Solve the Linear Programming Problem graphically:
Maximise $Z = 3x + 4y$
subject to the constraints: $x + y \leq 4, x \geq 0, y \geq 0.$

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 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 . . . . . . .

The corner points of the feasible region determined by the system of linear constraints are $(0,10), (5,5), (15,15), (0,20)$. 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 $(15,15)$ and $(0,20)$ is $\ldots \ldots$

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:

The maximum value of $z = 4x + y$ subject to the constraints $x + y \leq 5$, $2x + y \leq 7$, $3x + 2y \leq 11$, $x \geq 0$, $y \geq 0$ is: