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

The corner points of the feasible region are $(0, 6)$,$(3, 3)$,$(9, 9)$,and $(0, 12)$. What is the maximum value of the objective function $z = 6x + 12y$?

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

Corner points of the bounded feasible region for an $LP$ problem are $(0,4), (6,0), (12,0), (12,16)$ and $(0,10)$. Let $z = 8x + 12y$ be the objective function. Match the following:
$(i)$ Minimum value of $z$ occurs at $\ldots$
$(ii)$ Maximum value of $z$ occurs at $\ldots$
$(iii)$ Maximum of $z$ is $\ldots$
$(iv)$ Minimum of $z$ is $\ldots$

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 objective function of a Linear Programming Problem $(LPP)$ defined over a convex set attains its optimum value at: