Does Lagrange multiplier have solution if functions doesn't intersect

Question

I am trying to get intuition behind Lagrange multiplier and question that bothers me is: Does Lagrange multiplier have solution if two functions(main function and constraint) doesn't intersect.

Thank you

Answer 1

The main function and the constraint function are of two different forms. For instance the main function (to maximize or minimize) might be $$ f(x,y,z) = 3x + 3y^2 + \sin z $$ and the constraint function might be $$ g(x,y,z) = x + y + 4z \boldsymbol{ = 4} $$ Do you see the difference between the two? It doesn't make any sense to ask whether or not they intersect. The constraint function defines a set of points which you are considering. The function $f$ is just a function on those points, and not a constraint. It can't "intersect" with anything.

Answer 2

Your "main function" is a function $f:\ \Omega_f\to{\mathbb R}$, defined in some open set $\Omega_f\subset{\mathbb R}^n$. It "intersects" nothing. In addition, a constraint $g({\bf x})=0$ is given, where $g:\ \Omega'\to{\mathbb R}$ is another function, defined in some open set $\Omega_g\subset{\mathbb R}^n$. This constraint defines an $(n-1)$-dimensional hypersurface $S\subset\Omega_g$.

Lagrange's method deals with points ${\bf p}\in\Omega_f\cap S$, and declares, under which circumstances such a point is a conditionally stationary point of $f$, relative to the condition $g({\bf x})=0$.

These circumstances are the following: In the first place it should be the case that $$\nabla f({\bf p})=\lambda \>\nabla g({\bf p})$$ for some scalar $\lambda$, but there is also the technical condition that $\nabla g({\bf p})\ne{\bf 0}$ (meaning that ${\bf p}$ should be a regular point of $S$).