In a Linear Programming Problem ($L$.$P$.$P$.), the corner points of the feasible region determined by the constraints $3x - y \geq 6$, $x \leq 3$, $y \leq 2$, $y \geq 0$, and $x \geq 0$ are:

$A$ fruit grower can use two types of fertilizer in his garden,brand $P$ and brand $Q$. The amounts (in $kg$) of nitrogen,phosphoric acid,potash,and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least $240\,kg$ of phosphoric acid,at least $270\,kg$ of potash and at most $310\,kg$ of chlorine. If the grower wants to minimize the amount of nitrogen added to the garden,how many bags of each brand should be used? What is the minimum amount of nitrogen added in the garden (in $,kg$)?

	Brand $P$ ($kg$ per bag)	Brand $Q$ ($kg$ per bag)
Nitrogen	$3$	$3.5$
Phosphoric acid	$1$	$2$
Potash	$3$	$1.5$
Chlorine	$1.5$	$2$

The feasible region for a $LPP$ is shown in the following figure. Evaluate $Z = 4x + y$ at each of the corner points of this region. Find the minimum value of $Z$,if it exists.

The corner points of the bounded feasible region are $(0,1), (0,7), (2,7), (6,3), (6,0), (1,0)$. For the objective function $Z = 3x - y$:
$(i)$ At which point is $Z$ minimum?
$(ii)$ At which point is $Z$ maximum?
$(iii)$ The maximum value of $Z$ is $\ldots$
$(iv)$ The minimum value of $Z$ is $\ldots$

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

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: