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Question Number 54152 by ajfour last updated on 30/Jan/19

Commented by ajfour last updated on 30/Jan/19

$$GivenarcoflengthL.Findradius$$$$Rsuchthatsegmentareaisa$$$$\left(i\right)maximum\left(ii\right)minimum.$$

Commented by ajfour last updated on 30/Jan/19

$$A=\frac{R^2}{2}(\theta -\sin \theta ),where\theta =\frac{L}{R}$$$$\Rightarrow \frac{2A}{L^2}=\frac{1}{\theta }-\frac{\sin \theta }{{\theta }^2}$$$$\frac{dA}{d\theta }=0\Rightarrow -\frac{1}{{\theta }^2}-\frac{\cos \theta }{{\theta }^2}+\frac{2\sin \theta }{{\theta }^3}=0$$$$\Rightarrow \theta (1+\cos \theta )=2\sin \theta$$$$\Rightarrow \theta =0,\pi$$$$\Rightarrow R=\frac{L}{\pi }(formax.area)$$$$whileR\to \mathrm{\infty }(formin.area)$$$$\left(ThankstoMjSSir\right).$$

Commented by MJS last updated on 30/Jan/19

$$\mathrm{as}\mathrm{always}{\mathrm{you}}^{\prime }\mathrm{re}\mathrm{welcome}$$

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