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Question Number 170623 by cortano1 last updated on 27/May/22

	Answered by greougoury555 last updated on 28/May/22
	$d=(√(x^2 +81)) +(√((x−5)^2 +36)) d′=(x/( (√(x^2 +81)))) +((x−5)/( (√((x−5)^2 +36)))) =0 ⇒(x/( (√(x^2 +81)))) = ((5−x)/( (√((x−5)^2 +36)))) ⇒(x^2 /(x^2 +81)) =((25−10x+x^2 )/(x^2 −10x+61)) ⇒(x−3)(x−15)=0 { ((x=3)),((x=15)) :} when x=3⇒d=(√(90))+(√(40)) =3(√(10))+2(√(10)) d=5(√(10)) (min) ∴ Q(3,0) when x=15⇒d=(√(225+81))+(√(136))=5(√(34))$
	$d = \sqrt{x^{2} + 81} + \sqrt{{(x - 5)}^{2} + 36}$ $d^{'} = \frac{x}{\sqrt{x^{2} + 81}} + \frac{x - 5}{\sqrt{{(x - 5)}^{2} + 36}} = 0$ $\Rightarrow \frac{x}{\sqrt{x^{2} + 81}} = \frac{5 - x}{\sqrt{{(x - 5)}^{2} + 36}}$ $\Rightarrow \frac{x^{2}}{x^{2} + 81} = \frac{25 - 10 x + x^{2}}{x^{2} - 10 x + 61}$ $\Rightarrow (x - 3) (x - 15) = 0 {\begin{cases} x = 3 \\ x = 15 \end{cases}$ $w h e n x = 3 \Rightarrow d = \sqrt{90} + \sqrt{40} = 3 \sqrt{10} + 2 \sqrt{10}$ $d = 5 \sqrt{10} (m i n) ∴ Q (3, 0)$ $w h e n x = 15 \Rightarrow d = \sqrt{225 + 81} + \sqrt{136} = 5 \sqrt{34}$
	Answered by mr W last updated on 28/May/22

	$S = A Q + Q B = A Q + {Q B}^{'}$ $S_{m i n} = {A B}^{'} = \sqrt{5^{2} + {(9 + 6)}^{2}} = 5 \sqrt{10}$ $a t x = \frac{9}{9 + 6} \times 5 = 3$
	Commented by mr W last updated on 28/May/22

	Commented by nikif99 last updated on 28/May/22

	$T h i s r e m i n d s m e o f a g a m e o f m y$ $c h i l d h o o d . C h i l d r e n s t a r t f r o m p o i n t A,$ $t h e n m u s t t o u c h t h e ‘ ‘ {w a l l}^{″} O X a t$ $a n y w i s h i n g p o i n t, t h e n r u n t o$ $p o i n t B a s q u i c k l y a s p o s s i b l e .$
	Commented by mr W last updated on 28/May/22

	$n i c e g a m e!$ $t h e s h o r t e s t w a y i s t h a t w h i c h a$ $l i g h t r a y f o l l o w s w h e n t h e w a l l i s$ $a m i r r o r .$
	Commented by nikif99 last updated on 28/May/22

	$A n d t h e w i n n i n g b o y w a s w h o h a d a$ $s e n s e o f g e o m e t r y p l u s q u i c k l e g s .$
	Commented by mr W last updated on 28/May/22

	$h o w c a n t h e c h i l d r e n f i n d t h e i d e a l$ $p o i n t o n t h e w a l l, i f t h e y o n l y h a v e$ $a l o n g t a p e ?$
	Commented by nikif99 last updated on 29/May/22

	$T h e y h a v e t o c o n n e c t p o i n t s A a n d B$ $w i t h t h e t a p e, w h i l e t h e t a p e t o u c h e s$ $t h e w a l l, a n d m e a s u r i n g i f t h e l e n g t h$ $o f t h e t a p e i s m i n i m u m .$
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