Pronumeral question

**Naur** · June 29th, 2008, 11:47 PM

Another problem, this one's been digging at me for a week or two.

**Simplicity** · June 30th, 2008, 01:19 AM

Originally Posted by Naur

Another problem, this one's been digging at me for a week or two.

Yes, it should be $t^2$ . They have probably made a typo.

**Naur** · June 30th, 2008, 02:38 AM

But the answer fits with the formula, and my calculator also seems to agree.

**topsquark** · June 30th, 2008, 02:58 AM

Originally Posted by Naur

But the answer fits with the formula, and my calculator also seems to agree.

Given the form you wrote for N'(t) it should be a $t^2$ . Perhaps we should see the whole question.

-Dan

**Naur** · June 30th, 2008, 06:35 AM

There we go. It concerns finding the maximum or minimum stationary points by finding the derivative, and putting it =0.

**fobos3** · June 30th, 2008, 08:59 AM

Using the rules for differentiation:
$(500+100 e^{-t/12} t)'$
$500'+(100 e^{-t/12} t)'$
$100 \left(e^{-t/12} t\right)'$
$100(t(e^{-t/12})'+e^{-t/12}t')$
$100 \left(e^{-t/12} t \left(-\frac{t}{12}\right)'+e^{-t/12}\right)$
$100 \left(-\frac{1}{12} e^{-t/12} t t'+e^{-t/12}\right)$
$100 \left(e^{-t/12}-\frac{1}{12} e^{-t/12} t\right)$
$100 e^{-t/12}-\frac{25}{3} e^{-t/12} t$

**Naur** · June 30th, 2008, 05:12 PM

Err. Little confused.

**fobos3** · July 2nd, 2008, 02:07 AM

In my explanation I use:
a-constant
x,y-variables

Your function is $f(x)=500+100 e^{-t/12} t$
So $f'(x)=(500+100 e^{-t/12} t)'$
$f'(x)=500'+(100 e^{-t/12} t)'$ from (x+y)'=x'+y'
500'=0
$f'(x)=(100 e^{-t/12} t)'$
$f'(x)=100 \left(e^{-t/12} t\right)'$ from (ax)'=ax' (here $x=e^{-t/12} t$ )
$f'(x)=100(t(e^{-t/12})'+e^{-t/12}t')$ from (xy)'=x'y+xy' (here $x=e^{-t/12}$ and y=t)
t'=1
$f'(x)=100 \left(e^{-t/12} t \left(-\frac{t}{12}\right)'+e^{-t/12}\right)$ from $(e^{y})'=y'e^{y}$
$f'(x)=100 \left(-\frac{1}{12} e^{-t/12} t t'+e^{-t/12}\right)$ from $\left(\frac {x}{a}\right)'=\frac {1}{a}x'$
t'=1
$f'(x)=100 \left(e^{-t/12}-\frac{1}{12} e^{-t/12} t\right)$
$f'(x)=100e^{-t/12}-\frac{100}{12} e^{-t/12} t$
$f'(x)=100 e^{-t/12}-\frac{25}{3} e^{-t/12} t$

**Naur** · July 6th, 2008, 07:02 PM

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.

**fobos3** · July 7th, 2008, 10:37 PM

Ok. We have:
$f(t)=500+100e^{-t/12}t$
$f'(t)=e^{-t/12}-\frac{1}{12} e^{-t/12} t$
We have extremums in the points where f'(t)=0

$e^{-t/12}-\frac{1}{12} e^{-t/12} t=0$
$-\frac{1}{12} e^{-t/12} (-12+t)=0$

t-12=0
t=12

$-\frac{1}{12} e^{-t/12}=0$
$e^{-t/12}=0$
In this equation there is no solution for t therefore we have $t_{\max /\min }=12$

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)= $-\frac{1}{e^2}$ <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.

**Naur** · July 7th, 2008, 11:05 PM

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.

**fobos3** · July 8th, 2008, 03:28 AM

They have a typo. It should be $(-\frac {t}{12})'$ instead of $(-\frac {t}{12})$

**Naur** · July 8th, 2008, 10:03 PM

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

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