Solve the following L.P.P. graphically: Maximise Z =x+3 y subject to the constraints: x+2 y ≤ 200; x+ y ≤ 150; y ≤ 75; x, y ≥ 0

On plotting the graph of $\displaystyle x+2 \mathrm{y} \leq 200, x+\mathrm{y} \leq 150, \mathrm{y} \leq 75, \& x \geq 0, \mathrm{y} \geq 0$ we get the following graph and common shaded region is the region ABCDE . Now, Corner points of the common shaded region are $\displaystyle \mathrm{A}(0,75), B(50,75), C(100,50), D(150,0) \& E(0,0)$.Thus,

Corner points		Value of $\displaystyle \mathrm{Z}=x+3 \mathrm{y}$
\cline { $\displaystyle 1$ - $\displaystyle 1$ } \cline { $\displaystyle 3$ - $\displaystyle 3$ } $\displaystyle \mathrm{~A}(0,75)$		$\displaystyle 225$
$\displaystyle B(50,75)$		$\displaystyle 275$
$\displaystyle C(100,50)$		$\displaystyle 250$
$\displaystyle D(150,0)$		$\displaystyle 150$
$\displaystyle E(0,0)$		$\displaystyle 0$

So, Maximum Value of Z is $\displaystyle 275$ at $\displaystyle x=50 \& \mathrm{y}=75$.