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Question Number 15736 by ajfour last updated on 13/Jun/17

$$FromapointAonthecircum-$$$$ferenceofacircleofradius\mathit{r},a$$$$perpendicularAFisdroppedon$$$$atangenttothecircleatP.$$$$Findthemaximumpossible$$$$areaof\mathrm{\Delta }APF.$$

Answered by mrW1 last updated on 13/Jun/17

$$\mathrm{let}\theta =\mathrm{\angle }\mathrm{AOP}$$$$\mathrm{O}=\mathrm{center}\mathrm{of}\mathrm{circle}$$$$\mathrm{PF}=\mathrm{rsin}\theta$$$$\mathrm{AF}=\mathrm{r}-\mathrm{rcos}\theta$$$${\mathrm{A}}_{\mathrm{\Delta }\mathrm{APF}}=\frac{1}{2}\times \mathrm{rsin}\theta \times (\mathrm{r}-\mathrm{rcos}\theta )=\frac{{\mathrm{r}}^2}{2}\sin \theta (1-\cos \theta )$$$$=\frac{{\mathrm{r}}^2}{2}\mathrm{f}\left(\theta \right)$$$$\mathrm{with}\mathrm{f}\left(\theta \right)=\sin \theta (1-\cos \theta )$$$$\frac{\mathrm{df}}{\mathrm{d}\theta }=\cos \theta (1-\cos \theta )+{\sin }^2\theta =\cos \theta -{2\cos }^2\theta +1=0$$$$\cos \theta =\frac{1\pm \sqrt{1+8}}{4}=\frac{1\pm 3}{4}=1,-\frac{1}{2}$$$$\Rightarrow \theta =0°\left(\mathrm{not}\mathrm{what}\mathrm{we}\mathrm{need}\right)$$$$\Rightarrow \theta =120°$$$$\max .{\mathrm{A}}_{\mathrm{\Delta }\mathrm{APF}}=\frac{{\mathrm{r}}^2}{2}\times \sin 120\times (1-\cos 120)$$$$=\frac{{\mathrm{r}}^2}{2}\times \frac{\sqrt{3}}{2}\times (1+\frac{1}{2})$$$$=\frac{3\sqrt{3}{\mathrm{r}}^2}{8}$$

Commented by mrW1 last updated on 13/Jun/17

Commented by ajfour last updated on 13/Jun/17

$$greatsir,istherealsoaway$$$$ofgeometrytoarriveatthe$$$$resultwithouthavingto$$$$differentiate..$$

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