Solve the following Linear Programming Problem graphically:
Maximise $Z = 3x + 2y$
subject to the constraints:
$x + 2y \leq 10$
$3x + y \leq 15$
$x, y \geq 0$

The corner points of the feasible region determined by a system of linear constraints are $(0, 3)$, $(1, 1)$, and $(3, 0)$. If the objective function is $z = px + qy$, where $p, q > 0$, then the condition on $p$ and $q$ such that the minimum of $z$ occurs at both $(3, 0)$ and $(1, 1)$ is

The corner points of the feasible region are $A(0,0)$,$B(16,0)$,$C(8,16)$,and $D(0,24)$. The minimum value of the objective function $z = 300x + 190y$ is . . . . . . .

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

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 $\ldots$

For the objective function $Z = 4x + y$ subject to the constraints $x + y \leq 50$,$3x + y \leq 90$,$x \geq 0$,$y \geq 0$,whose corner points of the feasible region are $(0,0)$,$(30,0)$,$(20,30)$,and $(0,50)$,the maximum value of $Z$ is . . . . . . .