If a Linear Programming Problem $(L.P.P.)$ has optimum solutions at two consecutive corner points of the feasible region,then the $L.P.P.$ has:

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 corner points of the feasible region determined by the system of linear constraints are $(0,0), (0,40), (20,40), (60,20), (60,0)$. The objective function is $z=4x+3y$. Compare the quantity in Column $A$ and Column $B$.

Column	Value
$A$. Maximum of $z$	$300$
$B$. Constant value	$325$

Value of the objective function $Z = -50x + 20y$ subject to the constraints $2x - y \geq -5$,$3x + y \geq 3$,$2x - 3y \leq 12$,$x \geq 0$,$y \geq 0$. The corner points of the feasible region are $(0, 5)$,$(0, 3)$,$(1, 0)$,and $(6, 0)$. At which point is the value of $Z$ minimum?

Show that the minimum of $Z$ occurs at more than two points.
Maximize $Z = x + y$,subject to $x - y \leq -1$,$-x + y \leq 0$,$x, y \geq 0$.

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 $Z = qx + py$,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 . . . . . . .