Homework Help: Max of cont. fxn

1. Feb 4, 2010

Somefantastik

1. The problem statement, all variables and given/known data
max{nte-nt: t in [0,1]} = ?

2. Relevant equations

3. The attempt at a solution

if f(t) = nte-nt

f '(t)= ne-nt - te-nt

= e-nt(n-t)

set equal to zero to get critical points:

0 = = e-nt(n-t) => two critical points: t1 = n and t2 = -1/n;

plugging the cp's back into the original equation, we get

f(t1) = n2/en2, which converges to 0 as n-> infy

f(t2) = -e which converges to -e for all n.

What am I doing wrong?

2. Feb 4, 2010

JSuarez

Check your derivative again, especially the second factor.

3. Feb 4, 2010

Somefantastik

Yep, ok

so f ' (t) = ne-nt - n2te-nt

and 0 = e-nt(n-n2t)

=> e-nt = 0 && n(n-t)=0 <=> t = n

so t= n is my critical point?

let tc = t = n,
Plugging that back into the original,

f(tc) = n2e-n2

and I still get lim f(tc) = lim e-n2 (after l'hopital)

Can you see another mistake? I'm very sure the max of this function is not zero.

4. Feb 4, 2010

JSuarez

From what I see from the original statement, the function is defined only for $t \in \left[0,1\right]$, so does $t_c=n$ makes sense? What is the derivative's sign?

5. Feb 4, 2010

JSuarez

Just noticed something. Are you sure that you have the correct critical point? $n-n^{2}t=0$ gives t=?.

6. Feb 4, 2010

Somefantastik

No, but the other ones didn't seem relevant. e-nt = 0 doesn't make any sense.

7. Feb 4, 2010

JSuarez

See the other post.

8. Feb 4, 2010

Somefantastik

How embarrassing

then n-n2t = 0 => tc = 1/n

=> f(tc) = e-n

which still converges to zero as n -> inf

9. Feb 4, 2010

Somefantastik

unless, since we are considering the uniform norm $$\left\|\cdot\right\|_{\infty} = max\left{\{\left|f(t)\right| : t \in [0,1] \right\}$$

taking the absolute value of e-n would ensure that the max is infinite (sup)

?

10. Feb 4, 2010

JSuarez

But substituting t=1/n gives 1/e, not 1/(e^n). You should also check if the maximum doesn't occur at the extreme points of the interval (in this case t=1; t=0, gives 0); the derivative test misses those ones.

11. Feb 4, 2010

Somefantastik

Ok...I need sleep apparently. Thanks so much for your time.

12. Feb 4, 2010

JSuarez

We all do:zzz:. But I've checked it: there is a sequence of maxima approaching 0, and all equal to 1/e.