Another problem, this one's been digging at me for a week or two.
Using the rules for differentiation:
$\displaystyle (500+100 e^{-t/12} t)'$
$\displaystyle 500'+(100 e^{-t/12} t)'$
$\displaystyle 100 \left(e^{-t/12} t\right)'$
$\displaystyle 100(t(e^{-t/12})'+e^{-t/12}t')$
$\displaystyle 100 \left(e^{-t/12} t \left(-\frac{t}{12}\right)'+e^{-t/12}\right)$
$\displaystyle 100 \left(-\frac{1}{12} e^{-t/12} t t'+e^{-t/12}\right)$
$\displaystyle 100 \left(e^{-t/12}-\frac{1}{12} e^{-t/12} t\right)$
$\displaystyle 100 e^{-t/12}-\frac{25}{3} e^{-t/12} t$
In my explanation I use:
a-constant
x,y-variables
Your function is $\displaystyle f(x)=500+100 e^{-t/12} t$
So $\displaystyle f'(x)=(500+100 e^{-t/12} t)'$
$\displaystyle f'(x)=500'+(100 e^{-t/12} t)'$ from (x+y)'=x'+y'
500'=0
$\displaystyle f'(x)=(100 e^{-t/12} t)'$
$\displaystyle f'(x)=100 \left(e^{-t/12} t\right)'$ from (ax)'=ax' (here $\displaystyle x=e^{-t/12} t$)
$\displaystyle f'(x)=100(t(e^{-t/12})'+e^{-t/12}t')$ from (xy)'=x'y+xy' (here $\displaystyle x=e^{-t/12}$ and y=t)
t'=1
$\displaystyle f'(x)=100 \left(e^{-t/12} t \left(-\frac{t}{12}\right)'+e^{-t/12}\right)$ from $\displaystyle (e^{y})'=y'e^{y}$
$\displaystyle f'(x)=100 \left(-\frac{1}{12} e^{-t/12} t t'+e^{-t/12}\right)$ from $\displaystyle \left(\frac {x}{a}\right)'=\frac {1}{a}x'$
t'=1
$\displaystyle f'(x)=100 \left(e^{-t/12}-\frac{1}{12} e^{-t/12} t\right)$
$\displaystyle f'(x)=100e^{-t/12}-\frac{100}{12} e^{-t/12} t$
$\displaystyle f'(x)=100 e^{-t/12}-\frac{25}{3} e^{-t/12} t$
Okay, that all makes sense. That's as far as I got before the book confused me.
Could you please give an answer for the maximum population? Like I said, though it seems the book is doing it wrong, they have an answer that seems right.
Ps. Sorry I couldn't make it back before now, I had a friend down.
Ok. We have:
$\displaystyle f(t)=500+100e^{-t/12}t$
$\displaystyle f'(t)=e^{-t/12}-\frac{1}{12} e^{-t/12} t$
We have extremums in the points where f'(t)=0
$\displaystyle e^{-t/12}-\frac{1}{12} e^{-t/12} t=0$
$\displaystyle -\frac{1}{12} e^{-t/12} (-12+t)=0$
t-12=0
t=12
$\displaystyle -\frac{1}{12} e^{-t/12}=0$
$\displaystyle e^{-t/12}=0$
In this equation there is no solution for t therefore we have $\displaystyle t_{\max /\min }=12$
Now we just need to find if we hava a max or min at point t=12
We check the values of f'(t) for two points. One of them smaller than t ant the other larger.
f'(0)=1>0
f'(24)=$\displaystyle -\frac{1}{e^2}$<0
So the derivative at point t=0 is greater than 0, which means that the function is increasing. And at point t=24 is smaller than 0, which means the function is decreasing. In other words f(12)=max because first the function is increasing, it reaches the max and then begins to decrease.
Okay then. So the answer is 12....I still wonder how the book got the right answer by doing it wrong, and how my answer doesn't seem to fit. I mean, I'm doing it a different way to how you've done it, and I could swear it's all right except for that squared sign.
I'm still at a loss as to why my working isn't providing the right answer.
Care to take a look?
:O! Nevermind, I just worked it out. Sorry I didn't pick it up as quickly as I would've liked, it's just that we have different ways of explaining things.
I'll show you my page of working, it explains it all pretty well.
Thank you very much. I shall give you thanks