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Homework Help: Max of cont. fxn

  1. Feb 4, 2010 #1
    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. jcsd
  3. Feb 4, 2010 #2
    Check your derivative again, especially the second factor.
     
  4. Feb 4, 2010 #3
    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.
     
  5. Feb 4, 2010 #4
    From what I see from the original statement, the function is defined only for [itex]t \in \left[0,1\right][/itex], so does [itex]t_c=n[/itex] makes sense? What is the derivative's sign?
     
  6. Feb 4, 2010 #5
    Just noticed something. Are you sure that you have the correct critical point? [itex]n-n^{2}t=0[/itex] gives t=?.
     
  7. Feb 4, 2010 #6
    No, but the other ones didn't seem relevant. e-nt = 0 doesn't make any sense.
     
  8. Feb 4, 2010 #7
    See the other post.
     
  9. Feb 4, 2010 #8
    How embarrassing :redface:

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

    => f(tc) = e-n

    which still converges to zero as n -> inf
     
  10. Feb 4, 2010 #9
    unless, since we are considering the uniform norm [tex]\left\|\cdot\right\|_{\infty} = max\left{\{\left|f(t)\right| : t \in [0,1] \right\} [/tex]

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

    ?
     
  11. Feb 4, 2010 #10
    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.
     
  12. Feb 4, 2010 #11
    Ok...I need sleep apparently. Thanks so much for your time.
     
  13. Feb 4, 2010 #12
    We all do:zzz:. But I've checked it: there is a sequence of maxima approaching 0, and all equal to 1/e.
     
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