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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))

$${d}=\sqrt{{x}^{\mathrm{2}} +\mathrm{81}}\:+\sqrt{\left({x}−\mathrm{5}\right)^{\mathrm{2}} +\mathrm{36}} \\ $$$${d}'=\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{81}}}\:+\frac{{x}−\mathrm{5}}{\:\sqrt{\left({x}−\mathrm{5}\right)^{\mathrm{2}} +\mathrm{36}}}\:=\mathrm{0} \\ $$$$\Rightarrow\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{81}}}\:=\:\frac{\mathrm{5}−{x}}{\:\sqrt{\left({x}−\mathrm{5}\right)^{\mathrm{2}} +\mathrm{36}}} \\ $$$$\Rightarrow\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{81}}\:=\frac{\mathrm{25}−\mathrm{10}{x}+{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{61}} \\ $$$$\Rightarrow\left({x}−\mathrm{3}\right)\left({x}−\mathrm{15}\right)=\mathrm{0}\:\begin{cases}{{x}=\mathrm{3}}\\{{x}=\mathrm{15}}\end{cases} \\ $$$${when}\:{x}=\mathrm{3}\Rightarrow{d}=\sqrt{\mathrm{90}}+\sqrt{\mathrm{40}}\:=\mathrm{3}\sqrt{\mathrm{10}}+\mathrm{2}\sqrt{\mathrm{10}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{d}=\mathrm{5}\sqrt{\mathrm{10}}\:\left({min}\right)\:\therefore\:{Q}\left(\mathrm{3},\mathrm{0}\right) \\ $$$${when}\:{x}=\mathrm{15}\Rightarrow{d}=\sqrt{\mathrm{225}+\mathrm{81}}+\sqrt{\mathrm{136}}=\mathrm{5}\sqrt{\mathrm{34}} \\ $$$$ \\ $$

Answered by mr W last updated on 28/May/22

S=AQ+QB=AQ+QB′ S_(min) =AB′=(√(5^2 +(9+6)^2 ))=5(√(10)) at x=(9/(9+6))×5=3

$${S}={AQ}+{QB}={AQ}+{QB}' \\ $$$${S}_{{min}} ={AB}'=\sqrt{\mathrm{5}^{\mathrm{2}} +\left(\mathrm{9}+\mathrm{6}\right)^{\mathrm{2}} }=\mathrm{5}\sqrt{\mathrm{10}} \\ $$$${at}\:{x}=\frac{\mathrm{9}}{\mathrm{9}+\mathrm{6}}×\mathrm{5}=\mathrm{3} \\ $$

Commented by mr W last updated on 28/May/22

Commented by nikif99 last updated on 28/May/22

This reminds me of a game of my childhood. Children start from point A, then must touch the “wall” OX at any wishing point, then run to point B as quickly as possible.

$${This}\:{reminds}\:{me}\:{of}\:{a}\:{game}\:{of}\:{my} \\ $$$${childhood}.\:{Children}\:{start}\:{from}\:{point}\:{A}, \\ $$$${then}\:{must}\:{touch}\:{the}\:``{wall}''\:{OX}\:{at} \\ $$$${any}\:{wishing}\:{point},\:{then}\:{run}\:{to}\: \\ $$$${point}\:{B}\:{as}\:{quickly}\:{as}\:{possible}. \\ $$

Commented by mr W last updated on 28/May/22

nice game! the shortest way is that which a light ray follows when the wall is a mirror.

$${nice}\:{game}! \\ $$$${the}\:{shortest}\:{way}\:{is}\:{that}\:{which}\:{a} \\ $$$${light}\:{ray}\:{follows}\:{when}\:{the}\:{wall}\:{is} \\ $$$${a}\:{mirror}. \\ $$

Commented by nikif99 last updated on 28/May/22

And the winning boy was who had a sense of geometry plus quick legs.

$${And}\:{the}\:{winning}\:{boy}\:{was}\:{who}\:{had}\:{a} \\ $$$${sense}\:{of}\:{geometry}\:{plus}\:{quick}\:{legs}. \\ $$

Commented by mr W last updated on 28/May/22

how can the children find the ideal point on the wall, if they only have a long tape?

$${how}\:{can}\:{the}\:{children}\:{find}\:{the}\:{ideal}\: \\ $$$${point}\:{on}\:{the}\:{wall},\:{if}\:{they}\:{only}\:{have}\: \\ $$$${a}\:{long}\:{tape}? \\ $$

Commented by nikif99 last updated on 29/May/22

They have to connect points A and B with the tape, while the tape touches the wall, and measuring if the length of the tape is minimum.

$${They}\:{have}\:{to}\:{connect}\:{points}\:{A}\:{and}\:{B} \\ $$$${with}\:{the}\:{tape},\:{while}\:{the}\:{tape}\:{touches}\: \\ $$$${the}\:{wall},\:{and}\:{measuring}\:{if}\:{the}\:{length} \\ $$$${of}\:{the}\:{tape}\:{is}\:{minimum}. \\ $$

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