Minimize $Z = 13x - 15y$ subject to the constraints: $x + y \leq 7$,$2x - 3y + 6 \geq 0$,$x \geq 0$,$y \geq 0$.

The solution set of the constraints $x + 2y \geq 11$,$3x + 4y \leq 30$,$2x + 5y \leq 30$,$x \geq 0$,$y \geq 0$ includes the point:

The minimum value of $Z = 3x + 4y$ subject to the constraints $x + y \leq 4, x \geq 0, y \geq 0$ 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?

Solve the following linear programming problem graphically:
Minimise $Z = 200x + 500y$.......$(1)$
subject to the constraints:
$x + 2y \geqslant 10$.......$(2)$
$3x + 4y \leqslant 24$.....$(3)$
$x \geqslant 0, y \geqslant 0$......$(4)$

The corner points of the bounded feasible region are $(60,0), (120,0), (60,40), (40,20)$ and $(20,30)$. For the objective function $z=5x+10y$:
$(i)$ Maximum value of $z$.
$(ii)$ Minimum value of $z$.
$(iii)$ Maximum value of $z$ occurs at.
$(iv)$ Minimum value of $z$ occurs at.