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

Let $x$ and $y$ be optimal solutions of a Linear Programming $(LP)$ problem. Then,which of the following is true?

Show that the minimum of $Z$ occurs at more than two points.
Minimise and Maximise $Z = 5x + 10y$
subject to $x + 2y \leq 120, x + y \geq 60, x - 2y \geq 0, x, y \geq 0$.

For the $LP$ problem,maximize $z = 2x + 3y$,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 maximum value of $z$ is $\dots$

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:

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?