Lagrange multiplier derivative condition

Question

For the Lagrangian $\mathcal{L}(\mathbf{x}, \lambda) = f(\mathbf{x}) + \lambda g(\mathbf{x})$, I read that $\partial\mathcal{L}/\partial\lambda$ must equal $0$. Could someone please explain why?

Answer 1

The equation $\frac{\partial \mathcal L (x,\lambda)}{\partial \lambda} =0 $ just says that $g(x) = 0$. This must be satisfied, of course, if $x$ is a minimizer of $f$ subject to the constraint that $g$ is $0$.