Area of Region Bounded by Hyperbola and Lines

To find the area of the region defined by

$$\mathcal{R} = \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, \, y > \frac{1}{x}, \, 5x - 4y - 1 > 0, \, 4x + 4y - 17 < 0 \right\}$$

we first rewrite the inequalities for $y$ in terms of $x$:

1. $y > \frac{1}{x}$ (lower boundary)
2. $y < \frac{5x - 1}{4}$ (upper boundary for a certain range of $x$)
3. $y < \frac{17 - 4x}{4}$ (upper boundary for another range of $x$)

Step 1: Finding the Points of Intersection

1. Intersection of $y = \frac{5x - 1}{4}$ and $y = \frac{17 - 4x}{4}$:

   $$\frac{5x - 1}{4} = \frac{17 - 4x}{4} \implies 5x - 1 = 17 - 4x \implies 9x = 18 \implies x = 2$$

   At $x = 2$, the value of $y$ is:

   $$y = \frac{5(2) - 1}{4} = \frac{9}{4}$$

2. Intersection of $y = \frac{1}{x}$ and $y = \frac{5x - 1}{4}$:

   $$\frac{1}{x} = \frac{5x - 1}{4} \implies 5x^2 - x - 4 = 0 \implies (5x + 4)(x - 1) = 0$$

   Since $x > 0$, we have $x = 1$.

3. Intersection of $y = \frac{1}{x}$ and $y = \frac{17 - 4x}{4}$:

   $$\frac{1}{x} = \frac{17 - 4x}{4} \implies 4x^2 - 17x + 4 = 0 \implies (4x - 1)(x - 4) = 0$$

   The region of interest has the boundary at $x = 4$.

Step 2: Setting up the Integrals

The upper boundary of the region is:

$$y_{\text{upper}} = \begin{cases} \frac{5x - 1}{4}, & \text{for } 1 \le x \le 2 \\ \frac{17 - 4x}{4}, & \text{for } 2 \le x \le 4 \end{cases}$$

The lower boundary of the region across $x \in [1, 4]$ is:

$$y_{\text{lower}} = \frac{1}{x}$$

Thus, the total area $A$ is given by:

$$A = \int_{1}^{2} \left( \frac{5x - 1}{4} - \frac{1}{x} \right) dx + \int_{2}^{4} \left( \frac{17 - 4x}{4} - \frac{1}{x} \right) dx$$

This can be written as:

$$A = \int_{1}^{2} \frac{5x - 1}{4} \, dx + \int_{2}^{4} \frac{17 - 4x}{4} \, dx - \int_{1}^{4} \frac{1}{x} \, dx$$

Step 3: Evaluating the Integrals

1. First Integral:

   $$\int_{1}^{2} \frac{5x - 1}{4} \, dx = \left[ \frac{5x^2}{8} - \frac{x}{4} \right]_1^2 = \left( \frac{20}{8} - \frac{2}{4} \right) - \left( \frac{5}{8} - \frac{1}{4} \right) = 2 - \frac{3}{8} = \frac{13}{8}$$

2. Second Integral:

   $$\int_{2}^{4} \left( \frac{17}{4} - x \right) dx = \left[ \frac{17x}{4} - \frac{x^2}{2} \right]_2^4 = (17 - 8) - \left( \frac{17}{2} - 2 \right) = 9 - \frac{13}{2} = \frac{5}{2} = \frac{20}{8}$$

3. Third Integral:

   $$\int_{1}^{4} \frac{1}{x} \, dx = \left[ \log_e x \right]_1^4 = \log_e 4$$

Step 4: Summing the Results

$$A = \frac{13}{8} + \frac{20}{8} - \log_e 4 = \frac{33}{8} - \log_e 4$$

Thus, the correct option is (B).