Solve the following linear programming problem graphically:
Minimise $Z = 200x + 500y$.......$(1)$
subject to the constraints:
$x + 2y \geqslant 10$.......$(2)$
$3x + 4y \leqslant 24$.....$(3)$
$x \geqslant 0, y \geqslant 0$......$(4)$

Consider the following statements:
Statement $(I)$: In a $LPP$,the objective function is always linear.
Statement $(II)$: In a $LPP$,the linear inequalities on variables are called constraints.
Which of the following is correct?

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 $2x + y \leq 10$,$x + 3y \leq 15$,$x, y \geq 0$ are $(0,0)$,$(5,0)$,$(3,4)$,and $(0,5)$. Let $Z = qx + py$ where $p, q > 0$. The condition on $p$ and $q$ such that the maximum of $Z$ occurs at both $(3,4)$ and $(0,5)$ is:

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: