Derivation of Euler Lagrange Equation

I was reading on the derivation of the Euler Lagrange Equations (in the link: http://en.wikipedia.org/wiki/Euler%E2%80%93Lagrange_equation focusing on: "Derivation of one-dimensional Euler–Lagrange equation" ) and I cannot seem to understand the logic behind it.

So we begin by establishing a functional

$$L = L(x,y(x), y'(x))$$

And we are looking to find the minimal or maximal function that satisfies:

$$\int_{a}^{b} L(x,y(x), y'(x)) \space dx = H(x)$$

At this point we can consider such an optimal function which we denote as

$$G$$ and therefore note that any solution to the aforementioned will be of the form

$$G(x) + en(x)$$

where e is the magnitude of the error and $n(x)$ is an error function such that $n(a) = n(b) = 0$

At this point what exactly is it that we are trying to do? and more importantly, WHY?

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What you want to do is take the differential of the action and act upon it with the operator $\frac{\partial}{\partial e}$. The goal here is that $e$ is an infinitesimal change in the optimal function will not change the action. –  Chris K Apr 17 '14 at 15:47
Can you elaborate on that? The terms I don't exactly understand are 'action' for example intuitively when optimizing f(x): df/dx = 0 corresponds to a point where one cannot go any further with minimizing/maximizing a function. In that sense I am curious how solving $\frac{\partial G}{\partial e} = 0$ corresponds to a local optimum –  frogeyedpeas Apr 17 '14 at 15:47
First note that the action is the functional $H$, which maps functions to numbers... –  Chris K Apr 17 '14 at 15:51
This $H$ is what we want to derivate with respect to. –  Chris K Apr 17 '14 at 15:52
The goal is to extremise the functional; like for a function where you want $f'(x) = df/dx$, here the goal is $\delta H = \frac{\partial H}{\partial e} = 0$. –  Chris K Apr 17 '14 at 16:07