Self-student variational calculus - the maxima and minima simplest problem I am starting a self-taught study in Variational Calculus. The first thing I see in every book is the Euler-Lagrange equation.
However, what comes to my mind as the first and simplest question of Variational Calculus would be maxima and minima of a function. So I would expect the result of the derivative to be $0$ at the maxima and minima to be a particular result of the Euler-Lagrange equations.
But, I couldn't find this example anywhere (maybe because it's too simple).
Can you help me? Thank you!
 A: The Euler-Lagrange equation for the functional $F\colon X\to \mathbb R$, where $X$ is a Banach space, is given by
$$
F'(x)=0, $$
where $F'$ is the Fréchet derivative of $F$. Now, what is the Fréchet derivative? It is something that we do not need to define, many books are devoted to that. What we need to do is to observe that, when $X=\mathbb R$ (which is a Banach space), the Fréchet derivative reduces to the ordinary derivative of a function of one real variable. Therefore the Euler-Lagrange equation reduces to the familiar statement of calculus that a derivative must vanish at maxima and minima of a function.
ADDED. The comments argue that this answer does not really address the question. So, let me explain how does $F'(x_\star)=0$ follow from $x_\star$ being a maximizer for $F$. Note that the opposite statement does NOT hold; there is no way to prove that $x_\star$ is a maximizer by looking only at $F'$.  So, if the question is about that, the answer is a simple "no".
Suppose that $x_\star$ is a maximizer; this means that, however you take an $x\in X$, the incremental ratio
$$\tag{1}
\frac{F(x_\star + t x)-F(x_\star)}{t}
$$
is nonnegative for $t>0$ and nonpositive for $t<0$. This follows at once from observing that $F(x_\star + t x)\ge F(x_\star)$, since $x_\star$ is a maximizer.
Now, since $F$ is differentiable, the incremental ratio (1) has a limit as $t\to 0$. By the considerations above, this limit must be both nonnegative and nonpositive. So the limit is $0$, which concludes the proof.
