The region represented by the system of inequalities $x, y \geq 0, y \leq 6, x + y \leq 3$

The system of inequalities consists of:

$x \geq 0$

$y \geq 0$

$y \leq 6$

$x + y \leq 3$

All four inequalities must be satisfied simultaneously.

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The inequality $x \geq 0$ represents the region on or to the right of the y-axis.

The inequality $y \geq 0$ represents the region on or above the x-axis.

Together, $x \geq 0$ and $y \geq 0$ define the first quadrant.

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The inequality $y \leq 6$ represents the region on or below the horizontal line $y = 6$.

The inequality $x + y \leq 3$ represents the region on or below the line $x + y = 3$.

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To plot the boundary line $x + y = 3$:

When $x = 0$: $y = 3$, giving point $(0, 3)$

When $y = 0$: $x = 3$, giving point $(3, 0)$

The inequality $x + y \leq 3$ includes the region toward the origin.

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From $x + y \leq 3$, the maximum value of $y$ occurs when $x = 0$, giving $y = 3$.

Since $y \leq 3$ in the feasible region, the constraint $y \leq 6$ is automatically satisfied and does not restrict the region further.

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The feasible region forms a triangle with vertices:

$(0, 0)$ (origin)

$(3, 0)$ (x-axis intercept)

$(0, 3)$ (y-axis intercept)

All three vertices lie in the first quadrant.

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The triangular region is completely enclosed and can be contained within a circle.

Therefore, the region is bounded.

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The region is a bounded region in the first quadrant.