Compute Area of Bounded Region Defined by System of Inequalities

To determine the area of the region $R$ defined by the system of inequalities: $R = \left\{ (x, y) : 0 \le x \le \frac{9}{4}, \quad 0 \le y \le 1, \quad x \ge 3y, \quad x + y \ge 2 \right\}$

we first analyze the boundaries and identify the vertices of the region.

Step 1: Analyzing the 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. $x + y \ge 2 \implies y \ge 2 - x$

Notice that for $x \le \frac{9}{4}$, $y \le \frac{x}{3} \le \frac{9/4}{3} = \frac{3}{4} < 1$. Thus, the condition $y \le 1$ is automatically satisfied for all points in this region.

Step 2: Finding the Vertices of the Bounded Region

The bounded region is a quadrilateral with the following boundary lines:

1. $y = \frac{x}{3}$ (Upper boundary)
2. $y = 2 - x$ (Lower-left boundary)
3. $y = 0$ (Bottom boundary)
4. $x = \frac{9}{4}$ (Right vertical boundary)

Let's compute the intersection points (vertices) of these boundary lines:

• Intersection of $y = \frac{x}{3}$ and $y = 2 - x$: $\frac{x}{3} = 2 - x \implies \frac{4x}{3} = 2 \implies x = \frac{3}{2}$ $y = 2 - \frac{3}{2} = \frac{1}{2}$ This gives vertex $A\left(\frac{3}{2}, \frac{1}{2}\right)$.

• Intersection of $y = 2 - x$ and $y = 0$: $2 - x = 0 \implies x = 2$ This gives vertex $B(2, 0)$.

• Intersection of $x = \frac{9}{4}$ and $y = 0$: This gives vertex $C\left(\frac{9}{4}, 0\right)$.

• Intersection of $x = \frac{9}{4}$ and $y = \frac{x}{3}$: $y = \frac{1}{3} \left(\frac{9}{4}\right) = \frac{3}{4}$ This gives vertex $D\left(\frac{9}{4}, \frac{3}{4}\right)$.

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Step 3: Computing the Area via Definite Integrals

The region can be split into two sub-regions along $x$:

1. For $x \in \left[\frac{3}{2}, 2\right]$:

   • Upper boundary: $y = \frac{x}{3}$
   • Lower boundary: $y = 2 - x$

   $A_1 = \int_{3/2}^{2} \left( \frac{x}{3} - (2 - x) \right) dx = \int_{3/2}^{2} \left( \frac{4}{3}x - 2 \right) dx$ $A_1 = \left[ \frac{2}{3}x^2 - 2x \right]_{3/2}^{2} = \left( \frac{8}{3} - 4 \right) - \left( \frac{2}{3} \cdot \frac{9}{4} - 3 \right)$ $A_1 = -\frac{4}{3} - \left( \frac{3}{2} - 3 \right) = -\frac{4}{3} + \frac{3}{2} = \frac{1}{6}$

2. For $x \in \left[2, \frac{9}{4}\right]$:

   • Upper boundary: $y = \frac{x}{3}$
   • Lower boundary: $y = 0$

   $A_2 = \int_{2}^{9/4} \frac{x}{3} \, dx = \left[ \frac{x^2}{6} \right]_{2}^{9/4}$ $A_2 = \frac{1}{6} \left( \left(\frac{9}{4}\right)^2 - 2^2 \right) = \frac{1}{6} \left( \frac{81}{16} - 4 \right) = \frac{1}{6} \left( \frac{17}{16} \right) = \frac{17}{96}$

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Step 4: Total Area Calculation

Adding the two sub-areas together: $\text{Total Area} = A_1 + A_2 = \frac{1}{6} + \frac{17}{96} = \frac{16 + 17}{96} = \frac{33}{96} = \frac{11}{32}$

Thus, the area of the bounded region is $\frac{11}{32}$.

Correct Option: (A)