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Question Number 49935 by ajfour last updated on 12/Dec/18

Commented by ajfour last updated on 12/Dec/18

$F i n d m a x i m u m c o l o u r e d a r e a i f$ $l = 1 .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 12/Dec/18
	$eqn of l y−0=((b^2 −0)/(b−a))(x−a) y=(b^2 /(b−a))(x−a) 1=(√((b−a)^2 +(b^2 −0)^2 )) A area ∫_0 ^b x^2 dx+(b^2 /(b−a))∫_b ^a (x−a)dx A=(b^3 /3)+(b^2 /(b−a))∣((x^2 /2)−ax)∣_b ^a =(b^3 /3)+(b^2 /(b−a)){((a^2 /2)−a^2 )−((b^2 /2)−ab)} =(b^3 /3)+(b^2 /(b−a)){((a^2 −2a^2 −b^2 +2ab)/2)} =(b^3 /3)+(b^2 /(b−a))×{((−(b−a)^2 )/2)} =(b^3 /3)−((b^2 (b−a))/2) =((2b^3 −3b^3 +3ab^2 )/6)=((3ab^2 −b^3 )/6)=((b^2 (3a−b))/6) now 1=(b−a)^2 +b^4 (a−b)^2 =1−b^4 a=b+(√(1−b^4 )) A=(b^2 /6)×(3b+3(√(1−b^4 )) −b) =(b^2 /6)×(2b+3(√(1−b^4 )) ) (dA/db)=(b^2 /6)×{2+((3×−4b^3 )/(2(√(1−b^4 )) ))}+((2b)/6)(2b+3(√(1−b^4 )) ) 0=b{2−((6b^3 )/((√(1−b^4 )) ))}+4b+6(√(1−b^4 )) 0=((2b(√(1−b^4 )) −6b^4 +4b(√(1−b^4 )) +6(1−b^4 ))/) 6b(√(1−b^4 )) −12b^4 +6=0 b(√(1−b^4 )) −2b^4 +1=0 b^2 (1−b^4 )=4b^8 −4b^4 +1 4b^8 +b^6 −4b^4 −b^2 +1=0 to find b...wait...$
	$e q n o f l y - 0 = \frac{b^{2} - 0}{b - a} (x - a)$ $y = \frac{b^{2}}{b - a} (x - a)$ $1 = \sqrt{{(b - a)}^{2} + {(b^{2} - 0)}^{2}} A$ $a r e a \int_{0}^{b} x^{2} d x + \frac{b^{2}}{b - a} \int_{b}^{a} (x - a) d x$ $A = \frac{b^{3}}{3} + \frac{b^{2}}{b - a} ∣ (\frac{x^{2}}{2} - a x) ∣_{b}^{a}$ $= \frac{b^{3}}{3} + \frac{b^{2}}{b - a} {(\frac{a^{2}}{2} - a^{2}) - (\frac{b^{2}}{2} - a b)}$ $= \frac{b^{3}}{3} + \frac{b^{2}}{b - a} {\frac{a^{2} - 2 a^{2} - b^{2} + 2 a b}{2}}$ $= \frac{b^{3}}{3} + \frac{b^{2}}{b - a} \times {\frac{- {(b - a)}^{2}}{2}}$ $= \frac{b^{3}}{3} - \frac{b^{2} (b - a)}{2}$ $= \frac{2 b^{3} - 3 b^{3} + 3 {a b}^{2}}{6} = \frac{3 {a b}^{2} - b^{3}}{6} = \frac{b^{2} (3 a - b)}{6}$ $n o w 1 = {(b - a)}^{2} + b^{4}$ ${(a - b)}^{2} = 1 - b^{4}$ $a = b + \sqrt{1 - b^{4}}$ $A = \frac{b^{2}}{6} \times (3 b + 3 \sqrt{1 - b^{4}} - b)$ $= \frac{b^{2}}{6} \times (2 b + 3 \sqrt{1 - b^{4}})$ $\frac{d A}{d b} = \frac{b^{2}}{6} \times {2 + \frac{3 \times - 4 b^{3}}{2 \sqrt{1 - b^{4}}}} + \frac{2 b}{6} (2 b + 3 \sqrt{1 - b^{4}})$ $0 = b {2 - \frac{6 b^{3}}{\sqrt{1 - b^{4}}}} + 4 b + 6 \sqrt{1 - b^{4}}$ $0 = \frac{2 b \sqrt{1 - b^{4}} - 6 b^{4} + 4 b \sqrt{1 - b^{4}} + 6 (1 - b^{4})}{}$ $6 b \sqrt{1 - b^{4}} - 12 b^{4} + 6 = 0$ $b \sqrt{1 - b^{4}} - 2 b^{4} + 1 = 0$ $b^{2} (1 - b^{4}) = 4 b^{8} - 4 b^{4} + 1$ $4 b^{8} + b^{6} - 4 b^{4} - b^{2} + 1 = 0$ $t o f i n d b . . . w a i t . . .$
	Commented by ajfour last updated on 12/Dec/18

