So this is largely extracted from the wikipedia article on the above (http://en.wikipedia.org/wiki/Lagrange_multiplier) but hopefully I can make it clear.
I find lagrange multipliers easiest to understand in terms of vectors, so the concepts described apply mainly to functions of 2 variables but I imagine the concepts generalise readily to higher dimensions. So for the below discussion pretend the $\nabla$ is $(\partial/\partial x,\partial/\partial y)$
If you have a function $f(x,y)$ describing a surface over the x,y plane, one possible representation would be as a series of contour lines, curves of constant $f$. Similarly, any constraint on the variable $x,y$ can be rewritten as $g(x,y) = c$. This equation describes a single contour line of $g(x,y)$.
In order to find a minima of $f(x,y)$ subject to a constraint $g$ we are attempting to find the position on $g$ which lies lowest on f. If you calculate values of $f$ moving along the contour $g$, $f$ must decrease as you approach the minima, and increase afterwards. At the minima the change in $f$ must be zero. For this to be true the contour represented by $g$ must be parallel to a contour of $f$, such that infinitesimally any motion along $g$ is along the contour of $f$ and doesn't change $f$'s value. Similar considerations apply to maxima.
Now we need to translate the above statements into something mathematical: the quantity $\nabla f$ points along the direction of fastest change in $f$ (http://en.wikipedia.org/wiki/Gradient), so must be a vector normal to contours of $f$ (otherwise a component of the vector would lie along a contour, where by definition f does not change). Similarly the quantity $\nabla g$ is normal to contours of $g$, and hence to the contour given by our constraint $g(x,y)=c$. Finding the point where contours of $g$ and $f$ are tangential is now easy. We just need to find the points where the normals to $f$ and $g$ are parallel $\implies \nabla f = \lambda \nabla g$ where $\lambda \in R$. Note that if there are multiple minima or maxima this should just fall out of the maths.