1. **State the problem:** We have a triangle with two vertices fixed on the x-axis at points $(1,0)$ and $(5,0)$. The third vertex lies on the curve defined by $$y = \ln(2x) - \frac{1}{2}x + 5$$ for $$\frac{1}{2} \leq x \leq 8$$. We want to find the maximum area of this triangle. 2. **Formula for the area of a triangle with base on x-axis:** The base length is the distance between $(1,0)$ and $(5,0)$, which is $$5 - 1 = 4$$. The height is the y-coordinate of the third vertex on the curve at some $x$. Area formula: $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times y = 2y$$ 3. **Express area as a function of $x$:** $$A(x) = 2 \left( \ln(2x) - \frac{1}{2}x + 5 \right) = 2\ln(2x) - x + 10$$ 4. **Find critical points by differentiating $A(x)$:** $$A'(x) = 2 \times \frac{1}{2x} \times 2 - 1 = \frac{2}{x} - 1$$ Simplify: $$A'(x) = \frac{2}{x} - 1$$ 5. **Set derivative to zero to find maxima/minima:** $$\frac{2}{x} - 1 = 0 \implies \frac{2}{x} = 1 \implies x = 2$$ 6. **Check if $x=2$ is within the domain:** Given domain is $$\frac{1}{2} \leq x \leq 8$$, so $x=2$ is valid. 7. **Evaluate $A(x)$ at $x=2$ and endpoints to find maximum:** - At $x=2$: $$A(2) = 2\ln(4) - 2 + 10 = 2\ln(4) + 8$$ - At $x=\frac{1}{2}$: $$A\left(\frac{1}{2}\right) = 2\ln(1) - \frac{1}{2} + 10 = 0 - 0.5 + 10 = 9.5$$ - At $x=8$: $$A(8) = 2\ln(16) - 8 + 10 = 2\ln(16) + 2$$ 8. **Compare values:** - $A(2) = 2\ln(4) + 8$ - $A(\frac{1}{2}) = 9.5$ - $A(8) = 2\ln(16) + 2$ Since $\ln(4) \approx 1.386$, $2\ln(4) + 8 \approx 2.772 + 8 = 10.772$. $\ln(16) \approx 2.773$, so $2\ln(16) + 2 \approx 5.546 + 2 = 7.546$. The maximum area is at $x=2$ with value $$\boxed{2\ln(4) + 8}$$. **Final answer:** (C) $2 \ln 4 + 8$