# 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?

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I assume that the expression in the radical is meant to be the square of the derivative. That is, you are computing the ratio of area under the graph to arc length of the graph. – Arturo Magidin Feb 25 '11 at 0:52
@Arturo: yes there is a typo (thanks), but this is just an example. I'll update the question. – Eelvex Feb 25 '11 at 1:04
But if we add conditions $f(0)=f(a)=0$ the problem is no longer unbounded and we should be able to derive that the solution is a semicircle. I believe this is a subject for calculus of variations, which I never learned. – Ross Millikan Feb 25 '11 at 1:48
The specific question of a semicircle was already discussed at math.stackexchange.com/questions/4808/…, but this question seems more general. – Ross Millikan Feb 25 '11 at 1:59
definately take a look at "calculus of variations" by Gelfand (it's a small dover book, so super cheap) if you want an easy introduction to solving these type of problems (well posed ones anyway). – yoyo Feb 25 '11 at 16:18

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.

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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

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It helps a little :) but this is not exactly what I meant. – Eelvex Feb 25 '11 at 1:10

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.

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40 seconds apart! – Eric Naslund Feb 25 '11 at 0:56
@Eric: Yup, pretty much. – Arturo Magidin Feb 25 '11 at 0:56

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.

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A semicircle has $\frac{A}{L} = \frac{a}{4}$ so there is something wrong here... – Eelvex Feb 25 '11 at 17:12
Sorry. In my figure the chord length was $2a$ instead of $a$. – Christian Blatter Feb 26 '11 at 8:54

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. $π(f,a)= ca/a = c$

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π(f,a) will be maximized when we reduce its denominator to the lowest possible value. – Rajesh K Singh Jul 12 '12 at 2:03
Here f(x) is a constant function. – Rajesh K Singh Jul 12 '12 at 2:36
Did you notice that this is a really old question, and that there are other answers identical to yours? – huon Jul 12 '12 at 2:45