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Question Number 123528 by ajfour last updated on 26/Nov/20

Commented by ajfour last updated on 26/Nov/20

$F i n d m a x i m u m s h a d e d a r e a .$
	Answered by mr W last updated on 26/Nov/20

	Commented by ajfour last updated on 26/Nov/20
	$eq. of BC y=m(1−x) let y_D =q , x_D =1−(q/m) eq. of AD y=((q(x+1))/((2−(q/m)))) eq. of OC y=((−2mx)/((1−m^2 ))) For E(p , y_E ) ((q(p+1))/((2−(q/m))))=((−2mp)/(1−m^2 )) p=((−((q/(2−(q/m)))))/((q/((2−(q/m))))+((2m)/(1−m^2 )))) y_E =(((((2m)/(1−m^2 )))((q/(2−(q/m)))))/((q/((2−(q/m))))+((2m)/(1−m^2 )))) y_E = ((2qm)/(q(1−m^2 )+4m−2q)) 2△=y_D −y_E = q−((2qm)/(q(1−m^2 )+4m−2q)) .................................................. 2△ = ((2mq−q^2 (1+m^2 ))/(4m−q(1+m^2 ))) .................................................. ((∂(2△))/∂q)=(([2m−2q(1+m^2 )][4m−q(1+m^2 )]+(1+m^2 )[2mq−q^2 (1+m^2 )])/({4m−q(1+m^2 )]^2 ))=0 ⇒ 8m^2 −8mq(1+m^2 )+q^2 (1+m^2 )^2 =0 ............(I) & ((∂(2△))/∂m)=0 ⇒ (2q−2mq^2 )[4m−q(1+m^2 )] = (4−2qm)[2mq−q^2 (1+m^2 )] ⇒ 8mq−2q^2 (1+m^2 )−8m^2 q^2 +2mq^3 (1+m^2 )=8mq−4q^2 (1+m^2 ) −4m^2 q^2 +2mq^3 (1+m^2 ) ⇒ 4m^2 q^2 −2q^2 (1+m^2 )=0 ⇒ 3m^2 =2 ⇒ m=(√(2/3)) And from (I) 8((2/3))−8q(1+(2/3))(√(2/3)) +q^2 (1+(2/3))^2 =0 ⇒ ((25q^2 )/9)−((40(√2)q)/(3(√3)))+((16)/3)=0 ⇒ 25q^2 −40(√6)q+48=0 q=((4(√6))/5)−((√(96−48))/5) = ((4((√6)−(√3)))/5) q ≈ 0.57395 2△=((2mq−q^2 (1+m^2 ))/(4m−q(1+m^2 ))) △= (4/2)((((√6)−(√3))/5)){((2(√(2/3))−((20)/3)((((√6)−(√3))/5)))/(4(√(2/3))−((20)/3)((((√6)−(√3))/5))))} = ((2((√6)−(√3)))/5)(((2(√6)−4(√6)+4(√3))/(4(√6)−4(√6)+4(√3)))) = ((2((√6)−(√3)))/5)(((4(√3)+2(√6))/(4(√3)))) =((((√2)−1)(2(√3)+(√6)))/5) △ = ((√6)/5) sq. units ★ △≈ 0.489897949 % shaded area = ((40(√6))/π) ≈ 31.187872$
	$e q . o f B C$ $y = m (1 - x)$ $l e t y_{D} = q, x_{D} = 1 - \frac{q}{m}$ $e q . o f A D$ $y = \frac{q (x + 1)}{(2 - \frac{q}{m})}$ $e q . o f O C$ $y = \frac{- 2 m x}{(1 - m^{2})}$ $F o r E (p, y_{E})$ $\frac{q (p + 1)}{(2 - \frac{q}{m})} = \frac{- 2 m p}{1 - m^{2}}$ $p = \frac{- (\frac{q}{2 - \frac{q}{m}})}{\frac{q}{(2 - \frac{q}{m})} + \frac{2 m}{1 - m^{2}}}$ $y_{E} = \frac{(\frac{2 m}{1 - m^{2}}) (\frac{q}{2 - \frac{q}{m}})}{\frac{q}{(2 - \frac{q}{m})} + \frac{2 m}{1 - m^{2}}}$ $y_{E} = \frac{2 q m}{q (1 - m^{2}) + 4 m - 2 q}$ $2 △ = y_{D} - y_{E}$ $= q - \frac{2 q m}{q (1 - m^{2}) + 4 m - 2 q}$ $. