Question

N(t) = 10000 (8+t) e^-0.1t
I am trying to find the time when the population will be maximized. I know I have to find derivative equal to 0, I am not sure how to go about from there.
For my derivative I got:

10000 (-0.1 e^-0.1t) + (10000 (8+t)) (-0.1e^-o.1t))

Answer 1

\[\frac{d}{dt}N(t)=10000\left[\underbrace{(1)}_{f'}\underbrace{e^{-0.1t}}_{g}+\underbrace{(8+t)}_{f}\underbrace{(-0.1)e^{-0.1t}}_{g'}\right]\]

Answer 2

I don't know how to solve equal to 0 for my derivative. I get a weird answer.

Answer 3

\[10000\left( e^{-0.1t}+(8+t)(-0.1)e^{-0.1t}\right)=0\\
e^{-0.1t}+(8+t)(-0.1)e^{-0.1t}=0\\
e^{-0.1t}\left[ 1+(8+t)(-0.1)\right]=0, \,\,\,\text{by factoring }e^{-0.1t}\\
e^{-0.1t}[1-0.8-0.1t]=0\\\
e^{-0.1t}(0.2-0.1t)=0\]

You have a product of 2 factors, either product should result in 0:
But be careful, an exponential can never equal 0, so only the \(( 0.2-0.1t)\) factor can equal 0
\(0.2-0.1t=0\)
\(0.1t=0.2\)
\(t=2\)

Answer 4

Oh man thank you so much. I was stuck on this question forever. It makes sense!

Answer 5

:)

Answer 6

Do you tutor for math?

Answer 7

I have done some tutoring before