Question

Find the maximum of \(2x + 2\sqrt{x(1-x)}\) over \(0 \leq x \leq 1.\)

Answer 1

Olympiad or calculus question?

Answer 2

\[
f(x)=2x+2\sqrt{x(1-x)}\\
f(x)=2x+2\sqrt{\left(-x^2+x\right)}\\
f'(x)=2+\left(-x^2+x\right)^{-\frac{1}{2}}(-2x+1)
\]
\[
\begin{align*}
f'(x)&=0\\
2+\left(-x^2+x\right)^{-\frac{1}{2}}(-2x+1)&=0\\
\left(-x^2+x\right)^{-\frac{1}{2}}(-2x+1)&=-2\\
-2x+1&=-2\left(-x^2+x\right)^{\frac{1}{2}}\\
2x-1&=2\left(-x^2+x\right)^{\frac{1}{2}}\\
4x^2-4x+1&=4\left(-x^2+x\right)\\
8x^2-8x+1&=0\\
x=\frac{8+\sqrt{32}}{16}&\text{ or }x=\frac{8-\sqrt{32}}{16}\\
x=\frac{2+\sqrt{2}}{4}&\text{ or }x=\frac{2-\sqrt{2}}{4}\\
\end{align*}
\]
\[
\begin{align*}
f\left(\frac{2+\sqrt{2}}{4}\right)&=\frac{2+\sqrt{2}}{2}+2\sqrt{-\left(\frac{1}{4}+\frac{\sqrt{2}}{4}+\frac{1}{8}\right)+\frac{2+\sqrt{2}}{4}}\\
&=\frac{2+\sqrt{2}}{2}+2\sqrt{\frac{1}{8}}\\
&=\frac{2+\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\\
&=1+\sqrt{2}\\\\
f\left(\frac{2-\sqrt{2}}{4}\right)&=\frac{2-\sqrt{2}}{2}+2\sqrt{-\left(\frac{1}{4}-\frac{\sqrt{2}}{4}+\frac{1}{8}\right)+\frac{2-\sqrt{2}}{4}}\\
&=\frac{2-\sqrt{2}}{2}+2\sqrt{\frac{1}{8}}\\
&=\frac{2-\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\\
&=1-\sqrt{2}
\end{align*}\\
f\left(\frac{2+\sqrt{2}}{4}\right)=1+\sqrt{2}\text{ is the maximum.}
\]

Answer 3

Took me 40+ minutes since I am really tired.

Answer 4

it was a precalc question