The corner points of the bounded feasible region for an LPP are $(0,5), (6,6), (9,9)$ and $(0,12)$. If the objective function is $Z = px + qy; p,q>0$, then the condition on $p$ and $q$ so that the maximum of $Z$ occurs at $(9,9)$ and $(0,12)$ is

The maximum value of $z = 4x + 3y$, if the feasible region for an LPP is as shown below is:

In a linear programming problem, the constraints on decision variables $x$ and $y$ are $y-2x \leq 0$, $y \geq 0$, $0 \leq x \leq 5$. The feasible region of the above problem:

For the LPP: minimize $z = 6x + 3y$ subject to the constraints

$4x + y \geq 80$

$x + 5y \geq 115$

$3x + 2y \leq 150$

$x \geq 0, y \geq 0$ then the minimum value of z is

The corner points of the feasible region of a LPP are (0,2), (3,0), (6,0), (6,8) and (0,5). If the objective function is $z = 4x + 6y$, then the sum of maximum and minimum value of z is equal to:

If the corner points of the bounded feasible region of a Linear Programming Problem are (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5), then the minimum value of the objective function $z = 2x+3y$ occurs at

The corner points of the feasible region determined by the following 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 $z = px + qy$, where $p, q > 0$ condition on p and q so that maximum of z occurs at both (3, 4) and (0, 5) is :