To find the absolute minimum value of the function $f(x) = 10x(2 - \ln x)$ on the interval $[1, e^2]$, we follow these steps:

Step 1: Find the derivative of $f(x)$

We need to differentiate the function $f(x) = 10x(2 - \ln x)$.

First, apply the product rule: $f'(x) = \frac{d}{dx} [10x] \cdot (2 - \ln x) + 10x \cdot \frac{d}{dx} (2 - \ln x)$

$f'(x) = 10(2 - \ln x) + 10x \cdot (-\frac{1}{x})$

$f'(x) = 10(2 - \ln x) - 10$

$f'(x) = 20 - 10\ln x - 10$

$f'(x) = 10(1 - \ln x)$

Step 2: Find the critical points

Set the derivative equal to zero to find the critical points: $10(1 - \ln x) = 0$

$1 - \ln x = 0$

$\ln x = 1$

$x = e$

Step 3: Evaluate the function at the critical points and endpoints

We now evaluate $f(x)$ at the critical point $x = e$ and at the endpoints of the interval $[1, e^2]$.

1. At $x = 1$: $f(1) = 10 \cdot 1 \cdot (2 - \ln 1) = 10 \cdot 1 \cdot (2 - 0) = 20$

2. At $x = e$: $f(e) = 10 \cdot e \cdot (2 - \ln e) = 10 \cdot e \cdot (2 - 1) = 10e$

   Since $e \approx 2.718$, $10e \approx 27.18$.

3. At $x = e^2$: $f(e^2) = 10 \cdot e^2 \cdot (2 - \ln e^2) = 10 \cdot e^2 \cdot (2 - 2) = 0$

Step 4: Determine the absolute minimum

Among the values computed:

• $f(1) = 20$
• $f(e) \approx 27.18$
• $f(e^2) = 0$

The absolute minimum value of $f(x)$ on the interval $[1, e^2]$ is 0 at $x = e^2$.