1. **State the problem:** We are given the function $$f(n) = n \ln\left(\frac{10}{n}\right)$$ where $n$ is the original file size in MB, and we want to find the absolute extreme values (maximum and minimum) of $f(n)$ on the interval $[1, 10]$. 2. **Find the derivative:** To find extreme values, we first compute the derivative $f'(n)$. $$f(n) = n \ln\left(\frac{10}{n}\right) = n (\ln 10 - \ln n) = n \ln 10 - n \ln n$$ Using the product and chain rules: $$f'(n) = \ln 10 - \ln n - 1$$ 3. **Find critical points:** Set $f'(n) = 0$ to find critical points. $$\ln 10 - \ln n - 1 = 0 \implies \ln 10 - 1 = \ln n \implies n = e^{\ln 10 - 1} = \frac{10}{e}$$ Since $\frac{10}{e} \approx 3.678$ lies in $[1,10]$, this is a valid critical point. 4. **Evaluate $f(n)$ at critical points and endpoints:** - At $n=1$: $$f(1) = 1 \times \ln\left(\frac{10}{1}\right) = \ln 10 \approx 2.3026$$ - At $n=10$: $$f(10) = 10 \times \ln\left(\frac{10}{10}\right) = 10 \times \ln 1 = 0$$ - At $n=\frac{10}{e}$: $$f\left(\frac{10}{e}\right) = \frac{10}{e} \times \ln\left(\frac{10}{\frac{10}{e}}\right) = \frac{10}{e} \times \ln e = \frac{10}{e} \times 1 = \frac{10}{e} \approx 3.678$$ 5. **Determine absolute extrema:** - $f(1) \approx 2.3026$ - $f\left(\frac{10}{e}\right) \approx 3.678$ - $f(10) = 0$ The absolute maximum is approximately $3.678$ at $n=\frac{10}{e}$. The absolute minimum is $0$ at $n=10$. 6. **Answer the second question:** The students likely chose the interval $[1,10]$ MB because it represents a practical range of file sizes for their study, avoiding very small or very large files that may not be relevant or may behave differently in compression efficiency. **Final answers:** - Absolute maximum at $n=\frac{10}{e} \approx 3.678$ MB with value $f(n) \approx 3.678$ MB reduced. - Absolute minimum at $n=10$ MB with value $f(n) = 0$ MB reduced. - Interval $[1,10]$ MB chosen for practical relevance in file sizes.