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

In solving the $LP$ problem: "Minimize $z = 6x + 10y$ subject to $x \geq 6, y \geq 2, 2x + y \geq 10, x \geq 0, y \geq 0$." The redundant constraints are $....$

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

The corner points of the feasible region determined by the following system of linear inequalities: $2x + y \leq 10$,$x + 3y \leq 15$,$x, y \geq 0$ are $(0,0)$,$(5,0)$,$(3,4)$,and $(0,5)$. Let $Z = px + qy$,where $p, q > 0$. The condition on $p$ and $q$ so that the maximum of $Z$ occurs at both $(3,4)$ and $(0,5)$ 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	$325$

The corner points of the feasible region are $(0,10), (5,5), (15,15), (0,20)$. The maximum value of $Z = 3x + 9y$ is . . . . . . .