Consider the following L.P.P. Max. $z = 5x+2y$, subject to $-2x-3y≤-6,x-2y≤ 2,3x + 2y ≤ 12,-3x + 2y ≤ 3$ and $x, y ≥ 0$ then

Its feasible region is bounded and has 4 corner points

The correct answer is Option (1) → Its feasible region is bounded and has 4 corner points

Problem: Maximize $z=5x+2y$ subject to

$2x+3y\ge \;6,\quad x-2y\le 2,\quad 3x+2y\le 12,\quad -3x+2y\le 3,\quad x\ge0,\ y\ge0$.

Find corner points (intersections of boundary lines):

Solve the pairs of equalities:

$\begin{aligned} &2x+3y=6,\quad x-2y=2 \quad\Rightarrow\quad \left(\frac{18}{7},\frac{2}{7}\right)\\[4pt] &2x+3y=6,\quad -3x+2y=3 \quad\Rightarrow\quad \left(\frac{3}{13},\frac{24}{13}\right)\\[4pt] &x-2y=2,\quad 3x+2y=12 \quad\Rightarrow\quad \left(\frac{7}{2},\frac{3}{4}\right)\\[4pt] &3x+2y=12,\quad -3x+2y=3 \quad\Rightarrow\quad \left(\frac{3}{2},\frac{15}{4}\right) \end{aligned}$

Verify these points satisfy all constraints. All four do, so the feasible polygon has vertices

$(\tfrac{3}{13},\tfrac{24}{13}),\;(\tfrac{18}{7},\tfrac{2}{7}),\;(\tfrac{7}{2},\tfrac{3}{4}),\;(\tfrac{3}{2},\tfrac{15}{4})$.