TITLE: Alternative approaches to maximize $y=x\sqrt{100-x^2}$
QUESTION [5 upvotes]: I could find three good approaches to find maximum of the function $y=x\sqrt{100-x^2}$. I will explain them briefly :
First: Finding $x$ satisfies $y'=0$ then plugging it in the function.
Second: Using the substitution $x=10\sin\theta$ (or $x=10\cos\theta)$ for
$\theta\in(0,\frac{\pi}2)$ to get $y=100\sin\theta\cos\theta=50\sin(2\theta)$ hence the maximum is $50$.
Third: Using AM-GM inequality: It is obvious that maximum occurs for $x>0$ So we can rewrite $y$ as $y=\sqrt{x^2(100-x^2)}$ . Now the sum of $x^2$ and $100-x^2$ is $100$ so the maximum of product happens when $x^2=100-x^2$ or $x^2=50$  Hence $y_{\text{max}}=50$.
Just for fun, can you maximize $y=x\sqrt{100-x^2}$ with other approaches?

REPLY [2 votes]: $$x\sqrt{100-x^2}=A$$
$$\begin{align}&\implies \left(\frac Ax\right)^2=100-x^2,~x≠0 \\
&\implies \frac{A^2}{x^2}+x^2=100 \\
&\implies\left(\frac Ax-x\right)^2+2A=100 \\
&\implies \left(\frac Ax-x\right)^2=100-2A\\
&\implies 100-2A≥0\\
&\implies A≤50.\end{align}$$