Can I use Lagrangian multipliers with redundant constrains? For example, suppose I have the following problem:
Find the maximum of $F(x,y,z)$, subject to $f(x,y,z)=0$ and $g(x,y,z)=0$. But you also know that $f$ and $g$ are redundant, that is, there is an identity of the form $f(x,y,z) = H(g(x,y,z))$ valid for all $x,y,z$, where $H$ is another function. Now, in my particular problem, I don't know what $H$ is, or it is too complicated.
So my question is, is it still OK to use the method of Lagrange multipliers, introducing two Lagrange multipliers, one for each of the constrains $f,g$, even though they are redundant? Or is there some care I have to take?
If all constraints are linear, redundancy does not invalidate the validity of Lagrange multipliers (KKT conditions) for an optimum, but may result in non-uniqueness of the Lagrange multipliers, which could affect their interpretation as dual (constraint) prices.
If one or more constraints are nonlinear, and the constraint gradients are not linearly independent (full rank) at the optimum, then Lagrange multipliers (KKT conditions) could be invalid (may not hold) at the optimum.
See https://en.wikipedia.org/wiki/Karush%E2%80%93Kuhn%E2%80%93Tucker_conditions .