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

The constraints are:

1. $-2x - 3y \leq -6$
2. $x - 2y \leq 2$
3. $3x + 2y \leq 12$
4. $-3x + 2y \leq 3$
5. $x, y \geq 0$

Rewriting constraint (1) by multiplying by $-1$:

$2x + 3y \geq 6$

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The boundary lines are:

1. $2x + 3y = 6$
2. $x - 2y = 2$
3. $3x + 2y = 12$
4. $-3x + 2y = 3$
5. $x = 0$, $y = 0$

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Corner Point A: Intersection of $2x + 3y = 6$ and $x - 2y = 2$

From $x - 2y = 2$:

$x = 2 + 2y$

Substituting in $2x + 3y = 6$:

$2(2 + 2y) + 3y = 6$

$4 + 4y + 3y = 6$

$7y = 2$

$y = \frac{2}{7}$

$x = 2 + 2(\frac{2}{7}) = \frac{18}{7}$

Point A $= (\frac{18}{7}, \frac{2}{7})$

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Corner Point B: Intersection of $x - 2y = 2$ and $3x + 2y = 12$

Adding both equations:

$4x = 14$

$x = \frac{7}{2}$

From $x - 2y = 2$:

$\frac{7}{2} - 2y = 2$

$2y = \frac{3}{2}$

$y = \frac{3}{4}$

Point B $= (\frac{7}{2}, \frac{3}{4})$

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Corner Point C: Intersection of $3x + 2y = 12$ and $-3x + 2y = 3$

Adding both equations:

$4y = 15$

$y = \frac{15}{4}$

From $-3x + 2y = 3$:

$-3x + 2(\frac{15}{4}) = 3$

$-3x + \frac{15}{2} = 3$

$-3x = -\frac{9}{2}$

$x = \frac{3}{2}$

Point C $= (\frac{3}{2}, \frac{15}{4})$

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Corner Point D: Intersection of $-3x + 2y = 3$ and $2x + 3y = 6$

From $2x + 3y = 6$:

$x = \frac{6 - 3y}{2}$

Substituting in $-3x + 2y = 3$:

$-3[\frac{6 - 3y}{2}] + 2y = 3$

$\frac{-18 + 9y}{2} + 2y = 3$

$-18 + 9y + 4y = 6$

$13y = 24$

$y = \frac{24}{13}$

$x = \frac{6 - 3(\frac{24}{13})}{2} = \frac{6}{13}$

Point D $= (\frac{6}{13}, \frac{24}{13})$

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The feasible region is confined to the first quadrant with $x \geq 0$ and $y \geq 0$. The four constraints form a closed region connecting the four corner points to form a quadrilateral.

Therefore, the feasible region is bounded and has 4 corner points.