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

For linear programming problem maximize $z=2x+3y$

Subject to the constraints $x+y \leq 4$

$x \leq 2$

$x \geq 0, y \geq 0$

Maximum value of z is

The corner points of the bounded feasible region determined by the system of linear constraints are $(0, 0)$, $(5, 0)$, $(6, 5)$, $(6, 8)$, $(4, 10)$, $(0, 8)$. Let $Z = 3x - 4y$ be the objective function. The minimum value of Z occurs at

If the objective function for an LPP is max.$(z) = 300x + 700y$ and the corner points for the bounded feasible region are $(6,0)$ $(5,0)$ $(0,6)$ $(4,4)$ and $(0,4)$, then the maximum values of z occurs at :

For the LPP

Maximise $z = x + y$

subject to $x - y \leq -1$, $-x + y \leq 2$, $x, y \geq 0$, $z$ has :

Let the corner points of the bounded feasible region of an LPP determined by a set of constraints are A(1, 5), B(3, 2) and C(2, 1). If the objective function $z=2ax+by$, $a,b>0$ has maximum values at points A and B, and z has minimum value 8 at point C, then the value of $4a+b$ is:

The region represented by the system of inequalities $x, y \geq 0$ ; $2x + 3y \geq 4$ ; $x \geq 1$ is :