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?