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Question Number 122443 by ajfour last updated on 17/Nov/20

Commented by ajfour last updated on 17/Nov/20

$$Find{\left(\frac{r}{R}\right)}_{max}.$$

Answered by ajfour last updated on 17/Nov/20

$$letleftcornerangleof△be\theta .$$$$2R\cos \theta =p+q$$$$heighth=2R\cos \theta \sin \theta =p+r$$$$baseb=2R\cos {}^2\theta =q+r$$$$\Rightarrow 2R\cos \theta =2R\cos \theta \sin \theta -r$$$$+2R\cos {}^2\theta -r$$$$\Rightarrow \frac{r}{R}=\cos \theta \sin \theta -\cos \theta +\cos {}^2\theta$$$$=\cos \theta (\cos \theta +\sin \theta -1)$$$$\frac{d}{d\theta }\left(\frac{r}{R}\right)=-\sin {}^2\theta +\cos {}^2\theta +\sin \theta$$$$-2\cos \theta \sin \theta =0$$$$let\cos \theta =t$$$$\Rightarrow (1-t^2){(1-2t)}^2={(1-2t^2)}^2$$$$\Rightarrow (1-t^2)(4t^2-4t+1)=(4t^4-4t^2+1)$$$$\Rightarrow 8t^4-4t^3-7t^2+4t=0$$$$\Rightarrow t=0isnotappropriate$$$$\Rightarrow t^3-\frac{t^2}{2}-\frac{7t}{8}+\frac{1}{2}=0$$$$positivevaluesoftare$$$$t\approx 0.63111,\underset{-}{0.82694}$$$$correspindingvaluesof$$$$\left(\frac{r}{R}\right)=t(t+\sqrt{1-t^2}-1)$$$$are\approx 0.25674,\underset{-}{0.32187}$$$${\left(\frac{r}{R}\right)}_{max}\approx 0.32187$$$${\theta }_0\approx {\cos }^{-1}(0.82694)\approx 34.21432°$$

Commented by ajfour last updated on 17/Nov/20

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