In a Linear Programming Problem ($L$.$P$.$P$.), the corner points of the feasible region defined by the constraints $3x - y \geq 6$, $x \leq 3$, $y \leq 2$, $y \geq 0$, and $x \geq 0$ are:

The feasible region (shaded) for a $LPP$ is shown in the adjacent figure. Maximize $Z = 5x + 7y$.

For a Linear Programming Problem $(LPP)$,if the objective function is $Z = 4x + 3y$ and the corner points of the bounded feasible region are $(0,0), (25,5), (16,16),$ and $(5,24)$,then the maximum value of $Z$ occurs at the point . . . . . . .

The corner points of the bounded feasible region are $(0,0), (2,0), (4,2), (2,4)$ and $(0, \frac{10}{3})$. For the objective function $z = -x + 2y$:
$(i)$ Maximum value of $z$ is at $\ldots \ldots \ldots$
$(ii)$ Minimum value of $z$ is at $\ldots \ldots \ldots$
$(iii)$ The maximum value of $z$ is $\ldots \ldots \ldots$
$(iv)$ The minimum value of $z$ is $\ldots \ldots \ldots$

The corner points of the feasible region determined by the system of linear inequalities are $(0,3), (1,1)$ and $(3,0)$. Let $Z = px + qy$ where $p, q > 0$. Find the condition on $p$ and $q$ such that the minimum of $Z$ occurs at both $(3,0)$ and $(1,1)$.

Solve the following Linear Programming Problem graphically:
Minimise $Z = -3x + 4y$
Subject to the constraints:
$x + 2y \leq 8$
$3x + 2y \leq 12$
$x \geq 0, y \geq 0$