The corner points of the bounded feasible region determined by the system of linear constraints 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 the maximum of Z occurs at both (5, 0) and (3, 4) is

For an LPP: Maximize $z = 3x + 9y$, $x \geq 0, y \geq 0$, the feasible region OAB is shown in the figure, then the other constraints are

The linear inequalities satisfying the shaded feasible region given in the figure are

(A) $x \geq 0$, $y \geq 0$, $2x + y \geq 2$

(B) $x \geq 0$, $y \geq 0$, $2x + y \leq 2$

(C) $x \geq 0$, $y \geq 0$, $2x + y \geq 2$, $x + 2y \leq 8$, $x - y \leq 1$

(D) $x + 2y \geq 8$, $x - y \geq 1$

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Which one of the following set of constraints depicted the shaded region given below?

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

If $z = 3x + 4y$ be the objective function of a of a linear programming problem (LPP) and (3, 1), (2, 4), (0, 4), (5, 0) be corner points of the bounded feasible region. Then the maximum value of objective function is

If corner points of a feasible region are (0, 0), (2, 0), $\left(\frac{20}{19}, \frac{45}{19}\right)$ and (0, 3), then

(A) Maximum value of $z = 5x + 3y$ is 10

(B) Minimum value of $z = 5x + 3y$ is 0

(C) Maximum value of $z = 5x + 3y$ is $\frac{235}{19}$ and minimum value is 0

(D) Maximum value of $z = 5x + 3y$ is 10 and minimum value is 0

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