The corner points of the feasible region are $(0, 6)$,$(3, 3)$,$(9, 9)$,and $(0, 12)$. What is the maximum value of the objective function $z = 6x + 12y$?

The constraints $-x+y \leq 1, -x+3y \leq 9, x \geq 0, y \geq 0$ define a:

Solve the following problem graphically:
Minimise and Maximise $Z=3x+9y$......$(1)$
subject to the constraints:
$x+3y \leq 60$.....$(2)$
$x+y \geq 10$......$(3)$
$x \leq y$.......$(4)$
$x \geq 0, y \geq 0$......$(5)$

The corner points of the feasible region of the objective function $Z = 3x + 9y$ are $(0, 10)$,$(5, 5)$,$(15, 15)$,and $(0, 20)$. Then,the minimum value of $Z$ is:

Show that the minimum of $Z$ occurs at more than two points.
Minimise and Maximise $Z = 5x + 10y$
subject to $x + 2y \leq 120, x + y \geq 60, x - 2y \geq 0, x, y \geq 0$.

For the $LP$ problem,"Maximize $z = x + 4y$ subject to $3x + 6y \leq 6$,$4x + 8y \geq 16$ and $x \geq 0, y \geq 0$."