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

Commented by ajfour last updated on 30/Jan/19

Given arc of length L. Find radius R such that segment area is a (i)maximum (ii)minimum.

$${Given}\:{arc}\:{of}\:{length}\:{L}.\:{Find}\:{radius} \\ $$$${R}\:{such}\:{that}\:{segment}\:{area}\:{is}\:{a} \\ $$$$\:\:\:\:\left({i}\right){maximum}\:\:\left({ii}\right){minimum}. \\ $$

Commented by ajfour last updated on 30/Jan/19

A=(R^2 /2)(θ−sin θ) , where θ=(L/R) ⇒ ((2A)/L^2 )= (1/θ)−((sin θ)/θ^2 ) (dA/dθ) = 0 ⇒ −(1/θ^2 )−((cos θ)/θ^2 )+((2sin θ)/θ^3 ) = 0 ⇒ θ(1+cos θ)=2sin θ ⇒ θ= 0,π ⇒ R = (L/π) (for max. area) while R→∞ (for min. area) (Thanks to MjS Sir).

$${A}=\frac{{R}^{\mathrm{2}} }{\mathrm{2}}\left(\theta−\mathrm{sin}\:\theta\right)\:\:\:\:,\:{where}\:\theta=\frac{{L}}{{R}} \\ $$$$\Rightarrow\:\frac{\mathrm{2}{A}}{{L}^{\mathrm{2}} }=\:\frac{\mathrm{1}}{\theta}−\frac{\mathrm{sin}\:\theta}{\theta^{\mathrm{2}} } \\ $$$$\frac{{dA}}{{d}\theta}\:=\:\mathrm{0}\:\:\Rightarrow\:−\frac{\mathrm{1}}{\theta^{\mathrm{2}} }−\frac{\mathrm{cos}\:\theta}{\theta^{\mathrm{2}} }+\frac{\mathrm{2sin}\:\theta}{\theta^{\mathrm{3}} }\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\theta\left(\mathrm{1}+\mathrm{cos}\:\theta\right)=\mathrm{2sin}\:\theta \\ $$$$\Rightarrow\:\theta=\:\mathrm{0},\pi \\ $$$$\Rightarrow\:\:{R}\:=\:\frac{{L}}{\pi}\:\:\left({for}\:{max}.\:{area}\right) \\ $$$${while}\:\:\:{R}\rightarrow\infty\:\:\left({for}\:{min}.\:{area}\right) \\ $$$$\left({Thanks}\:{to}\:{MjS}\:{Sir}\right). \\ $$

Commented by MJS last updated on 30/Jan/19

as always you′re welcome

$$\mathrm{as}\:\mathrm{always}\:\mathrm{you}'\mathrm{re}\:\mathrm{welcome} \\ $$

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