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Question Number 84655 by ajfour last updated on 14/Mar/20

Commented by ajfour last updated on 14/Mar/20

$$Ifareaof△ABCisequalto$$$$maxtrapeziumareaBCED,$$$$findrintermsofellipse$$$$parametersaandb.$$

Answered by mr W last updated on 15/Mar/20

Commented by mr W last updated on 15/Mar/20

$$let\alpha ={\cos }^{-1}\frac{h}{b}$$$$BC=2a\sin \alpha$$$${\left[BCED\right]}_{max}=\frac{ab}{2}[3\sin \left(\frac{2\alpha }{3}\right)-\sin \left(2\alpha \right)]$$$$\left[ABC\right]=a(b+h)\sin \alpha$$$$\frac{ab}{2}[3\sin \left(\frac{2\alpha }{3}\right)-\sin \left(2\alpha \right)]=a(b+h)\sin \alpha$$$$3\sin \left(\frac{2\alpha }{3}\right)-\sin \left(2\alpha \right)=2(1+\frac{h}{b})\sin \alpha$$$$let\lambda =\frac{h}{b}$$$$\Rightarrow \alpha ={\cos }^{-1}\lambda$$$$\Rightarrow 3\sin \left(\frac{2\alpha }{3}\right)=2(2\lambda +1)\sin \alpha$$$$\Rightarrow \lambda \approx 0.121555$$$$$$$$AC=\sqrt{a^2(1-{\lambda }^2)+b^2{(1+\lambda )}^2}$$$$ab(1+\lambda )\sin \alpha =r(a\sin \alpha +\sqrt{a^2(1-{\lambda }^2)+b^2{(1+\lambda )}^2}$$$$b(1+\lambda )=r[1+\sqrt{1+\frac{(1+\lambda )b^2}{(1-\lambda )a^2}}]$$$$\Rightarrow \frac{r}{b}=\frac{1+\lambda }{1+\sqrt{1+\frac{(1+\lambda )b^2}{(1-\lambda )a^2}}}$$$$examples:$$$$\frac{b}{a}=1\Rightarrow \frac{r}{b}=0.447032$$$$\frac{b}{a}=\frac{2}{3}\Rightarrow \frac{r}{b}=0.498031$$$$\frac{b}{a}=\frac{1}{2}\Rightarrow \frac{r}{b}=0.522003$$$$\frac{b}{a}=2\Rightarrow \frac{r}{b}=0.323100$$

Commented by ajfour last updated on 15/Mar/20

$$Siri{couldn}^{\prime }tgetyourlogicfor$$$${\left[BCED\right]}_{max},kindlyenlighten..$$

Commented by mr W last updated on 15/Mar/20

Commented by mr W last updated on 15/Mar/20

$$whenwirshrinktheellipseinx-$$$$directioninratio\frac{b}{a}toacircle,the$$$$ratiooftheareasfromthetriangle$$$$ABCtotrapeziumBCEDremains$$$$unchanged,i.e.$$$$\frac{\left[ABC\right]}{\left[BCED\right]}=\frac{\left[{AB}^{\prime }{C}^{\prime }\right]}{\left[B^{\prime }{C}^{\prime }{E}^{\prime }{D}^{\prime }\right]}$$$$forthecirclewithradiusb,the$$$$trapezium{B}^{\prime }{C}^{\prime }{E}^{\prime }{D}^{\prime }ismaximum,$$$$when{B}^{\prime }{D}^{\prime }=D^{\prime }{E}^{\prime }=E^{\prime }{C}^{\prime }\left(proofomitted\right).$$$$$$$$\cos \alpha =\frac{h}{b}\Rightarrow \alpha ={\cos }^{-1}\frac{h}{b}$$$$\beta =\frac{2\alpha }{3}$$$${\left[B^{\prime }{C}^{\prime }{E}^{\prime }{D}^{\prime }\right]}_{max}=3\times \frac{b^2\sin \beta }{2}-\frac{b^2\sin 2\alpha }{2}$$$$=\frac{b^2}{2}[3\sin \left(\frac{2\alpha }{3}\right)-\sin 2\alpha ]$$$${\left[BCED\right]}_{max}=\frac{a}{b}\times {\left[B^{\prime }{C}^{\prime }{E}^{\prime }{D}^{\prime }\right]}_{max}$$$$=\frac{ab}{2}(3\sin \left(\frac{2\alpha }{3}\right)-\sin 2\alpha )$$$$$$$$B^{\prime }{C}^{\prime }=2b\sin \alpha$$$$BC=\frac{a}{b}\times B^{\prime }{C}^{\prime }=2a\sin \alpha$$

Commented by ajfour last updated on 15/Mar/20

$$Thankssirfortheexplanation,$$$$ishallbethoroughlythroughit$$$$bytomorrow..$$

Answered by ajfour last updated on 15/Mar/20

$$C(a\cos \theta ,-b\sin \theta );$$$$E(a\cos \varphi ,-b\sin \varphi ).$$$$say{△}_{ABC}=S,{Area}_{BCED}=T$$$$S=Tand\frac{\partial T}{\partial \varphi }=0gives$$$$\cos (\varphi -\theta )+\cos 2\varphi -\sin 2\theta$$$$+\sin (\varphi -\theta )=0\mathrm{\&}$$$$\sin (\varphi -\theta )+\frac{\sin 2\varphi }{2}-\sin 2\theta -\cos \theta =0$$$$butiknownothowtoobtain$$$$\theta and\varphi fromthesetwoeqs.Sir..$$

Commented by mr W last updated on 16/Mar/20

$$A=ab\cos \theta (1+\sin \theta )$$$$S=ab(\cos \theta +\cos \varphi )(\sin \varphi -\sin \theta )$$$$$$$$\frac{dS}{d\varphi }=0$$$$(\cos \theta +\cos \varphi )\cos \varphi -\sin \varphi (\sin \varphi -\sin \theta )=0$$$$\cos \theta \cos \varphi +{\cos }^2\varphi +\sin \varphi \sin \theta -{\sin }^2\varphi =0$$$$\Rightarrow \cos (\varphi -\theta )+\cos 2\varphi =0$$$$\Rightarrow \varphi -\theta +2\varphi =\pi$$$$\Rightarrow \varphi =\frac{\pi +\theta }{3}$$$$$$$$A=S$$$$\cos \theta (1+\sin \theta )=(\cos \theta +\cos \varphi )(\sin \varphi -\sin \theta )$$$$\cos \theta +\sin \theta \cos \theta =\sin \varphi \cos \theta +\cos \varphi \sin \varphi -\sin \theta \cos \theta -\cos \varphi \sin \theta$$$$\sin (\varphi -\theta )+\frac{\sin 2\varphi }{2}-\sin 2\theta -\cos \theta =0$$$$\sin \left(\frac{\pi -2\theta }{3}\right)+\frac{1}{2}\sin \left(\frac{2\pi +2\theta }{3}\right)-\sin 2\theta -\cos \theta =0$$$$\Rightarrow \frac{3}{2}\sin \left(\frac{\pi -2\theta }{3}\right)-\sin 2\theta -\cos \theta =0$$$$\Rightarrow \theta \approx 6.981874°$$$$\Rightarrow \frac{h}{b}=\sin \theta \approx 0.121555(=\lambda )$$

Commented by mr W last updated on 16/Mar/20

$$theresultisthesameasmine.$$

Commented by ajfour last updated on 16/Mar/20

$$Thankyousir,beautifully$$$$handled!$$

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