Answer
Absolute maximum value is at $f(e)=10e$
Absolute minimum value is at $f(e^2)=0$
Work Step by Step
Let us consider $y=f(x)=10x (2-\ln x)=20 x-10 \ln x$
Now, $f'(x)=10(1-\ln x)$
Now, $f'(e)= 10(1-\ln e)=10(1-1)0$
Thus, at $x=e; f'(x)=0$
This shows a critical point.
Now, $f(e)= 10e (2-\ln e)=10e$
So, absolute maximum value is at $f(e)=10e$
and
$f(e^2)= 10e^2(2-\ln e^2)=10e^2(2-2)=0$
So, absolute minimum value is at $f(e^2)=0$
