The feasible region of a linear programming problem is bounded. The corresponding objective function is Z= 3x-4y. The objective function attains

The optimal solution of the Linear Programming problem

Maximize $Z = 3x_1 + 5x_2$,

s.t. $3x_1 + 2x_2 \leq 18$

$x_1 \leq 4$

$x_2 \leq 6$

$x_1 \geq 0, x_2 \geq 0$ is

The corner points of the feasible region associated with the LPP: Maximise $Z = px + qy$, $p,q > 0$ subject to $2x + y \leq 10$, $x + 3y \leq 15$, $x, y \geq 0$ are (0, 0), (5, 0), (3, 4) and (0, 5). If optimum value occurs at both (3, 4) and (0, 5), then

If the objective function for a LPP is $Z=3x+5y$ and the corner points of the bounded feasible region are (0, 0), (6, 0), (3, 5) and (0, 3), then the value of $|max(Z) - min(Z)|$ is:

The corner points of the feasible region determined by $x+y \leq 8, 2 x+y \geq 8, x \geq 0, y \geq 0$ are $A(0,8), B(4,0)$ and $C(8,0)$. If the objective function $Z=a x+$ by has its maximum value on the line segment $A B$, then the relation between $a$ and $b$ is :

The optimal value of the objective function of the LPP, Minimize $Z = 3x - 2y$ subject to constraints $x + y \ge 10$, $3x + 5y \ge 15$, $x \ge 0$, $y \ge 0$, is equal to:

Consider the following L.L.P.

Minimize z = 30x - 30y + 1800;

subject to x + y ≤ 30, x ≤ 15, y ≤ 20, x + y ≥ 15 and x, y ≥ 0.

Then it attains its optimal value at the point