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:

For a Linear Programming $(LP)$ problem,the objective function is $z = 3x + 2y$. The coordinates of the corner points of the bounded feasible region are $A(3, 3)$,$B(20, 3)$,$C(20, 10)$,$D(18, 12)$,and $E(12, 12)$. The minimum value of $z$ 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:

An oil company has two depots $A$ and $B$ with capacities of $7000 \, L$ and $4000 \, L$ respectively. The company is to supply oil to three petrol pumps,$D, E$ and $F$ whose requirements are $4500 \, L, 3000 \, L$ and $3500 \, L$ respectively. The distances (in $km$) between the depots and the petrol pumps are given in the following table:

From/To	$A$	$B$
$D$	$7$	$3$
$E$	$6$	$4$
$F$	$3$	$2$

Assuming that the transportation cost of $10 \, L$ of oil is $Rs. \, 1$ per $km$,how should the delivery be scheduled in order that the transportation cost is minimum? What is the minimum cost?

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 minimum value of $Z = 2x + 3y$ for the system of linear constraints: $2x + 4y \leq 12$,$x + y \leq 3$,$x \geq 0$,and $y \geq 0$ is . . . . . . .