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?
 A: 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!
A: Square both sides to get $y^2 = x^2 (100-x^2) \implies -x^4+100x^2-y^2=0$. This is a quadratic in $x^2$: when $\Delta = 0$, $100^2 - 4(-1)(-y^2) = 0 \implies y = ±50$.
The maximum value is $y = 50$ as considering the negative branch, $f(-x) = -f(x)$, hence the maximum of the positive branch is the same as the negative branch.
A: 
$\,\\\;\;\;\;\;\;\;\; 2\,S \;=\; x \cdot \sqrt{100-x^2} \;=\; h \cdot 10 \;\le\; 5 \cdot 10$
A: $$x\sqrt{100-x^2}=A$$
$$\begin{align}&\implies \left(\frac Ax\right)^2=100-x^2,~x≠0 \\
&\implies \frac{A^2}{x^2}+x^2=100 \\
&\implies\left(\frac Ax-x\right)^2+2A=100 \\
&\implies \left(\frac Ax-x\right)^2=100-2A\\
&\implies 100-2A≥0\\
&\implies A≤50.\end{align}$$
