Solve the following linear programming problem graphically: Maximise Z =2 x+3 y subject to the constraints: x+y ≤ 6; x ≥ 2; y ≤ 3; x, y ≥ 0

On plotting the graph of $\displaystyle x+y \leq 6, x \geq 2, y \leq 3, \& x \geq 0, y \geq 0$ we get the following graph and common shaded region is the region ABCD . Now, Corner points of the common shaded region are $\displaystyle \mathrm{A}(2,3), B(3,3), C(6,0) \& D(2,0)$.Thus,

Corner points		Value of $\displaystyle \mathrm{Z}=2 x+3 y$
\cline { $\displaystyle 1$ - $\displaystyle 1$ } \cline { $\displaystyle 3$ - $\displaystyle 3$ } $\displaystyle \mathrm{~A}(2,3)$		$\displaystyle 13$
$\displaystyle B(3,3)$		$\displaystyle 15$
$\displaystyle C(6,0)$		$\displaystyle 12$
$\displaystyle D(2,0)$		$\displaystyle 4$

So, Maximum Value of Z is $\displaystyle 15$ at $\displaystyle x=3, y=3$.