The corner points of the feasible region of an $LPP$ are $(0,2), (3,0), (6,0), (6,8)$ and $(0,5)$. Then the minimum value of $z = 4x + 6y$ occurs at:

The feasible solution for a $LPP$ is shown in the figure. Let $z=3x-4y$ be the objective function. The maximum value of $Z$ occurs at $......$

The solution set of the constraints $x + 2y \geq 11$,$3x + 4y \leq 30$,$2x + 5y \leq 30$,$x \geq 0$,$y \geq 0$ includes the point:

The constraints $-x_{1} + x_{2} \leq 1$,$-x_{1} + 3x_{2} \leq 9$,and $x_{1}, x_{2} \geq 0$ define:

Solve the Linear Programming Problem graphically:
Minimise $Z = 3x + 5y$
subject to the constraints:
$x + 3y \geq 3$
$x + y \geq 2$
$x, y \geq 0$

The corner points of the feasible region determined by a system of linear constraints are $(0, 3)$, $(1, 1)$, and $(3, 0)$. If the objective function is $z = px + qy$, where $p, q > 0$, then the condition on $p$ and $q$ such that the minimum of $z$ occurs at both $(3, 0)$ and $(1, 1)$ is