The maximum value of $Z = 60x + 10y$ whose corner points are $(10, 0)$,$(2, 4)$,$(1, 5)$,and $(0, 8)$ is . . . . . . .

The feasible region for an $LPP$ is shown in the figure. Let $z = 3x - 4y$ be the objective function. The minimum value of $Z$ is:

In a Linear Programming Problem ($L$.$P$.$P$.), the corner points of the feasible region determined by the constraints $3x - y \geq 6$, $x \leq 3$, $y \leq 2$, $y \geq 0$, and $x \geq 0$ are:

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:

The feasible region represented by the constraints $y - 2x \leq 4$, $x + y \geq 5$, $x \leq 4$, $y \geq 2$, and $x, y \geq 0$ is

The maximum value of $Z=x+3y$ subject to the constraints $2x+y \leq 20$,$x+2y \leq 20$,$x \geq 0$,$y \geq 0$ is: