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Lagrange multipliers

  1. Jun 21, 2005 #1
    I'm not entirely sure what the english terms are for some of the things i'm about to say but i hope it's clear what I mean exactly. I'n my handbook the theorom is said to be:

    Say G is a part (wich is open) of R^n, f and g are functions from G to R (f:G->R, g:G->R) and both are differentiable (and the differential is continues). If f(p)=0, Df(p)#0 and g has an extremum on p (f^-1(0)) then there is a delta (element of R) for wich you can write: Dg(p)=delta*Df(p)

    But if g has an extremum on p then wouldn't Dg(p) = 0?

    thx in advance
  2. jcsd
  3. Aug 27, 2009 #2


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    Not really. Under the given conditions, the implicit function theorem guarantees that the expression [tex]f(p)=f(p_1,p_2)=0[/tex] defines a function [tex]y=y(x), x\in (p_1-\epsilon, p_1+\epsilon)=I[/tex] (or [tex]x=x(y)[/tex], but choose the former for convenience) . So an extremum for g on [tex]f^{-1}(0)[/tex] is in fact the extremum of -say- [tex]g(x,y(x)),x\in I[/tex].

    So [tex]\frac{d}{dx}[g(x,y(x))][/tex] is zero at [tex]p[/tex] and not necessarily [tex]\frac{\partial g}{\partial x}(p)=0=\frac{\partial g}{\partial y}(p)[/tex]
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