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$ $2 △ = \frac{2 m q - q^{2} (1 + m^{2})}{4 m - q (1 + m^{2})}$ $. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$ $\frac{\partial (2 △)}{\partial q} = \frac{[2 m - 2 q (1 + m^{2})] [4 m - q (1 + m^{2})] + (1 + m^{2}) [2 m q - q^{2} (1 + m^{2})]}{{{4 m - q (1 + m^{2})]}^{2}} = 0$ $\Rightarrow$ $8 m^{2} - 8 m q (1 + m^{2}) + q^{2} {(1 + m^{2})}^{2} = 0$ $. . . . . . . . . . . . (I) &$ $\frac{\partial (2 △)}{\partial m} = 0$ $\Rightarrow (2 q - 2 {m q}^{2}) [4 m - q (1 + m^{2})]$ $= (4 - 2 q m) [2 m q - q^{2} (1 + m^{2})]$ $\Rightarrow$ $8 m q - 2 q^{2} (1 + m^{2}) - 8 m^{2} q^{2}$ $+ 2 {m q}^{3} (1 + m^{2}) = 8 m q - 4 q^{2} (1 + m^{2})$ $- 4 m^{2} q^{2} + 2 {m q}^{3} (1 + m^{2})$ $\Rightarrow$ $4 m^{2} q^{2} - 2 q^{2} (1 + m^{2}) = 0$ $\Rightarrow 3 m^{2} = 2$ $\Rightarrow m = \sqrt{\frac{2}{3}}$ $A n d f r o m (I)$ $8 (\frac{2}{3}) - 8 q (1 + \frac{2}{3}) \sqrt{\frac{2}{3}}$ $+ q^{2} {(1 + \frac{2}{3})}^{2} = 0$ $\Rightarrow \frac{25 q^{2}}{9} - \frac{40 \sqrt{2} q}{3 \sqrt{3}} + \frac{16}{3} = 0$ $\Rightarrow 25 q^{2} - 40 \sqrt{6} q + 48 = 0$ $q = \frac{4 \sqrt{6}}{5} - \frac{\sqrt{96 - 48}}{5} = \frac{4 (\sqrt{6} - \sqrt{3})}{5}$ $q \approx 0.57395$ $2 △ = \frac{2 m q - q^{2} (1 + m^{2})}{4 m - q (1 + m^{2})}$ $△ = \frac{4}{2} (\frac{\sqrt{6} - \sqrt{3}}{5}) {\frac{2 \sqrt{\frac{2}{3}} - \frac{20}{3} (\frac{\sqrt{6} - \sqrt{3}}{5})}{4 \sqrt{\frac{2}{3}} - \frac{20}{3} (\frac{\sqrt{6} - \sqrt{3}}{5})}}$ $= \frac{2 (\sqrt{6} - \sqrt{3})}{5} (\frac{2 \sqrt{6} - 4 \sqrt{6} + 4 \sqrt{3}}{4 \sqrt{6} - 4 \sqrt{6} + 4 \sqrt{3}})$ $= \frac{2 (\sqrt{6} - \sqrt{3})}{5} (\frac{4 \sqrt{3} + 2 \sqrt{6}}{4 \sqrt{3}})$ $= \frac{(\sqrt{2} - 1) (2 \sqrt{3} + \sqrt{6})}{5}$ $△ = \frac{\sqrt{6}}{5} s q . u n i t s ★$ $△ \approx 0.489897949$ $% s h a d e d a r e a = \frac{40 \sqrt{6}}{π} \approx 31.187872$
	Commented by mr W last updated on 26/Nov/20

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