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:

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

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 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:

The shaded region in the given figure is a graph of $.....$

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