	$N o w i t s c o r r e c t t h i s f a r, S i r!$
	Commented by tanmay.chaudhury50@gmail.com last updated on 12/Dec/18

	$y e s y o u a r e r i g h t . . . l e t m e r e c t i f y$
	Commented by MJS last updated on 12/Dec/18

	$no ‘ ‘ {beautiful}^{″} solution for b$ $b \approx . 9266792684 \lor b \approx . 6693949264$ $\Rightarrow a \approx 1.439100107 \lor a \approx 1.563383595$
	Answered by mr W last updated on 12/Dec/18

	$e q n . o f p a r a b o l a : y = {c x}^{2} w i t h c = 1$ $f r o m (h, k) t o (a, 0) :$ $k = {c h}^{2}$ ${(a - h)}^{2} + k^{2} = l^{2}$ $a^{2} - 2 a h + h^{2} + c^{2} h^{4} = l^{2}$ $2 a \frac{d a}{d h} - 2 h \frac{d a}{d h} - 2 a + 2 h + 4 c^{2} h^{3} = 0$ $\Rightarrow \frac{d a}{d h} = 1 - \frac{2 c^{2} h^{3}}{a - h}$ $A = \frac{h k}{3} + \frac{(a - h) k}{2} = \frac{1}{6} (3 a - h) {c h}^{2}$ $\frac{d (6 A / c)}{d h} = 2 h (3 a - h) + h^{2} (3 \frac{d a}{d h} - 1) = 0$ $2 (3 a - h) + h (3 \frac{d a}{d h} - 1) = 0$ $\frac{2 a}{h} + \frac{d a}{d h} - 1 = 0$ $\frac{2 a}{h} - \frac{2 c^{2} h^{3}}{a - h} = 0$ $\frac{a}{h} = \frac{c^{2} h^{3}}{a - h}$ $a^{2} - h a - c^{2} h^{4} = 0$ $\Rightarrow a = \frac{h}{2} [1 + \sqrt{1 + {(2 c h)}^{2}}]$ $a^{2} - 2 a h + h^{2} + c^{2} h^{4} = l^{2}$ $4 c^{2} h^{4} + h^{2} [1 - \sqrt{1 + {(2 c h)}^{2}}] = 2 l^{2}$ $4 c^{2} h^{2} + 1 - \sqrt{1 + {(2 c h)}^{2}} = \frac{8 c^{2} l^{2}}{4 c^{2} h^{2}}$ $w i t h λ = {(2 c h)}^{2}$ $1 + λ - \sqrt{1 + λ} = \frac{8 c^{2} l^{2}}{λ}$ $w i t h c = 1, l = 1$ $1 + λ - \sqrt{1 + λ} = \frac{8}{λ}$ $l e t t = \sqrt{1 + λ}$ $(t^{2} - 1) (t^{2} - t) = 8$ $t^{4} - t^{3} - t^{2} + t - 8 = 0$ $\Rightarrow λ = 3.4349$ $\Rightarrow h = \frac{\sqrt{λ}}{2} = 0.9267$ $\Rightarrow a = \frac{0.9267}{2} [1 + \sqrt{1 + 3.4349}] = 1.4391$ $\Rightarrow A = \frac{1}{6} (3 \times 1.4391 - 0.9267) \times 0 . 9267^{2} = 0.4853$
	Commented by ajfour last updated on 12/Dec/18

	$T h a n k s S i r, b e a u t i f u l w a y!$
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