The feasible solution for a Linear Programming Problem $(LPP)$ is shown in the figure. Let $z = 3x - 4y$ be the objective function. The value of (Maximum value of $z$ + Minimum value of $z$) is equal to $....$

The constraints $-x_{1} + x_{2} \leq 1$,$-x_{1} + 3x_{2} \leq 9$,and $x_{1}, x_{2} \geq 0$ define:

For the $LP$ problem,minimize $z = 2x + 3y$,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 minimum value of $z$ is:

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

The corner points of the feasible region determined by the 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 $q$ such that the maximum of $Z$ occurs at both $(3,4)$ and $(0,5)$ is:

The corner points of the feasible region determined by the system of linear constraints are $(2, 72)$,$(15, 20)$,and $(40, 15)$. Let $Z = 6x + 3y$ be the objective function. The minimum value of $Z$ occurs at: