The corner points of the feasible region determined by the system of linear constraints are $(0,10), (5,5), (15,15), (0,20)$. 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 $(15,15)$ and $(0,20)$ is $\ldots \ldots$

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

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

Maximize the function $Z = 11x + 7y$,subject to the constraints:
$x \leq 3, y \leq 2, x \geq 0, y \geq 0$

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 . . . . . . .