Find the extrema of $f(x,y)=max(x,y)$ constrained to $\mathscr A=\{(x,y) ∈ \mathbb R^2 ∣ x^2+y^2=1\}$ I know the maxima occurs at $f(x,y) = 1$ at the points $(1,0)$ and $(0,1)$... but I can't seem to find all the minima.
I've come to the conclusion that the minima occurs at a point $(a,a)$ where $a ∈ \mathbb R_<0$, such that $2a^2=1\rightarrow a=\frac{\sqrt 2}{2} \lor a=-\frac{\sqrt 2}{2}$.
Does the minima occur at ($-\frac{\sqrt 2}{2}, -\frac{\sqrt 2}{2}$)?

The original curve can be expressed by the following equation in polar coordinates $(\rho,\theta)$:
$$
\rho=1.
$$
$(1)$ When $y\ge x$, there is $\theta\in[\frac{\pi}{4},\frac{3\pi}{4}]$, we have $f(x,y)=y=\text{sin}(\theta)$, so $\min f(x,y)=-\frac{\sqrt{2}}{2}$ when $\theta=\frac{3\pi}{4}$ or $x=y=-\frac{\sqrt{2}}{2}$.
$(2)$ When $y\le x$, there is $\theta\in[-\frac{3\pi}{4},\frac{\pi}{4}]$, we have $f(x,y)=x=\text{cos}(\theta)$, so $\min f(x,y)=-\frac{\sqrt{2}}{2}$ when $\theta=\frac{-3\pi}{4}$ or $x=y=-\frac{\sqrt{2}}{2}$.
To sum up, the minima occurs at $\left(-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)$ under the rectangular coordinate system $(x,y)$.
