Question Number 54152 by ajfour last updated on 30/Jan/19
Commented by ajfour last updated on 30/Jan/19
$${Given}\:{arc}\:{of}\:{length}\:{L}.\:{Find}\:{radius} \\ $$$${R}\:{such}\:{that}\:{segment}\:{area}\:{is}\:{a} \\ $$$$\:\:\:\:\left({i}\right){maximum}\:\:\left({ii}\right){minimum}. \\ $$
Commented by ajfour last updated on 30/Jan/19
$${A}=\frac{{R}^{\mathrm{2}} }{\mathrm{2}}\left(\theta−\mathrm{sin}\:\theta\right)\:\:\:\:,\:{where}\:\theta=\frac{{L}}{{R}} \\ $$$$\Rightarrow\:\frac{\mathrm{2}{A}}{{L}^{\mathrm{2}} }=\:\frac{\mathrm{1}}{\theta}−\frac{\mathrm{sin}\:\theta}{\theta^{\mathrm{2}} } \\ $$$$\frac{{dA}}{{d}\theta}\:=\:\mathrm{0}\:\:\Rightarrow\:−\frac{\mathrm{1}}{\theta^{\mathrm{2}} }−\frac{\mathrm{cos}\:\theta}{\theta^{\mathrm{2}} }+\frac{\mathrm{2sin}\:\theta}{\theta^{\mathrm{3}} }\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\theta\left(\mathrm{1}+\mathrm{cos}\:\theta\right)=\mathrm{2sin}\:\theta \\ $$$$\Rightarrow\:\theta=\:\mathrm{0},\pi \\ $$$$\Rightarrow\:\:{R}\:=\:\frac{{L}}{\pi}\:\:\left({for}\:{max}.\:{area}\right) \\ $$$${while}\:\:\:{R}\rightarrow\infty\:\:\left({for}\:{min}.\:{area}\right) \\ $$$$\left({Thanks}\:{to}\:{MjS}\:{Sir}\right). \\ $$
Commented by MJS last updated on 30/Jan/19
$$\mathrm{as}\:\mathrm{always}\:\mathrm{you}'\mathrm{re}\:\mathrm{welcome} \\ $$