1. Not finding help here? Sign up for a free 30min 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!

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


    User Avatar

    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]
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Lagrange multipliers
  1. Lagrange Multipliers. (Replies: 0)

  2. Lagrange Multipliers (Replies: 10)

  3. Lagrange multiplier (Replies: 1)

  4. Lagrange multiplier (Replies: 6)