<s>[INST] 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?)
 [/INST] Substitutions work much faster .
Let $\thinspace x=a+b,\thinspace y=a-b$, then you have :
$$\begin{align}&x^2+y^2=1-xy\\
\iff
&3a^2+b^2=1\end{align}$$
This leads to the following :
$\underline{\rm {Global\thinspace\thinspace maximum}}$
$$\begin{align}x^2+y^2&=2(a^2+b^2)\\
&=2(a^2+1-3a^2)\\
&=2-4a^2\leq 2\end{align}$$
Equality occurs iff, when $a=0,\thinspace |b|=1$, which corresponds to $\left(x,y\right)=\left(\pm 1,\mp 1\right)\thinspace . $
$\underline {\rm {Global\thinspace\thinspace minimum }}$
$$\begin{align}x^2+y^2&=2(a^2+b^2)\\
&=2\left(\frac {1-b^2}{3}+b^2\right)\\
&=\frac 23\left(2b^2+1\right)\geq\frac 23\end{align}$$
Equality occurs iff, when $b=0,\thinspace |a|=\frac {\sqrt 3}{3}$, which corresponds to $\left(x,y\right)=\left(\pm \frac {\sqrt 3}{3},\pm \frac {\sqrt 3}{3}\right).$
This completes the answer.
</s>