# Lagrange multipliers

1. Jun 21, 2005

### allistair

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?

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