Question 2

The region is bounded by:

1. $y > \frac{1}{x}$
2. $y < \frac{5x - 1}{4}$ (from $5x - 4y - 1 > 0$)
3. $y < \frac{17 - 4x}{4}$ (from $4x + 4y - 17 < 0$)

Find intersection points:

• Between $y = \frac{1}{x}$ 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 $x > 0$, $x = 1$.
• Between $y = \frac{1}{x}$ and $y = \frac{17 - 4x}{4}$: $\frac{1}{x} = \frac{17 - 4x}{4} \implies 4x^2 - 17x + 4 = 0 \implies (4x - 1)(x - 4) = 0$. Intersection at $x = \frac{1}{4}$ and $x = 4$.
• Between $y = \frac{5x - 1}{4}$ and $y = \frac{17 - 4x}{4}$: $\frac{5x - 1}{4} = \frac{17 - 4x}{4} \implies 9x = 18 \implies x = 2$.

The region lies between $x = 1$ and $x = 4$. For $x \in [1, 2]$, upper bound is $\frac{5x-1}{4}$. For $x \in [2, 4]$, upper bound is $\frac{17-4x}{4}$. The lower bound is $\frac{1}{x}$.

Area $A = \int_1^2 \left( \frac{5x - 1}{4} - \frac{1}{x} \right) dx + \int_2^4 \left( \frac{17 - 4x}{4} - \frac{1}{x} \right) dx$ $A = \left[ \frac{5x^2}{8} - \frac{x}{4} - \log_e x \right]_1^2 + \left[ \frac{17x}{4} - \frac{x^2}{2} - \log_e x \right]_2^4$ $A = \left( (\frac{20}{8} - \frac{2}{4} - \log_e 2) - (\frac{5}{8} - \frac{1}{4} - 0) \right) + \left( (17 - 8 - \log_e 4) - (\frac{34}{4} - 2 - \log_e 2) \right)$ $A = (2 - \log_e 2 - \frac{3}{8}) + (9 - 2\log_e 2 - 6.5 + \log_e 2)$ $A = (\frac{13}{8} - \log_e 2) + (\frac{5}{2} - \log_e 2) = \frac{13 + 20}{8} - 2\log_e 2 = \frac{33}{8} - \log_e 4$.