Function that maximizes a function 
Let's say we have a real, continuous,
  positive function f(x) for which we
  define the quantity:
$$\pi(f,a) = \frac{\int_0^a f(x)
 dx}{\int_0^a \sqrt{1+\left(\frac{df(x)}{dx}
 \right)^2 }dx}$$
we want to find the function f that
  maximizes $\pi$ for a given $a$.

In general how do we attack problems of this kind: find $f$ such that  $\mathrm{F}(f)$ is maximum? Are there any constrains that guarantee that there is an analytical solution? How could the problem above be modified to have a solution?
 A: It looks like you are wanting to maximize the integral of $f$ with respect to its arclength. (there appears to be a typo on the bottom, but I could be wrong)  However, this quantity is unbounded.  
Consider the constant function $f(x)=c$, and let $a>0$.  Then $$\pi(f,a)=\frac{\int_0^a cdx}{\int_0^a\sqrt{1+0^2}dx}=\frac{ac}{a}=c$$
Taking $c$ to be as large as we want we see there is no maximum, and $\pi(f,a)$ is unbounded.
Hope that helps
A: You are computing the ratio of area under the graph of $f(x)$ to arc length, from $x=0$ to $x=a$. The arc length is invariant under up-shifts, but the area under the graph is not. For any $g(x)$ and any $r\gt 0$, the difference between $\pi(g+r,a)$ and $\pi(g,a)$ is proportional to $r$, and so you have that $\pi(f,a)$ is unbounded. There is no maximum for any class of functions that is closed under adding constants.
A: I don't know very much about this, but it seems like the relevant field is Calculus of Variations.  For a fixed $a$, $\pi(f,a)$ is a functional of f.  The continuity of the functional would depend on what space of functions you are optimizing over.  I read a few chapters of Calculus of Variations by Gelfand a while back and I found it quite accessible.
A: Add the condition $f(0)=f(a)=0$. The graph of the optimal $f$ has to be a circular arc of a certain angle $2\alpha$, $\ 0\leq\alpha\leq{\pi\over2}$, for otherwise one could increase the area $A$ without changing the length $L$. Expressing $A$ and $L$ using the variable $\alpha$ we get
$${A\over L}={a\over 4} ({1\over\sin\alpha}-{\cos\alpha\over\alpha}),$$
 and this assumes its maximal value ${a\over 4}$ for $\alpha={\pi\over2}$, so that we again arrive at a semicircle.
A: For $π(f,a)$ to be maximum the term $df(x)/dx = 0$
i.e.$ f(x) = c$
i.e. $\int_0^a \sqrt{1 + (\frac{df(x)}{dx})^2} dx = \int_0^a dx$
i.e. $π(f,a)= ca/a = c$
