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 for minimizing the function $z = x + y$ under an $L$.$P$.$P$. with constraints $x + y \geqslant 2$,$x + 2y \leqslant 8$,$y \leqslant 3$,$x, y \geqslant 0$ is

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:

Consider a $LPP$ given by minimize $Z = 6x + 10y$. Subject to $x \geq 6, y \geq 2, 2x + y \geq 10, x \geq 0, y \geq 0$. Redundant constraints in this $LPP$ are $....$

$z = 30x - 30y + 1800$ is an objective function. The corner points of the feasible region are $(15, 0), (15, 15), (10, 20), (0, 20),$ and $(0, 15)$. $z$ has the minimum value at $\ldots$ point.

In the following figure,the feasible region (shaded) for a $LPP$ is shown. Determine the maximum and minimum value of $Z=x+2y$.