The objective function $z = 4x + 3y$ can be maximised subject to the constraints $3x + 4y \leq 24$, $8x + 6y \leq 48$, $x \leq 5$, $y \leq 6$, $x \geq 0$, $y \geq 0$ :

The corner points of the feasible region determined by the system of linear constraints are (0, 10), (5, 5), (15, 15), (0, 20). Let $Z = ax + by$, where a, b > 0. Condition on a and b so that the maximizing value of Z occurs at both the points (15, 15) and (0, 20) is:

For the feasible reason of a LPP as shown, if the equation of OA and BC are $y - 2x = 0$ and $y - 2x = 4$ respectively than constraints for LPP are

For the LPP, Min $Z = 6x + 10y$ subject to $x \geq 6, y \geq 3, 2x + y \geq 10, x \geq 0, y \geq 0$, redundant constraint is :

The minimum value of $Z = 30x + 10y$ subject to the constraints $x + 2y \leq 30, 3x + y \geq 30, 4x + 3y \geq 60, x, y \geq 0$ is

If the objective function $z=10(x-7y+190)$ and the corner points of the bounded feasible region are A(0,4), B(0,5), C(3,5), D(5,3), E(5,0), F(4,0), then the minimum value of z will be:

Maximum value of Z equals to 40x + 50y subject to constraints $3x + y \leq 9$, $x + 2y \leq 8$, $x, y \geq 0$ occurs at