#### Question

The region represented by the inequation system *x*, *y* ≥ 0, *y* ≤ 6, *x* + *y* ≤ 3 is

unbounded in first quadrant

unbounded in first and second quadrants

bounded in first quadrant

none of these

#### Solution

bounded in first quadrant

Converting the given inequations into equations, we obtain

\[y = 6, x + y = 3, x = 0 \text{ and }y = 0\] *y* = 6 is the line passing through (0, 6) and parallel to the X axis.The region below the line *y* = 6 will satisfy the given inequation.

The line *x* + *y* = 3 meets the coordinate axis at *A*(3, 0) and *B*(0, 3). Join these points to obtain the line *x* + *y* =3.

Clearly, (0, 0) satisfies the inequation *x* + *y* ≤ 3. So, the region in *xy*-plane that contains the origin represents the solution set of the given equation.

Region represented by *x* ≥ 0 and* y* ≥ 0:

Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations.

These lines are drawn using a suitable scale.The shaded region represents the feasible region of the given LPP, which is bounded in the first quadrant