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Question Number 214372 by ajfour last updated on 06/Dec/24

Commented by ajfour last updated on 06/Dec/24

$$Findmaximumr.Outerfigureisa$$$$rhombus.Circleisinscribedinan$$$$isoscelestriangleasshownabove.$$

Answered by a.lgnaoui last updated on 07/Dec/24

$$\mathrm{on}\mathrm{peut}\mathrm{verifier}\mathrm{facilement}\mathrm{que}\mathrm{le}\mathrm{rayon}\mathbf{r}\mathbf{est}$$$$\mathbf{maximum}\mathrm{lorsqaue}\mathbf{y}=\mathbf{x}=45°$$$$\mathrm{alors}\mathrm{AB}=\frac{\mathrm{a}}{2}\tan 45=\frac{\mathbf{r}}{\mathbf{a}/2}=\frac{2\mathbf{r}}{\mathbf{a}}$$$$\mathbf{alors}\mathbf{r}=\frac{\mathbf{a}}{2}$$$$$$

Commented by a.lgnaoui last updated on 07/Dec/24

Commented by mr W last updated on 07/Dec/24

$$non-senseagain.thecircleshould$$$$beinscribedintheisosceles$$$$triangle,notintherhombus!$$

Answered by mr W last updated on 07/Dec/24

Commented by mr W last updated on 07/Dec/24

$$0<\theta ⩽\alpha <\frac{\pi }{2}$$$$OB=\frac{r+\frac{r}{\sin \theta }}{\cos \theta }=\frac{(1+\sin \theta )r}{\sin \theta \cos \theta }$$$$\frac{OB}{\sin (\pi -2\alpha )}=\frac{a}{\sin (\pi -2\alpha +\alpha -\theta )}$$$$\frac{(1+\sin \theta )r}{\sin \theta \cos \theta \sin 2\alpha }=\frac{a}{\sin (\theta +\alpha )}=\frac{a}{\sin \theta \cos \alpha +\cos \theta \sin \alpha }$$$$\Rightarrow \frac{r}{2a}=\frac{\sin \theta \sin \alpha }{(1+\sin \theta )(\tan \theta +\tan \alpha )}$$$$\mathrm{\Phi }=\ln \left(\frac{r}{2a}\right)=\ln \left(\sin \theta \right)+\ln \left(\sin \alpha \right)-\ln (1+\sin \theta )-\ln (\tan \theta +\tan \alpha )$$$$r_{max}\iff {\mathrm{\Phi }}_{max}$$$$\frac{\partial \mathrm{\Phi }}{\partial \alpha }=\frac{1}{\tan \alpha }-\frac{1}{(\tan \theta +\tan \alpha ){\cos }^2\alpha }=0$$$$\tan \theta =(\frac{1}{{\cos }^2\alpha }-1)\tan \alpha ={\tan }^3\alpha$$$$\Rightarrow \tan \alpha =\sqrt[3]{\tan \theta }$$$$\frac{\partial \mathrm{\Phi }}{\partial \theta }=\frac{1}{\tan \theta }-\frac{\cos \theta }{1+\sin \theta }-\frac{1}{(\tan \theta +\tan \alpha ){\cos }^2\theta }=0$$$$(\frac{1}{\tan \theta }-\frac{\cos \theta }{1+\sin \theta }){\cos }^2\theta =\frac{1}{\tan \theta +\tan \alpha }$$$$\Rightarrow {\cos }^2\theta (\frac{1}{\tan \theta }-\frac{\cos \theta }{1+\sin \theta })=\frac{1}{\tan \theta +\sqrt[3]{\tan \theta }}$$$$\Rightarrow \theta \approx 0.5984(34.286°)$$$$\Rightarrow \alpha \approx 0.7218(41.356°)\ne 45°$$$$\Rightarrow \frac{r_{max}}{a}\approx 0.304840051439✓$$$$examplewitha=4seebelow$$

Commented by mr W last updated on 07/Dec/24

Commented by ajfour last updated on 08/Dec/24

$$Thanksforyoursolution.Iam$$$$tryingstillsometimes..$$

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