If $a^2x^4+b^2y^4=c^6$, then the maximum value of $xy$ Simply subsisting the value of $y$ in $xy$ will not work, but I don’t know any other method to solve it. Can I get a hint?
 A: Applying AM-GM inequality for $2$ variables (which, if you're not familiar with)
$$ (a-b)^2 \geq 0 \Rightarrow a^2+b^2 \geq 2ab $$
$$ (ax^2)^2 + (by^2)^2 \geq 2abx^2y^2 \Rightarrow \frac{c^6}{2ab} \geq x^2y^2 $$
Therefore the maximum value is $\frac{c^3}{\sqrt{2ab}}$ which is obtained when $ax^2 
 = by^2 $
(Thanks to Reinhard Meier for spotting an error I made)
A: By AM-GM $$c^6\geq2\sqrt{a^2b^2x^4y^4}=2|ab|x^2y^2.$$
Thus, for $ab\neq0$ $$xy\leq|xy|\leq\frac{|c^3|}{\sqrt{2|ab|}}.$$
The equality occurs for $a^2x^4=b^2y^4$ and $xy\geq0$.
Also, we need to understand, what happens for $ab=0$.
A: Method 1
You could use the parametric equation of ellipse to solve it.
Let $p=x^2,q=y^2$ then $x^4 = p^2,y^4=y^2$, the equation turns to
$\frac{p^2}{c^6/a^2} + \frac{q^2}{c^6/b^2}=1$, which is the ellipse function. Thus
$p = \frac{c^3}{a} \cos\theta$
$q = \frac{c^3}{b} \sin\theta,\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]$
Since $xy = \sqrt{pq}$, $\max xy = \sqrt{\frac{c^6}{2ab}}$
the maximum is taken when $\theta=2k\pi+ \frac{\pi}{2},k\in\mathbb{Z}$
Method 2
Use the AM-GM inequality:
$c^6 = (ax^2)^2 + (by^2)^2 \ge 2ab(xy)^2$
thus
$xy \le \sqrt{\frac{c^6}{2ab}}$
equal is taken when $a x^2 = b y^2$.
