The feasible region for a linear programming problem is shown in the figure.

Let $Z = 3x - 4y$ be the objective function, then minimum of Z occurs at the point

While solving an LPP, the corner points of a bounded feasible region were found to be (0,0), (1, 9), (3, 7), (4, 5.5) and (10, 0). The objective function $Z = 4x + y$ will maximise at:

For the graph of a linear programming problem(LPP) given below. If $\min z = 3x+5y$ is the objective function and shaded portion in the figure is the feasible region of the LPP, then constraints other than $x, y \geq 0$ are

The corner points of the bounded feasible region determined by the system of linear constraints of an LPP are (0, 0), (0, 8), (5, 0) and (4, 10). Let $z = 3x - 4y$ be the objective function, then the minimum value of $z$ is:

The corner points of feasible region are:

A. (0, 240)

B. (0, 0)

C. (300, 0)

D. (120, 180)

E. (180, 120)

Choose the correct answer from the options given below:

The number of corner points of the feasible region determined by the constraints $2x - y \geq 0, x - 3y + 3 \geq 0, x \geq 0, y \geq 0$ is

The corner points of the feasible region determined by inequalities of LPP are $(4, 10)$, $(6, 8)$ and $(6, 5)$. Let $z = 3x + 4y$ be the objective function. Then the sum of maximum value of z and minimum value of z is :