[Math] Lagrange multiplier when constraint are not linearly dependent.

Question

To optimize a function $f:\mathbb R^n\longrightarrow \mathbb R$ under constraint $g_1,g_2$ we need to check that $\nabla g_1$ and $\nabla g_2$ are linearly independent. But what happen if $\nabla g_1$ and $\nabla g_2$ are linearly dependent ? It mean that $$\min_{g_1=0, g_2=0}f=\min_{g_1=0}f\ \ ?$$
Or we can't apply Lagrange multiplier ?

Answer 1

No, you can't just drop a constraint in this case. Basically, the solution will be found amongst the feasible points where either the gradient of the Lagrange function is zero or the gradients of constraints are linearly dependent. You will have to find all the feasible points where the gradients of the constraints are not linearly independent and check whether the solution is among them.