# lagrange multipliers fails

I am looking for a certain counter example. Assume a $C^1$ function $f$ is to be optimized with respect to a $C^1$ constraint $g=0$, and we have at a point $(x,y)$, the existence of a lagrange multiplier $\lambda$ with \begin{align} &\nabla f(x,y)=\lambda\nabla g(x,y)\\ &\nabla g(x,y)\neq0\\ \end{align} But $f$ fails to have an extremum at this point with respect to $g=0$

thanks

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But the extremum occurs at (0,0).. where g has a singular point.. so this is not a counter example. – mark May 27 '13 at 9:20
You're right, it's a maximum. OK, let me find something else then. Shouldn't be too hard. – Raskolnikov May 27 '13 at 9:23
How about $f(x,y)=x^3+y^3$ and $g(x,y)=x$? – Raskolnikov May 27 '13 at 9:25
@Raskolnikov This $g$ doesn't define a compact manifold, hence you can't apply Lagrange method. – TZakrevskiy May 27 '13 at 9:30
Thank you Raskolnikov – mark May 27 '13 at 9:31

Let $f(x,y):=x^3+y$ and $g(x,y):=y$. Then $\nabla f(0,0)=\nabla g(0,0)=(0,1)$, but $f(x,0)-f(0,0)=x^3$ assumes both signs in the immediate neighborhood of $(0,0)$.