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Question Number 71255 by ajfour last updated on 13/Oct/19

Commented by MJS last updated on 13/Oct/19

$$\mathrm{the}\mathrm{maximum}\mathrm{area}\mathrm{depends}\mathrm{on}a$$$$\mathrm{the}\mathrm{area}\mathrm{for}\theta =90°\mathrm{is}\frac{\pi }{4}b\mathrm{and}\mathrm{this}\mathrm{value}\mathrm{is}$$$$\mathrm{again}\mathrm{reached}\mathrm{st}\theta =45°.\mathrm{this}\mathrm{is}\mathrm{only}\mathrm{possible}$$$$\mathrm{if}a=\sqrt{2}b\mathrm{at}\mathrm{least}.\mathrm{for}b⩽a⩽\sqrt{2}b\mathrm{the}\mathrm{maximum}$$$$\mathrm{area}\mathrm{is}\frac{\pi }{4}b;\mathrm{for}a>\sqrt{2}b\mathrm{the}\mathrm{maximum}\mathrm{is}\mathrm{at}r=a$$$$$$

Commented by ajfour last updated on 13/Oct/19

$$thankssir,letmeseeififollow$$$$itallinentirety.$$

Answered by ajfour last updated on 13/Oct/19

$$letsectorradiusx.$$$$x\sin \theta =b$$$$A=\frac{x^2\theta }{2}\Rightarrow A=\frac{b^2\theta }{2\sin {}^2\theta }$$$$\frac{dA}{d\theta }=\frac{b^2\{\sin {}^2\theta -\theta \sin 2\theta \}}{2\sin {}^4\theta }$$$$=\frac{b^2(1-\frac{2\theta }{\tan \theta })}{2\sin {}^2\theta }$$$$\frac{dA}{d\theta }=0\Rightarrow 2\theta =\tan \theta$$$${\theta }_c=1.16456rad$$$$\frac{d\left(2\theta \right)}{d\theta }=2\mathrm{\&}\frac{d\tan \theta }{d\theta }=1+4{\theta }^2$$$$for\theta >{\theta }_c\frac{dA}{d\theta }>0$$$$henceaminimumat\theta ={\theta }_c$$$$maximumsectorarea$$$$A_{max}=\frac{a^2}{2}{\tan }^{-1}\frac{b}{a}.$$

Commented by ajfour last updated on 13/Oct/19

$$someoneansweritplease.$$

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