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Question Number 30299 by ajfour last updated on 19/Feb/18

Commented by ajfour last updated on 19/Feb/18

$F i n d m a x i m u m a r e a o f$ $q u a d r i l a t e r a l A B C D g i v e n a, r .$
Commented by ajfour last updated on 20/Feb/18

Commented by ajfour last updated on 20/Feb/18

$A = 2 (\frac{1}{2} r^{2} \cos α \sin α) +$ $4 (\frac{1}{2} r^{2} \cos ϕ \sin ϕ) + 2 (\frac{1}{2} r^{2} \cos θ \sin θ)$ $A = \frac{r^{2}}{2} (\sin 2 α + 2 \sin 2 ϕ + \sin 2 θ)$ $a l s o α + θ + 2 ϕ = π \Rightarrow \frac{d (2 ϕ)}{d θ} = - 1$ $\frac{d A}{d θ} = \frac{r^{2}}{2} (- 2 \cos 2 ϕ + 2 \cos 2 θ) = 0$ $\Rightarrow θ = ϕ$ $h e n c e 3 θ = π - α$ $A_{m a x} = \frac{r^{2}}{2} (\sin 2 α + 3 \sin 2 θ)$ $\sin α = \frac{a}{2 r}, \cos α = \frac{\sqrt{4 r^{2} - a^{2}}}{2 r}$ $\sin 2 α = \frac{a \sqrt{4 r^{2} - a^{2}}}{2 r^{2}}$ $A_{m a x} = \frac{a}{4} \sqrt{4 r^{2} - a^{2}}$ $+ \frac{3 r^{2}}{2} \sin (\frac{2 π}{3} - \frac{2}{3} \sin^{- 1} \frac{a}{2 r}) .$
Commented by ajfour last updated on 20/Feb/18

$I h a v e a s s u m e d, w i t h o u t p r o o f$ $t h a t q u a d r i l a t e r a l s h a l l b e a n$ $i s o s c e l e s t r a p e z i u m .$
Commented by mrW2 last updated on 20/Feb/18

$I n t h e n a t u r e, t h e r e f o r e a l s o i n$ $m a t h e m a t i c s, o p t i m u m h a s a l w a y s$ $s y m m e t r y .$
	Answered by mrW2 last updated on 19/Feb/18

	Commented by mrW2 last updated on 19/Feb/18
	$sin (θ/2)=(a/(2r)) ⇒θ=2 sin^(−1) (a/(2r))=constant sin θ=2×(a/(2r))×(√(1−(a^2 /(4r^2 ))))=(a/r)(√(1−(a^2 /(4r^2 )))) α+β+γ=ϕ=2π−θ=constant A_1 =A_(AOBCD) =(r^2 /2)(sin α+sin β+sin γ) =(r^2 /2)[sin α+sin β+sin (ϕ−α−β)] (∂A_1 /∂α)=(r^2 /2)[cos α−cos (ϕ−α−β)]=0 ⇒cos α=cos (ϕ−α−β)=cos γ ⇒α=γ (∂A_1 /∂β)=(r^2 /2)[cos β−cos (ϕ−α−β)]=0 ⇒cos β=cos (ϕ−α−β)=cos γ ⇒α=β=γ=(ϕ/3)=(1/3)(2π−2 sin^(−1) (a/(2r)))=((2π)/3)−(2/3)sin^(−1) (a/(2r)) sin α=sin β=sin γ=(1/2)[(√3) cos ((2/3)sin^(−1) (a/(2r)))+sin ((2/3)sin^(−1) (a/(2r)))] max. A_(ABCD) =(r^2 /2){3sin α+sin θ} =(r^2 /2){(3/2)[(√3) cos ((2/3)sin^(−1) (a/(2r)))+sin ((2/3)sin^(−1) (a/(2r)))]+(a/r)(√(1−(a^2 /(4r^2 ))))} =((3r^2 )/4){(√3) cos ((2/3)sin^(−1) (a/(2r)))+sin ((2/3)sin^(−1) (a/(2r)))+((2a)/(3r))(√(1−(a^2 /(4r^2 ))))}$
	$\sin \frac{θ}{2} = \frac{a}{2 r}$ $\Rightarrow θ = 2 \sin^{- 1} \frac{a}{2 r} = c o n s t a n t$ $\sin θ = 2 \times \frac{a}{2 r} \times \sqrt{1 - \frac{a^{2}}{4 r^{2}}} = \frac{a}{r} \sqrt{1 - \frac{a^{2}}{4 r^{2}}}$ $α + β + γ = φ = 2 π - θ = c o n s t a n t$ $A_{1} = A_{A O B C D} = \frac{r^{2}}{2} (\sin α + \sin β + \sin γ)$ $= \frac{r^{2}}{2} [\sin α + \sin β + \sin (φ - α - β)]$ $\frac{\partial A_{1}}{\partial α} = \frac{r^{2}}{2} [\cos α - \cos (φ - α - β)] = 0$ $\Rightarrow \cos α = \cos (φ - α - β) = \cos γ$ $\Rightarrow α = γ$ $\frac{\partial A_{1}}{\partial β} = \frac{r^{2}}{2} [\cos β - \cos (φ - α - β)] = 0$ $\Rightarrow \cos β = \cos (φ - α - β) = \cos γ$ $\Rightarrow α = β = γ = \frac{φ}{3} = \frac{1}{3} (2 π - 2 \sin^{- 1} \frac{a}{2 r}) = \frac{2 π}{3} - \frac{2}{3} \sin^{- 1} \frac{a}{2 r}$ $\sin α = \sin β = \sin γ = \frac{1}{2} [\sqrt{3} \cos (\frac{2}{3} \sin^{- 1} \frac{a}{2 r}) + \sin (\frac{2}{3} \sin^{- 1} \frac{a}{2 r})]$ $m a x . A_{A B C D} = \frac{r^{2}}{2} {3 \sin α + \sin θ}$ $= \frac{r^{2}}{2} {\frac{3}{2} [\sqrt{3} \cos (\frac{2}{3} \sin^{- 1} \frac{a}{2 r}) + \sin (\frac{2}{3} \sin^{- 1} \frac{a}{2 r})] + \frac{a}{r} \sqrt{1 - \frac{a^{2}}{4 r^{2}}}}$ $= \frac{3 r^{2}}{4} {\sqrt{3} \cos (\frac{2}{3} \sin^{- 1} \frac{a}{2 r}) + \sin (\frac{2}{3} \sin^{- 1} \frac{a}{2 r}) + \frac{2 a}{3 r} \sqrt{1 - \frac{a^{2}}{4 r^{2}}}}$
	Commented by ajfour last updated on 20/Feb/18

	$T h a n k y o u v e r y m u c h S i r .$
	Commented by mrW2 last updated on 20/Feb/18

	$I s a w y o u r e a r l i e r c o m m e n t s, b u t I$ $h a d n o t i m e t o a n s w e r .$ $B a s i c a l l y o n e c a n a l s o u s e o n l y o n e v a r i a b l e$ $t o s o l v e t h e q u e s t i o n . T h e n w e s h o u l d$ $g e t a n e q n . l i k e \cos (a - θ) + \cos 2 θ = 0,$ $n o t 2 \cos (a - θ) + \cos 2 θ = 0 a s y o u g o t .$ $P l s t e l l m e m o r e a b o u t y o u r m e t h o d .$
	Commented by ajfour last updated on 20/Feb/18

	$y e s S i r, y o u a r e a b s o l u t e l y r i g h t,$ $i h a v e f o u n d m y e r r o r . T h a n k y o u .$
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