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zmudz
 one year ago
Find the maximum of \(2x + 2\sqrt{x(1x)}\) over \(0 \leq x \leq 1.\)
zmudz
 one year ago
Find the maximum of \(2x + 2\sqrt{x(1x)}\) over \(0 \leq x \leq 1.\)

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thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1Olympiad or calculus question?

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1\[ f(x)=2x+2\sqrt{x(1x)}\\ f(x)=2x+2\sqrt{\left(x^2+x\right)}\\ f'(x)=2+\left(x^2+x\right)^{\frac{1}{2}}(2x+1) \] \[ \begin{align*} f'(x)&=0\\ 2+\left(x^2+x\right)^{\frac{1}{2}}(2x+1)&=0\\ \left(x^2+x\right)^{\frac{1}{2}}(2x+1)&=2\\ 2x+1&=2\left(x^2+x\right)^{\frac{1}{2}}\\ 2x1&=2\left(x^2+x\right)^{\frac{1}{2}}\\ 4x^24x+1&=4\left(x^2+x\right)\\ 8x^28x+1&=0\\ x=\frac{8+\sqrt{32}}{16}&\text{ or }x=\frac{8\sqrt{32}}{16}\\ x=\frac{2+\sqrt{2}}{4}&\text{ or }x=\frac{2\sqrt{2}}{4}\\ \end{align*} \] \[ \begin{align*} f\left(\frac{2+\sqrt{2}}{4}\right)&=\frac{2+\sqrt{2}}{2}+2\sqrt{\left(\frac{1}{4}+\frac{\sqrt{2}}{4}+\frac{1}{8}\right)+\frac{2+\sqrt{2}}{4}}\\ &=\frac{2+\sqrt{2}}{2}+2\sqrt{\frac{1}{8}}\\ &=\frac{2+\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\\ &=1+\sqrt{2}\\\\ f\left(\frac{2\sqrt{2}}{4}\right)&=\frac{2\sqrt{2}}{2}+2\sqrt{\left(\frac{1}{4}\frac{\sqrt{2}}{4}+\frac{1}{8}\right)+\frac{2\sqrt{2}}{4}}\\ &=\frac{2\sqrt{2}}{2}+2\sqrt{\frac{1}{8}}\\ &=\frac{2\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\\ &=1\sqrt{2} \end{align*}\\ f\left(\frac{2+\sqrt{2}}{4}\right)=1+\sqrt{2}\text{ is the maximum.} \]

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1Took me 40+ minutes since I am really tired.

zmudz
 one year ago
Best ResponseYou've already chosen the best response.0it was a precalc question
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