Show that the minimum of $Z$ occurs at more than two points.
Maximize $Z = -x + 2y$,subject to the constraints:
$x \geq 3, x + y \geq 5, x + 2y \geq 6, y \geq 0$

(A) The feasible region determined by the constraints $x \geq 3, x + y \geq 5, x + 2y \geq 6,$ and $y \geq 0$ is unbounded.
The corner points of the feasible region are $A(6, 0), B(4, 1),$ and $C(3, 2)$.
Evaluating $Z = -x + 2y$ at these corner points:

Corner Point	$Z = -x + 2y$
$A(6, 0)$	$Z = -6 + 2(0) = -6$
$B(4, 1)$	$Z = -4 + 2(1) = -2$
$C(3, 2)$	$Z = -3 + 2(2) = 1$

Since the feasible region is unbounded,we check if $Z = -6$ is the minimum value by graphing the inequality $-x + 2y < -6$.
The line $-x + 2y = -6$ passes through $(6, 0)$ and $(4, -1)$. The region $-x + 2y < -6$ does not share any common points with the feasible region.
Thus,the minimum value of $Z$ is $-6$,which occurs at the point $A(6, 0)$. However,the question asks to show that the minimum occurs at more than two points. Re-evaluating the objective function $Z = -x + 2y$ with constraints,it is observed that the minimum value $-6$ is attained at all points on the line segment connecting $(6, 0)$ and $(8, 1)$ if the constraints were different,but here the minimum is unique at $A(6, 0)$. Given the prompt's instruction,we conclude the minimum value is $-6$.