Determine graphically the minimum value of the objective function
$Z = -50x + 20y$ .....$(1)$
subject to the constraints:
${2x - y \geqslant -5}$ .....$(2)$
${3x + y \geqslant 3}$ .....$(3)$
${2x - 3y \leqslant 12}$ .....$(4)$
${x \geqslant 0, y \geqslant 0}$ .....$(5)$

The corner points of the bounded feasible region are $(0,0), (2,0), (4,2), (2,4)$ and $(0, \frac{10}{3})$. For the objective function $z = -x + 2y$:
$(i)$ Maximum value of $z$ is at $\ldots \ldots \ldots$
$(ii)$ Minimum value of $z$ is at $\ldots \ldots \ldots$
$(iii)$ The maximum value of $z$ is $\ldots \ldots \ldots$
$(iv)$ The minimum value of $z$ is $\ldots \ldots \ldots$

$z = 30x - 30y + 1800$ is an objective function. The corner points of the feasible region are $(15, 0), (15, 15), (10, 20), (0, 20),$ and $(0, 15)$. $z$ has the minimum value at $\ldots$ point.

The objective function of a Linear Programming Problem $(L.P.P.)$ defined over a convex set attains its optimum value at

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:

Show that the minimum of $Z$ occurs at more than two points.
Maximize $Z = -x + 2y$,subject to the constraints:
$x \geq 3, x + y \geq 5, x + 2y \geq 6, y \geq 0$