The corner points of the feasible region determined by the following system of linear inequalities: $2x + y \leq 10$,$x + 3y \leq 15$,$x, y \geq 0$ are $(0,0)$,$(5,0)$,$(3,4)$,and $(0,5)$. Let $Z = px + qy$,where $p, q > 0$. The condition on $p$ and $q$ so that the maximum of $Z$ occurs at both $(3,4)$ and $(0,5)$ 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

$A$ 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	$1.5$	$2$

For a Linear Programming $(LP)$ problem,the objective function is $z = 3x + 2y$. 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 minimum value of $z$ is . . . . . . .

The point at which the maximum value of $Z = 3x + 2y$ subject to the constraints $x + 2y \leq 2$,$x \geq 0$,$y \geq 0$ occurs is $.....$

The corner points of the feasible region determined by the system of linear inequalities are $(0,3), (1,1)$ and $(3,0)$. Let $Z = px + qy$ where $p, q > 0$. Find the condition on $p$ and $q$ such that the minimum of $Z$ occurs at both $(3,0)$ and $(1,1)$.