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I am trying to find a function prescribed in polar coordinates $r = f(\theta)$ that maximizes the following quantity

$$\frac{\int_0^{2\pi}r^3\cos\theta\, d\theta}{\int_0^{2\pi}r^4\, d\theta}$$

subject to the constraint $r \leq R \quad \forall \theta \in [0,2\pi)$.

I don't even know where to begin to be honest. I went through a couple of books on calculus of variations. I haven't seen anything that comes close to the problem I have at hand.

gebruiker
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Calculon
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  • Differentiate the given fraction. – hjpotter92 Aug 27 '14 at 09:31
  • @hjpotter92 differentiate with respect to what? $r$ is a function of $\theta$, not a constant. – Calculon Aug 27 '14 at 09:33
  • @Surb If you read the sentence of the question, you will see that it has already been pointed out. It is common practice to drop the $x$ from the notation $y(x)$ once it has been made clear that $y$ is a function of $x$. – Calculon Aug 27 '14 at 10:51

1 Answers1

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Clearly the maximum is $+\infty$. Consider for example $r=f(\theta)=\epsilon |\cos(\theta/2)|$. Then $$ \frac{\int_0^{2\pi}r^3\cos\theta d\theta}{\int_0^{2\pi}r^4 d\theta}=\frac{32}{15\pi\epsilon} $$ and this tends to $+\infty$ as $\epsilon$ goes to $ 0^+$.

Omran Kouba
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