1. Understanding the Region

The given region is defined by the set of inequalities:

1. $0 \le x \le \frac{9}{4}$
2. $0 \le y \le 1$
3. $x \ge 3y \implies y \le \frac{x}{3}$
4. $x + y \ge 2 \implies y \ge 2 - x$

We need to find the area of the region that satisfies all these conditions simultaneously.

2. Visualizing the Region and Finding Vertices

Let's sketch the boundary lines:

• $y = x/3$ (a line through the origin)
• $y = 2 - x$ (a line with intercepts at (2,0) and (0,2))
• $x = 9/4$ (a vertical line)
• $y = 0$ (the x-axis)

The region is bounded from above by $y = x/3$ and from below by $y = \max(0, 2-x)$.

Let's find the key intersection points (vertices) of the boundary lines that define our area:

• Intersection of $y = x/3$ and $y = 2-x$: $\frac{x}{3} = 2 - x \implies x = 6 - 3x \implies 4x = 6 \implies x = \frac{3}{2}$ The corresponding y-value is $y = \frac{1}{3} \left(\frac{3}{2}\right) = \frac{1}{2}$. So, point A is $(3/2, 1/2).

• Intersection of $y = 2-x$ and $y=0$: $0 = 2 - x \implies x = 2$ So, point D is $(2, 0).

The lower boundary of the region is defined by $y = 2-x$ for $x \in [3/2, 2]$ and by $y=0$ for $x \in [2, 9/4]$. The upper boundary is $y=x/3$. The integration will be done from $x = 3/2$ to $x = 9/4$. We need to split the integral at $x=2$ because the lower boundary function changes.

3. Setting up the Area Integral

The area A can be calculated by integrating the difference between the upper and lower boundary functions with respect to $x$. We split the integral into two parts based on the change in the lower boundary function.

$A = \int_{3/2}^{2} (y_{upper} - y_{lower}) dx + \int_{2}^{9/4} (y_{upper} - y_{lower}) dx$

• For the interval $[3/2, 2]$, $y_{upper} = x/3$ and $y_{lower} = 2-x$.
• For the interval $[2, 9/4]$, $y_{upper} = x/3$ and $y_{lower} = 0$.

So the integral becomes: $A = \int_{3/2}^{2} \left( \frac{x}{3} - (2-x) \right) dx + \int_{2}^{9/4} \left( \frac{x}{3} - 0 \right) dx$ $A = \int_{3/2}^{2} \left( \frac{4x}{3} - 2 \right) dx + \int_{2}^{9/4} \frac{x}{3} dx$

4. Evaluating the Integrals

Part 1: $\int_{3/2}^{2} \left( \frac{4x}{3} - 2 \right) dx = \left[ \frac{4}{3} \frac{x^2}{2} - 2x \right]_{3/2}^{2} = \left[ \frac{2x^2}{3} - 2x \right]_{3/2}^{2}$ $= \left( \frac{2(2)^2}{3} - 2(2) \right) - \left( \frac{2(3/2)^2}{3} - 2(3/2) \right)$ $= \left( \frac{8}{3} - 4 \right) - \left( \frac{2(9/4)}{3} - 3 \right) = \left( \frac{8-12}{3} \right) - \left( \frac{3}{2} - 3 \right)$ $= -\frac{4}{3} - \left( -\frac{3}{2} \right) = -\frac{4}{3} + \frac{3}{2} = \frac{-8+9}{6} = \frac{1}{6}$

Part 2: $\int_{2}^{9/4} \frac{x}{3} dx = \left[ \frac{x^2}{6} \right]_{2}^{9/4}$ $= \frac{1}{6} \left[ \left(\frac{9}{4}\right)^2 - (2)^2 \right] = \frac{1}{6} \left( \frac{81}{16} - 4 \right)$ $= \frac{1}{6} \left( \frac{81 - 64}{16} \right) = \frac{1}{6} \left( \frac{17}{16} \right) = \frac{17}{96}$

5. Calculating Total Area

Now, we add the results from the two parts: $A = \frac{1}{6} + \frac{17}{96}$ To add these fractions, we find a common denominator, which is 96. $A = \frac{1 imes 16}{6 imes 16} + \frac{17}{96} = \frac{16}{96} + \frac{17}{96} = \frac{33}{96}$ Simplifying the fraction by dividing the numerator and denominator by 3: $A = \frac{11}{32}$

6. Conclusion

The area of the given region is $\frac{11}{32}$. This corresponds to option A.