The corner points of the feasible region are $(0, 6)$,$(3, 3)$,$(9, 9)$,and $(0, 12)$. What is the maximum value of the objective function $z = 6x + 12y$?

The corner points of the bounded feasible region are $(0,0), (2,0), (4,2), (2,4)$ and $(0, \frac{10}{3})$. For the objective function $z = -x + 2y$:
$(i)$ Maximum value of $z$ is at $\ldots \ldots \ldots$
$(ii)$ Minimum value of $z$ is at $\ldots \ldots \ldots$
$(iii)$ The maximum value of $z$ is $\ldots \ldots \ldots$
$(iv)$ The minimum value of $z$ is $\ldots \ldots \ldots$

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

The coordinates of the corner points of the bounded feasible region are $(0, 0), (0, 40), (20, 40), (60, 20), (60, 0)$. The maximum of the objective function $z = 40x + 30y$ is . . . . . . .

The minimum value of $Z = 3x + 4y$ subject to the constraints $x + y \leq 4, x \geq 0, y \geq 0$ is . . . . . . .

For the linear programming constraints $x+2y \geq 10$,$3x+4y \leq 24$,$x \geq 0$,and $y \geq 0$,which of the following is not a corner point of the feasible region?