Math Question and Answers


Question and Answers Forum

All Questions Topic List
Geometry Questions
Previous in All Question Next in All Question
Previous in Geometry Next in Geometry
Question Number 166007 by mr W last updated on 11/Feb/22

Commented by mr W last updated on 12/Feb/22

$f i n d t h e m a x i m u m c o l o u r e d a r e a$ $i n t e r m s o f a, b, c .$ $[r e l a t e d t o Q 165965 f r o m a j f o u r s i r]$
	Answered by mr W last updated on 12/Feb/22

	Commented by mr W last updated on 12/Feb/22
	$say the fourth side on the wall has the length d=variable x. from the bretschneider′s formula for the area of a general quadrilateral with sides a, b, c, d we can see that the area of the quadrilateral is maximum, when the quadrilateral is cyclic, as the diagram above shows. the area is A=(√((s−a)(s−b)(s−c)(s−d))) with s=((a+b+c+d)/2) let Φ=A^2 Φ=(s−a)(s−b)(s−c)(s−d) ln Φ=ln (s−a)+ln (s−b)+ln (s−c)+ln (s−d) ((Φ′)/Φ)=(1/2)((1/(s−a))+(1/(s−b))+(1/(s−c))−(1/(s−d))) such that A is maximum, Φ′=(dΦ/dd)=0 (1/(s−a))+(1/(s−b))+(1/(s−c))−(1/(s−d))=0 (1/(s−a))+(1/(s−b))=(1/(s−d))−(1/(s−c)) ((2s−(a+b))/(s^2 −(a+b)s+ab))=((d−c)/(s^2 −(c+d)s+cd)) 2s^3 −(a+b+c+3d)s^2 +2d(a+b+c)s+abc−(ab+bc+ca)d=0 d^3 −(a^2 +b^2 +c^2 )d−2abc=0 ⇒d=2(√((a^2 +b^2 +c^2 )/3)) sin {(π/3)+(1/3)sin^(−1) [abc((3/(a^2 +b^2 +c^2 )))^(3/2) ]}$
	$s a y t h e f o u r t h s i d e o n t h e w a l l h a s t h e$ $l e n g t h d = v a r i a b l e x .$ $f r o m t h e {b r e t s c h n e i d e r}^{'} s f o r m u l a f o r$ $t h e a r e a o f a g e n e r a l q u a d r i l a t e r a l$ $w i t h s i d e s a, b, c, d w e c a n s e e t h a t t h e$ $a r e a o f t h e q u a d r i l a t e r a l i s m a x i m u m,$ $w h e n t h e q u a d r i l a t e r a l i s c y c l i c, a s$ $t h e d i a g r a m a b o v e s h o w s .$ $t h e a r e a i s$ $A = \sqrt{(s - a) (s - b) (s - c) (s - d)}$ $w i t h s = \frac{a + b + c + d}{2}$ $l e t Φ = A^{2}$ $Φ = (s - a) (s - b) (s - c) (s - d)$ $\ln Φ = \ln (s - a) + \ln (s - b) + \ln (s - c) + \ln (s - d)$ $\frac{Φ^{'}}{Φ} = \frac{1}{2} (\frac{1}{s - a} + \frac{1}{s - b} + \frac{1}{s - c} - \frac{1}{s - d})$ $s u c h t h a t A i s m a x i m u m, Φ^{'} = \frac{d Φ}{d d} = 0$ $\frac{1}{s - a} + \frac{1}{s - b} + \frac{1}{s - c} - \frac{1}{s - d} = 0$ $\frac{1}{s - a} + \frac{1}{s - b} = \frac{1}{s - d} - \frac{1}{s - c}$ $\frac{2 s - (a + b)}{s^{2} - (a + b) s + a b} = \frac{d - c}{s^{2} - (c + d) s + c d}$ $2 s^{3} - (a + b + c + 3 d) s^{2} + 2 d (a + b + c) s + a b c - (a b + b c + c a) d = 0$ $d^{3} - (a^{2} + b^{2} + c^{2}) d - 2 a b c = 0$ $\Rightarrow d = 2 \sqrt{\frac{a^{2} + b^{2} + c^{2}}{3}} \sin {\frac{π}{3} + \frac{1}{3} \sin^{- 1} [a b c {(\frac{3}{a^{2} + b^{2} + c^{2}})}^{\frac{3}{2}}]}$
	Commented by mr W last updated on 12/Feb/22

	$f u r t h e r s t u d y s h o w s t h a t i n t h i s c a s e$ $t h e c o r r e s p o n d i n g m a x i m u m a r e a i s$ $w h e n t h e s i d e s a, b, c a r e t h e c h o r d s$ $o n a s e m i c i r c l e w i t h d i a m e t e r d w h i c h$ $c a n b e c a l c u l a t e d a c c o r d i n g t o t h e$ $f o r m u l a a b o v e .$
	Commented by mr W last updated on 12/Feb/22

	Commented by mr W last updated on 12/Feb/22
	Bretschneider's formula: https://en.m.wikipedia.org/wiki/Bretschneider%27s_formula
	Commented by mr W last updated on 12/Feb/22

Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum