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Question Number 41157 by ajfour last updated on 02/Aug/18

Commented by ajfour last updated on 02/Aug/18

$I f a r c l e n g t h i s c o n s t a n t, e q u a l$ $t o l, a n d s e g m e n t a r e a a m a x i m u m,$ $f i n d t h e r a d i u s .$
Commented by MJS last updated on 02/Aug/18

$I came to the conclusion that$ $r = \frac{l}{π} \Rightarrow area = \frac{l^{2}}{2 π}$ ${didn}^{'} t calculate it but it seems obvious to me$
Commented by ajfour last updated on 03/Aug/18

$y e s s i r, s e m i c i r c l e .$
	Answered by MrW3 last updated on 02/Aug/18

	$θ = \frac{l}{r}$ $A = \frac{r^{2}}{2} (θ - \sin θ) = \frac{r^{2}}{2} (\frac{l}{r} - \sin \frac{l}{r}) = \frac{l r}{2} - \frac{r^{2}}{2} \sin \frac{l}{r}$ $\frac{d A}{d r} = \frac{l}{2} - r \sin \frac{l}{r} - \frac{r^{2}}{2} \cos \frac{l}{r} (- \frac{l}{r^{2}})$ $= \frac{l}{2} - r \sin \frac{l}{r} + \frac{l}{2} \cos \frac{l}{r} = 0$ $\Rightarrow \frac{l}{r} (1 + \cos \frac{l}{r}) = 2 \sin \frac{l}{r}$ $\Rightarrow (\frac{l}{2 r} \cos \frac{l}{2 r} - \sin \frac{l}{2 r}) \cos \frac{l}{2 r} = 0$ $\Rightarrow \cos \frac{l}{2 r} = 0 \Rightarrow \frac{l}{2 r} = \frac{π}{2} \Rightarrow r = \frac{l}{π}$ $\Rightarrow \frac{l}{2 r} \cos \frac{l}{2 r} - \sin \frac{l}{2 r} = 0$ $\Rightarrow \tan \frac{l}{2 r} = \frac{l}{2 r} \Rightarrow n o s o l u t i o n w i t h \frac{l}{2 r} < π$
	Answered by tanmay.chaudhury50@gmail.com last updated on 03/Aug/18
	$l=rθ segment area=((r^2 θ)/2)−(1/2)×b×h sin(θ/2)=((b/2)/r) b=2rsin(θ/2) cos(θ/2)=(h/r) h=rcos(θ/2) segment area=((r^2 θ)/2) −(1/2).2rsin(θ/2).rcos(θ/2) =((r^2 θ)/2)−(r^2 /2)sinθ=(r^2 /2)(θ−sinθ) A=(r^2 /2)(θ−sinθ) A=(l^2 /2)(((θ−sinθ)/θ^2 )) (dA/dθ)=(l^2 /2){((θ^2 (1−cosθ)−(θ−sinθ)2θ)/θ^4 )} (dA/dθ)=0=θ^2 −θ^2 cosθ−2θ^2 +2θsinθ θ−θcosθ−2θ+2sinθ=0 2sinθ−θcosθ=θ 2sin(θ/2).cos(θ/2)−θ(2cos^2 (θ/2))=0 cos(θ/2)(2sin(θ/2)−2θcos(θ/2))=0 cos(θ/2)=0 θ=Π tanθ−θ=0 segment area=(r^2 /2)(Π−sinΠ) =((Πr^2 )/2)atθ=Π l=rθ r=(l/θ) r=(l/Π)$
	$l = r θ$ $s e g m e n t a r e a = \frac{r^{2} θ}{2} - \frac{1}{2} \times b \times h$ $s i n \frac{θ}{2} = \frac{\frac{b}{2}}{r} b = 2 r s i n \frac{θ}{2}$ $c o s \frac{θ}{2} = \frac{h}{r} h = r c o s \frac{θ}{2}$ $s e g m e n t a r e a = \frac{r^{2} θ}{2} - \frac{1}{2} . 2 r s i n \frac{θ}{2} . r c o s \frac{θ}{2}$ $= \frac{r^{2} θ}{2} - \frac{r^{2}}{2} s i n θ = \frac{r^{2}}{2} (θ - s i n θ)$ $A = \frac{r^{2}}{2} (θ - s i n θ)$ $A = \frac{l^{2}}{2} (\frac{θ - s i n θ}{θ^{2}})$ $\frac{d A}{d θ} = \frac{l^{2}}{2} {\frac{θ^{2} (1 - c o s θ) - (θ - s i n θ) 2 θ}{θ^{4}}}$ $\frac{d A}{d θ} = 0 = θ^{2} - θ^{2} c o s θ - 2 θ^{2} + 2 θ s i n θ$ $θ - θ c o s θ - 2 θ + 2 s i n θ = 0$ $2 s i n θ - θ c o s θ = θ$ $2 s i n \frac{θ}{2} . c o s \frac{θ}{2} - θ (2 {c o s}^{2} \frac{θ}{2}) = 0$ $c o s \frac{θ}{2} (2 s i n \frac{θ}{2} - 2 θ c o s \frac{θ}{2}) = 0$ $c o s \frac{θ}{2} = 0 θ = Π$ $t a n θ - θ = 0$ $s e g m e n t a r e a = \frac{r^{2}}{2} (Π - s i n Π) = \frac{Π r^{2}}{2} a t θ = Π$ $l = r θ r = \frac{l}{θ} r = \frac{l}{Π}$
	Commented by behi83417@gmail.com last updated on 03/Aug/18

	$t h e r e i s a t y p o i n l i n e \neq 4 f r o m e n d .$ $c o s \frac{θ}{2} = 0 \Rightarrow θ = π .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 03/Aug/18

	$y e s t r u e . . . l e t m e r e c t i f y . . .$
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