The corner points of the feasible region are $(0, 6)$,$(3, 3)$,$(9, 9)$,and $(0, 12)$. What is the maximum value of the objective function $z = 6x + 12y$?

The corner points of the feasible region of the objective function $Z = 3x + 9y$ are $(0, 10)$,$(5, 5)$,$(15, 15)$,and $(0, 20)$. Then,the minimum value of $Z$ 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:

$z = 30x - 30y + 1800$ 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.

The corner points of the feasible region determined by the system of linear constraints are $(0,10), (10,15), (15,25), (0,30)$. 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 $(15,25)$ and $(0,30)$ is . . . . . . .

$A$ feasible solution to an $LP$ problem . . . . . . .