Maximize and minimize $Z = 3x - 4y$ subject to the constraints $x - 2y \leq 0$,$-3x + y \leq 4$,$x - y \leq 6$,and $x, y \geq 0$.

(A) We have to maximize and minimize $Z = 3x - 4y$ subject to the constraints:
$x - 2y \leq 0$
$-3x + y \leq 4$
$x - y \leq 6$
$x, y \geq 0$
These inequalities define a feasible region as shown in the figure. The feasible region is unbounded with corner points $O(0, 0)$,$A(12, 6)$,and $B(0, 4)$.

Corner points	Value of $Z = 3x - 4y$
$O(0, 0)$	$0$
$A(12, 6)$	$3(12) - 4(6) = 36 - 24 = 12$
$B(0, 4)$	$3(0) - 4(4) = -16$

Since the feasible region is unbounded,we must check if the minimum and maximum values exist.
$1$. For the minimum value: We check the inequality $3x - 4y < -16$. The open half-plane defined by $3x - 4y < -16$ has common points with the feasible region. Therefore,$Z$ has no minimum value.
$2$. For the maximum value: We check the inequality $3x - 4y > 12$. The open half-plane defined by $3x - 4y > 12$ has no common points with the feasible region. Therefore,the maximum value of $Z$ is $12$ at point $A(12, 6)$.