Maximize $x^2 + y^2$ subject to $x^2 + y^2 = 1 - xy$ The equation $x^2 + y^2 = 1 - xy$ represents an ellipse. I am trying to show that its major axis is along $y=-x$ and find the vertex. To find the vertex I tried to find the vector in the ellipse with the greatest norm, which is equivalent to maximizing $x^2 + y^2$ subject to $x^2 + y^2 = 1 - xy$. How can I approach this?
(The maximum value of the objective function $x^2 + y^2$ seems to be 2. If I start out by assuming this I am able to show that the function cannot exceed this value. But this is kind of doing it in reverse as I start by evaluating the function at $y=-x=1.$ This also doesn't prove that $y=-x$ is the only optimal solution. How would I do it without "guessing" the solution first?)

Using polar coordinates, then
$ x = r \cos \theta $
$ y = r \sin \theta $
So we want to maximize $ r^2 = x^2 + y^2 $ subject to $ r^2 = 1 - \dfrac{1}{2} r^2 \sin(2 \theta) $
Hence,
$ r^2 = \dfrac{1}{1 + \frac{1}{2} \sin(2 \theta) } $
Clearly the maximum is when $ \sin(2 \theta) = -1 $, which corresponds to $ \theta = \dfrac{3 \pi}{4} $ and $ \theta = \dfrac{7 \pi}{4} $
At which points , $ r^2 = 2 $ (and this is the maximum value) and for the first value of $\theta$, we get
$ x = -1 , y = 1 $
And for the second value of $\theta$ we get
$ x = 1, y = -1 $
