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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?

J. M. ain't a mathematician
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Eelvex
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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 00:52
  • @Arturo: yes there is a typo (thanks), but this is just an example. I'll update the question. – Eelvex Feb 25 '11 at 01:04
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    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 01:48
  • The specific question of a semicircle was already discussed at http://math.stackexchange.com/questions/4808/why-circle-encloses-largest-area, but this question seems more general. – Ross Millikan Feb 25 '11 at 01:59
  • @Ross: In this problem, the arc is variable not given. I think the solution for when $f(0) = f(a) = 0$ is a $\delta$ function, giving $\pi(f,a) = \frac{1}{2}a$. – Eelvex Feb 25 '11 at 02:06
  • The arc length of any approximation to a delta function is quite long as it rises very high, so I think $\pi$ will approach zero for them. We know the semicircle maximizes the area to arclength ratio-there are links at the other question – Ross Millikan Feb 25 '11 at 03:48
  • @Ross: this is not the case here. You can easily check that semicircle gives $\pi_c = a/4$, "semisquare" also $\pi_s = a/4$ and in fact, any polygon has $\pi_n = a/4$. – Eelvex Feb 25 '11 at 04:15
  • I mean regular polygon. – Eelvex Feb 25 '11 at 04:23
  • @Eelvex: I see what you mean. I think you mean half a regular polygon(in analogy with the semicircle and half square.) – Ross Millikan Feb 25 '11 at 04:33
  • @Ross: Yes, half-regular-polygon :) – Eelvex Feb 25 '11 at 06:37
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    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

5 Answers5

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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

Eric Naslund
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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.

Arturo Magidin
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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.

Henry B.
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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. $π(f,a)= ca/a = c$

HOLYBIBLETHE
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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.

Christian Blatter
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