Let $\mathcal C$ the curve in the plane given by the polar equation $$ r=1+\cos\vartheta. $$ Find the maximum and the minimum of $f(x,y)=\max\{x,y\}$ on $C$.

First question: how can we apply Weierstrass theorem? In other words, is the set $\mathcal C$ bounded? (Suppose so, otherwise the exercise could be meaningless!) How can we prove it?

I cannot express the equation of the curve in cartesian coordinates. I start considering the function $$ g(r, \vartheta)=f(r\cos{\vartheta}, r\sin{\vartheta}) = \max\{r\cos{\vartheta}, r\sin{\vartheta}\} = \begin{cases} r\cos{\vartheta} & \text{if } 0 \le \vartheta \le \frac{\pi}{4} \vee \frac{5\pi}{4} \le \vartheta \le 2\pi \quad (\bmod 2\pi)\\ r\sin{\vartheta} & \text{otherwise} \end{cases} $$

The restriction of $g$ to $\mathcal C$ is then $$ g \mid_{\mathcal C} (r, \vartheta) = \begin{cases} \cos{\vartheta} + \cos^2{\vartheta} & \text{if } 0 \le \vartheta \le \frac{\pi}{4} \vee \frac{5\pi}{4} \le \vartheta \le 2\pi \quad (\bmod 2\pi)\\ \sin{\vartheta}+\sin{\vartheta}\cos{\vartheta} & \text{otherwise} \end{cases} $$

which does not depend on $r$. So I can study this function as a one-variable function: in particular $$ g'(\vartheta) = \begin{cases} -\sin{\vartheta} - 2\sin{\vartheta}\cos{\vartheta} & \text{if } 0 < \vartheta < \frac{\pi}{4} \vee \frac{5\pi}{4} < \vartheta < 2\pi \quad (\bmod 2\pi)\\ \cos{\vartheta}+\cos{2\vartheta} & \text{otherwise} \end{cases} $$

What do you think? I am a ittle stuck because I do not like the expression of $g'$: how would you conclude the exercise? I would find the zeroes of $g'$ and evaluate the function on that points. But I'm afraid this is somehow wrong...

What do you think?

  • 1
    $\begingroup$ Regarding boundedness: In polar coordinates it is easy to see if a set is bounded, because all you have to do is look at the coordinate $r$. The set is bounded iff $r$ is bounded on it. $\endgroup$ – Giuseppe Negro Sep 16 '12 at 13:05
  • $\begingroup$ Consider $\vartheta\to\max\{x(\vartheta),y(\vartheta)\}$. This is a continuos function on the compact interval $[0,2\pi]$, hence has a maximum and a minimum. $\endgroup$ – Hagen von Eitzen Sep 16 '12 at 13:10
  • $\begingroup$ @GiuseppeNegro Thank you very much for your comment, good observation. I'll keep it in mind. $\endgroup$ – Romeo Sep 16 '12 at 13:19

If $f(x, y) = \max\{x, y\}$ then $f(\theta) = r(\theta)\max\{\sin \theta, \cos \theta\}$. To find maximum and minimum of $f$, you should find the maximum and minimum of $f_x(\theta) = r(\theta) \sin \theta = \sin \theta + \sin \theta \cos \theta$ and $f_y(\theta) = r(\theta) \cos \theta = \cos \theta + \cos^2 \theta$. Then pick the larger or smaller of each pair. Those are found using the usual method of differentiating.

Note that $f_x' = \cos \theta + \cos 2\theta = 2\cos \frac{3\theta}{2} \cos \frac{\theta}{2}$ and $f_y' = -\sin \theta (1+2\cos \theta)$. Now the roots should be obvious. You only need to pick the right ones.

Good luck.


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