Corner points of the feasible region for an $\operatorname{LPP}$ are $(0,2), (3,0), (6,0), (6,8)$ and $(0,5)$. Let $F = 4x + 6y$ be the objective function. Find the value of $\text{Maximum of } F - \text{Minimum of } F$.

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 $Z = px + qy$,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:

Determine the maximum value of $Z=11 x+7 y$ subject to the constraints:
$2 x+y \leq 6, x \leq 2, x \geq 0, y \geq 0$

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

Consider a $LPP$ given by minimize $Z = 6x + 10y$. Subject to $x \geq 6, y \geq 2, 2x + y \geq 10, x \geq 0, y \geq 0$. Redundant constraints in this $LPP$ are $....$

The vertices of the feasible region determined by some linear constraints are $(0,2), (1,1), (3,3), (1,5)$. Let $Z = px + qy$ where $p, q > 0$. The condition on $p$ and $q$ so that the maximum of $Z$ occurs at both the points $(3,3)$ and $(1,5)$ is . . . . . . .