1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Local/Global Optimization

  1. Apr 29, 2008 #1
    1) http://www.geocities.com/asdfasdf23135/advcal28.JPG

    From the assumptions, I think that the mean value theorem and/or the extreme value theorem may be helpful in this problem, but I can't figure out how to apply them to reach the conclusion. Could someone please give me some general hints? Very much appreciated!:smile:
  2. jcsd
  3. Apr 29, 2008 #2


    User Avatar

    Actually it looks like you want to use Rolle's Theorem (which is a specific case of the mean value theorem anyway). Since f(x)=0 for all x then between any x1 and x2 you have a point c such that f'(c)=0.
  4. Apr 29, 2008 #3


    User Avatar
    Science Advisor

    No, "Rolle's theorem" applies to functions on R, not Rn. You can use the "Extreme Value theorem" and "mimic" the proof or Rolle's theorem. There are 3 possibliities:
    1) There are positive values of f(x,y) inside the boundary
    2) There are negative values of f(x,y) inside the boundary
    3) There are neither positive nor negative values of f(x,y) inside the boundary

    What does the extreme value theorem tell you in each of those cases?
  5. Apr 29, 2008 #4
    Thanks! Your hints are helpful!

    By extreme value theorem (EVT), since f is continuous and the closure of D is compact, there exists an absolute max value and an absolute min value on the closure of D.

    If f(x) is identically 0 on closure of D, then any a on D will do.
    If f(x) is not identically 0 on closure of D, since f(x)=0 on boundary of D, we must have either f(x)>0 for some x on D or f(x)<0 for some x on D
    Let's consider the case f(x)<0 for some x on D. Absolute min value must be <0 and since f(x)=0 on boundary of D, this min value must occur on the open set D. Say f(a), a E D is the absolute min value.

    Now, does this imply that f(a) is a local min and that grad f(a)=0? Why or why not?
  6. May 1, 2008 #5
    Continuing with the last post:

    f(a), the absolute min, is attained on the "open" set D. Since it is "open", it contains none of its boundary points, so "a" must be a critical point with grad f(a) = 0, am I right?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook