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Local/Global Optimization

  • Thread starter kingwinner
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  • #1
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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:
 

Answers and Replies

  • #2
exk
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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.
 
  • #3
HallsofIvy
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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?
 
  • #4
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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?
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?
 
  • #5
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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?
 

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