Alternative approaches to maximize $y=x\sqrt{100-x^2}$ 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?

Let $f(x) = x\sqrt{100-x^{2}}$. Then $x\in[-10,10]$.
In order to find the maximum value, we can consider that $x\geq 0$ (otherwise $f(x) \leq 0$).
For such values of $x$, one has that
\begin{align*}
f(x) & = x\sqrt{100 - x^{2}}\\\\
& = \sqrt{100x^{2} - x^{4}}\\\\
& = \sqrt{2500 - (2500 - 100x^{2} + x^{4})}\\\\
& = \sqrt{2500 - (50-x^{2})^{2}}
\end{align*}
Since the square root function is strictly increasing, $f(x)$ attains its maximum when $x^{2} = 50$.
Hopefully this helps!
