follows.

Clearly, the minimum value of Z is 300 at the points (60,0)