Question

A flu has hit a city and the percentage of a population with the flu, t days after the disease arrives is approximated by f(t)=10te^(-t/8) for 0<=t>=40.
After how many days is the percentage of the population with the flu a maximum?

Answer 1

do you mean, \(\large f(t)=10te^{-t/8}, \,\,\,\,\,\,\,\,0\le t \le 40\)

Answer 2

You can find the first derivative:
\[ \large f'(t)=10e^{-t/8}+10xe^{-t/8}\left( \frac{-1}{8}\right)=10e^{-t/8}-1.25xe^{-t/8}\]
Set the derivative equal to 0:
\[\large 10e^{-t/8}-1.25te^{-t/8}=0\\
\large e^{-t/8}(10-1.25t)=0\\
\large \implies e^{-t/8}=0 \text{ or } 10-1.25t =0\]

But an exponential function can never equal 0, so you find that \(10-1.25t = 0 \implies t =8\)
So you find the maximum by plugging \(t=8\) in \(f(t)\)

Now, do you really have a maximum? verify if \(f''(8)<0 \)

Also, you must check your boundary conditions. Verify the value of \(f(0)\) and \(f(40)\). If any of these exceed \(f(8)\), that will be your max.