The programming problem

Max $Z = 2x + 3y$ subject to the conditions $0 \leq x \leq 3, 0 \leq y \leq 4$ is :

The number of feasible solutions of the linear programming problem as

Maximum $z=15x+30y$ subject to the constraints

$3x+y \leq 12$

$x+2y \leq 10$

$x \geq 0, y \geq 0$

is

The corner points of the bounded feasible region determined by $x+3y\le60, x+y\ge10, x\le y, x\ge0, y\ge0$ are A(0, 10), B(5, 5), C(15, 15) and D(0, 20). If the objective function $z=ax+by$ has its maximum value on the line segment CD, then the relation between $a$ and $b$ is:

The maximum value of $z=5x+3y$ subjected to the constraints $2x+3y \ge 60, 2x+y \le 40; x,y \ge 0$ is:

The solution of LPP

max.(z) = $5x + 3y$ subject to

$2x + y \leq 6$

$x + y \leq 4$

$x \geq 0, y \geq 0$, is

For LPP: Maximize $z = 2x + 3y$ subject to the constraints

$x + y \geq 2$, $x + 2y \geq 3$, $x \geq 0$, $y \geq 0$,

which of the following graph represents the feasible region of the above LPP as shaded portion?

If $5x + y \leq 100$, $x + y \leq 60$, $x \geq 0$, $y \geq 0$. Then one of the corner points of the feasible region is :