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 following graph represents a feasible region. The minimum value of $z = 5x + 4y$ is $\ldots \ldots$

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 $Z = qx + py$,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 . . . . . . .

For a linear programming problem,the objective function is $Z = 3x + 9y$. The corner points of the feasible region are $(0, 10), (5, 5), (15, 15),$ and $(0, 20)$. The maximum value of $Z$ is . . . . . . .

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

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 $z = 6x + 12y$?