Lagrange Multipliers for two constraints, degenerate case

Question

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

Lagrange multipliers with two constraints work well if the gradients $\nabla g$ and $\nabla h$ are linearly independent since this implies that in the intersection of level surfaces for $f$ and $g$ we get a regular curve. Without the requirement of linear independence of $\nabla g$ and $\nabla h$ strange things may happen:

Any closed set $K$ in $\mathbb{R}^n$ is a zero set of a smooth nonnegative function. That is a well known fact. Take a smooth function $\tilde{g}:\mathbb{R}^2\to\mathbb{R}_+$ with the zero set $K$ (so $\nabla\tilde{g}=0$ on $K$) and define $g(x,y,z)=z-\tilde{g}(x,y)$, $h(x,y,z)=z$. Then the constraint $g(x,y,z)=h(x,y,z)=0$ is satisfied by the set $K\times\{0\}\subset\mathbb{R}^3$ only so:

Find max/min of $f(x,y,z)$ subject to the constraint $g(x,y,z)=h(x,y,z)=0$,

is equivalent to

Find max/min of $f(x,y,0)$ on $K$.

Since $K$ can be an arbitrary closed set in $\mathbb{R}^2$, one cannot get any multiplier type conditions. Observe that on the set $K\times \{ 0\}$ the gradients of $g$ and $h$ are equal and hence linearly dependent $\nabla g=\nabla h=\langle 0,0,1\rangle$ on $K$.

Answer 2

If $\nabla g$ and $\nabla h$ are linearly independent at a given point, then the Linear Independence Constraint Qualification (LICQ) is satisfied, and presuming that $f$, $g$, and $h$ are all continuously differentiable, the Karush–Kuhn–Tucker (KKT) conditions are necessary (must hold) for a local minimum or local maximum.

If $\nabla g$ and $\nabla h$ are not linearly independent, and no other constraint qualification holds, then KKT is not necessary, and need not be satisfied at a local minimum or local maximum (i.e., Lagrange Multipliers satisfying KKT may not exist).