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 $q$ such that the minimum of $z$ occurs at both $(3,0)$ and $(1,1)$ is:

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:

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 $\ldots \ldots \ldots$

Show that the minimum of $Z$ occurs at more than two points.
Minimise and Maximise $Z = x + 2y$
subject to $x + 2y \geq 100, 2x - y \leq 0, 2x + y \leq 200; x, y \geq 0$.

The coordinates of the corner points of the bounded feasible region are $(0, 0), (0, 40), (20, 40), (60, 20), (60, 0)$. The maximum of the objective function $z = 40x + 30y$ is . . . . . . .

The feasible region for a $LPP$ is shown in the figure. Find the maximum value of $Z=11x+7y$.