For a linear programming problem,the objective function is $Z = 3x + 9y$. The corner points of the feasible region are $(0, 10), (5, 5), (15, 15),$ and $(0, 20)$. The maximum value of $Z$ is . . . . . . .

The minimum value of $Z = 3x + 4y$ subject to the constraints $x + y \leq 4, x \geq 0, y \geq 0$ is . . . . . . .

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

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 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 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 $(1,1)$.