Lagrange Multipliers for two constraints, degenerate case

Question

The original post.

I was suggested to ask the question here.

To optimize $f(x,y,z)$ subject to $g(x,y,z)=h(x,y,z)=0$, we use the Lagrange Multiplier method and solve \begin{equation*} \nabla f=\lambda \nabla g+\mu\nabla h,\quad g=0,\quad h=0. \end{equation*} Geometrically, $\nabla f$ must lie on the normal plane spanned by $\nabla g$ and $\nabla h$. However, it can happen that $\nabla g$ is parallel to $\nabla h$ at certain points, and hence they cannot span the normal plane. In this case, does $\nabla f$ have to be parallel to $\nabla g$ to be a critical point? If yes, how to explain it? If no, can it happen that some critical points are missing? Thanks.

Answer 1

Consider the following example: We are looking for the min of the function $$f(x,y,z):=y$$ under the constraints $$G(x,y,z):=x^6-z=0,\qquad H(x,y,z):=y^3-z=0\ .\tag{1}$$ It is easy to see that $(1)$ describes the set $$S:=\bigl\{(x,y,z)\bigm|x\in{\mathbb R}, \ y=x^2, \ z=x^6\bigr\}\ ,$$ so that it is sufficient to consider the pullback $$\psi(x):=f(x,x^2,x^6)=x^2\qquad(-\infty<x<\infty)\ .$$ The function $\psi$ is minimal at $x=0$, hence the minimum of $f$ on $S$ is taken at the origin ${\bf 0}$.

If we try to do the same problem with Lagrange's method we get nothing. The reason is the following: The tangent space $T$ of $S$ at ${\bf 0}$ is the $x$-axis. But $$\nabla G(0,0,0)=\nabla H(0,0,0)=(0,0,-1)$$ do not span the full orthogonal complement of $T$. Note that in the proof of Lagrange's method it is assumed that the two surfaces $G=0$ and $H=0$ intersect transversally at each point ${\bf p}\in S$, but this is not the case at ${\bf 0}$.