Minimum and maximum $z = 5 x + 2 y$ subject to the following constraints:
$x - 2 y \leq 2$
$3 x + 2 y \leq 12$
$- 3 x + 2 y \leq 3$
$x \geq 0, y \geq 0$

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Solution

Point $A$
$3 x + 2 y = 12$ and $x - 2 y = 2$
$3 (2 + 2 y) + 2 y = 12$
$=> 6 + 8 y = 12, y = \frac{3}{4}$ , $x = \frac{7}{2}$
Point $B$
$- 3 x + 2 y = 3$ and $3 x + 2 y = 12$
$4 y = 15$ ,
$y = \frac{15}{4}$
Maximum value of $z = 5 \times \frac{7}{0} + 2 \times \frac{15}{4} = 25$
Minimum value of $z = 5 \times 0 + 2 \times 0 = 0$ .

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