Does Lagrange multiplier have solution if functions doesn't intersect 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
 A: 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.
A: 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 $S$).
