The feasible region represented by the constraints $y - 2x \leq 4$, $x + y \geq 5$, $x \leq 4$, $y \geq 2$, and $x, y \geq 0$ is

The corner points of the feasible region determined by the system of linear constraints are $(0,0), (0,40), (20,40), (60,20), (60,0)$. The objective function is $z=4x+3y$. Compare the quantity in Column $A$ and Column $B$.

Column	Value
$A$. Maximum of $z$	$300$
$B$. Constant value	$325$

The corner points of the feasible region determined by the following system of linear inequalities: $2x + y \leq 10$,$x + 3y \leq 15$,$x, y \geq 0$ are $(0,0)$,$(5,0)$,$(3,4)$,and $(0,5)$. Let $Z = qx + py$,where $p, q > 0$. The condition on $p$ and $q$ so that the maximum of $Z$ occurs at both $(3,4)$ and $(0,5)$ is . . . . . . .

For the $LP$ problem,maximize $z = 2x + 3y$,the coordinates of the corner points of the bounded feasible region are $A(3, 3), B(20, 3), C(20, 10), D(18, 12),$ and $E(12, 12)$. The maximum value of $z$ is $\dots$

The corner points of the feasible region determined by a system of linear constraints are $(0, 3), (1, 1)$ and $(3, 0)$. If the objective function is $z = px + qy$ where $p, q > 0$, then the condition on $p$ and $q$ such that the minimum of $z$ occurs at both $(3, 0)$ and $(1, 1)$ is . . . . . . .

Corner points of the bounded feasible region for an $LP$ problem are $(0,4), (6,0), (12,0), (12,16)$ and $(0,10)$. Let $z = 8x + 12y$ be the objective function. Match the following:
$(i)$ Minimum value of $z$ occurs at $\ldots$
$(ii)$ Maximum value of $z$ occurs at $\ldots$
$(iii)$ Maximum of $z$ is $\ldots$
$(iv)$ Minimum of $z$ is $\ldots$