Solve the Linear Programming Problem graphically:
Minimise $Z = 3x + 5y$
subject to the constraints:
$x + 3y \geq 3$
$x + y \geq 2$
$x, y \geq 0$

Corner points of the feasible region determined by the system of linear constraints are $(0,3)$,$(1,1)$ and $(3,0)$. Let $Z = px + qy$,where $p, q > 0$. The condition on $p$ and $q$ so that the minimum value of $Z$ occurs at $(3,0)$ and $(1,1)$ is . . . . . . .

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	$325$

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 $\text{Maximum of } F - \text{Minimum of } F$.

Minimize $Z = 13x - 15y$ subject to the constraints: $x + y \leq 7$,$2x - 3y + 6 \geq 0$,$x \geq 0$,$y \geq 0$.