If $a^2x^4+b^2y^4=c^6$, then the maximum value of $xy$

Question

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?

Answer 1

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 $ab=0$.

Answer 2

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)

Answer 3

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 $a x^2 = b y^2$.