If $a^2{x}^4+b^2{y}^4=c^6$ then the maximum value of xy is---

$\begin{array}{l}{a}^2{x}^4+b^2{y}^4=c^6\to 1\\ AM\ge GM\text{ for }{a}^2{x}^4\text{ and }{b}^2{y}^4\\ \Rightarrow \frac{a^2{x}^4+b^2{y}^4}{2}\ge \sqrt{a^2{x}^4\cdot b^2{y}^4}\\ \Rightarrow \frac{c^6}{2}\ge ab\cdot x^2{y}^2\\ \Rightarrow xy\le \frac{c^3}{2\sqrt{ab}}\end{array}$
$\therefore$max. value of xy is $\frac{c^3}{2\sqrt{ab}}$