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 :

The feasible region for a linear programming problem is shown in the figure.

Let $Z = 3x - 4y$ be the objective function, then minimum of Z occurs at the point

The maximum value of $z = 3x + 4y$ subject to the constraints

$x + y \geq 2, 2x + 3y \leq 6, x \geq 0, y \geq 0$ is

For the L.P.P. Maximize z = 10x + 6y subjected to

3x + y ≤ 12,

2x + 5y ≤ 34,

x, y ≥ 0.

Then the feasible region represented by system of inequalities is

The feasible region for an LPP is shown by shaded region in the figure. Then the minimum value of Z = 11x + 7y is

The corner points of the bounded feasible region for an LPP are (0,4), (4,4), (6,6), (0,12). If the objective function is $Z = px + qy, p > 0, q > 0$, then the condition on p and q so that maximum of Z occurs at (6,6) and (0,12) is

If $z = 5x + 8y$ is the objective function of a LPP and (0, 0), (3, 1), (2, 4), (0, 3), (5, 0) are corner points of the bounded feasible region, then the maximum value of the objective function is