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Question Number 186302 by ajfour last updated on 03/Feb/23

Commented by ajfour last updated on 03/Feb/23

$$Canitbe=\frac{1}{6}?$$

Commented by Frix last updated on 04/Feb/23

$$\mathrm{The}\mathrm{maximum}\mathrm{area}\mathrm{is}\mathrm{reached}\mathrm{when}\mathrm{the}$$$$\mathrm{left}\mathrm{vertice}\left(\mathrm{with}\mathrm{the}\mathrm{right}\mathrm{angle}\right)\mathrm{is}\mathrm{located}$$$$\mathrm{on}\mathrm{the}\mathrm{horizontal}\mathrm{diameter}\mathrm{of}\mathrm{the}\mathrm{circle}.$$$$\mathrm{The}\mathrm{maximum}\mathrm{area}\mathrm{is}\frac{1}{6}.$$

Commented by Frix last updated on 04/Feb/23

$$\mathrm{Sorry}\mathrm{I}\mathrm{calculated}\mathrm{with}r=1\Rightarrow$$$$\mathrm{maximum}\mathrm{area}=2\mathrm{and}\mathrm{then}\mathrm{forgot}\mathrm{a}\mathrm{factor}...$$

Answered by ajfour last updated on 04/Feb/23

Commented by ajfour last updated on 04/Feb/23

$$A=p\sin \theta (p\cos \theta +\sqrt{R^2-p^2\sin {}^2\theta })$$$$herep=2R,hence$$$$\frac{A}{p^2}\left(\theta \right)=\sin \theta \cos \theta +\sin \theta \sqrt{\frac{1}{4}-\sin {}^2\theta }$$$$\frac{d}{d\theta }\left(\frac{A}{4R^2}\right)=\cos 2\theta +\cos \theta \sqrt{\frac{1}{4}-\sin {}^2\theta }$$$$-\frac{\sin {}^2\theta \cos \theta }{\sqrt{\frac{1}{4}-\sin {}^2\theta }}=0$$$$\Rightarrow \cos {}^{2}2\theta (\frac{1}{4}-\sin {}^2\theta )$$$$=\cos {}^2\theta (2\sin {}^2\theta -\frac{1}{4})$$$$say\sin {}^2\theta =s$$$$\Rightarrow {(1-2s)}^2(\frac{1}{4}-s)=(1-s)(2s-\frac{1}{4})$$$$\Rightarrow s=\frac{1}{5}\Rightarrow \sin \theta =\frac{1}{\sqrt{5}},\cos \theta =\frac{2}{\sqrt{5}}$$$$Hence$$$$\frac{A_{max}}{4R^2}=\frac{2}{5}+\frac{1}{\sqrt{5}}\sqrt{\frac{1}{4}-\frac{1}{5}}=\frac{1}{2}$$$$and2R\sqrt{3}=1$$$$\Rightarrow A_{max}=\frac{1}{2}\left(4R^2\right)=\frac{1}{2}\left(\frac{1}{3}\right)=\frac{1}{6}$$

Commented by mr W last updated on 05/Feb/23

$$nicesolutionsir!$$

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