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 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:

For the $LP$ problem,"Maximize $z = x + 4y$ subject to $3x + 6y \leq 6$,$4x + 8y \geq 16$ and $x \geq 0, y \geq 0$."

The corner points of the feasible region determined by a 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 $q$ such that the minimum of $Z$ occurs at both $(3,0)$ and $(1,1)$ is $.....$

Show that the minimum of $Z$ occurs at more than two points.
Minimise and Maximise $Z = 5x + 10y$
subject to $x + 2y \leq 120, x + y \geq 60, x - 2y \geq 0, x, y \geq 0$.