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Question Number 202056 by mr W last updated on 19/Dec/23
Commented by mr W last updated on 19/Dec/23
the lake with center at O has a radius r (r=1 km). the shortest distances from the villages A and B to the lake are a and b respectively (a=4 km, b=2 km). a pump station should be build at C to supply the villages with water. find the minimum length of pipes for this project. (θ=60°).
$${the}\:{lake}\:{with}\:{center}\:{at}\:{O}\:{has}\:{a}\:{radius} \\ $$$${r}\:\left({r}=\mathrm{1}\:{km}\right).\:{the}\:{shortest}\:{distances}\: \\ $$$${from}\:{the}\:{villages}\:{A}\:{and}\:{B}\:{to}\:{the}\: \\ $$$${lake}\:{are}\:{a}\:{and}\:{b}\:{respectively}\:\left({a}=\mathrm{4}\:{km},\right. \\ $$$$\left.{b}=\mathrm{2}\:{km}\right).\:{a}\:{pump}\:{station}\:{should}\:{be} \\ $$$${build}\:{at}\:{C}\:{to}\:{supply}\:{the}\:{villages} \\ $$$${with}\:{water}.\:{find}\:{the}\:{minimum} \\ $$$${length}\:{of}\:{pipes}\:{for}\:{this}\:{project}. \\ $$$$\left(\theta=\mathrm{60}°\right). \\ $$
Commented by a.lgnaoui last updated on 19/Dec/23
Commented by a.lgnaoui last updated on 19/Dec/23
ΔOAB AB^2 =OA^2 +OB^2 −2OA.OBcos 60 =25+9−15=34−15=19 ⇒AB=(√(19)) AB^2 =AC^2 +CB^2 (∡ACB=90) [Courte distance entre A B] ΔOAC ∡DOC=x AC^2 =OC^2 +AD^2 −2.OC.ADcos x =26−10cos x ⇒ AC=(√(26−10cos x)) ΔzOBC BC^2 =OC^2 +OB^2 −2.OC.OBcos (60−x) =10−6(cos 60cos x+sin 60sin x) =10−3(cos x+(√3) sin x) BC=(√(10−3(cos x+(√3) sin x))) AC^2 +BC^2 =AB^2 19= 26−10cos x+10−3cos x−3(√3) sin x 13cos x+3(√3) sin x−36=−19 13cos x+3(√3) sin x−17=0 posonz z=cos x 13z+3(√(3−3z^2 ))−17=0 (√(3−3z^2 )) =(((17−13 z)/9))^2 27−27z^2 =17^2 +169z^2 −442z 169z^2 −442z+262=0 z=cos x=0,90 ⇒ x=24,77° Longeur minimale est L=AC+BC= AC=( (√(26−10cos 24,77)) ) =4,113396 BC=((√(10−3(cos 24,77+(√3) sin 24,77)) ) =2,258083 L=4,113396+2,258083 L=6,371 Km
$$\Delta\boldsymbol{\mathrm{OAB}}\:\:\:\:\: \\ $$$$\boldsymbol{\mathrm{AB}}^{\mathrm{2}} =\boldsymbol{\mathrm{OA}}^{\mathrm{2}} +\boldsymbol{\mathrm{OB}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{\mathrm{OA}}.\boldsymbol{\mathrm{OB}}\mathrm{cos}\:\mathrm{60} \\ $$$$=\mathrm{25}+\mathrm{9}−\mathrm{15}=\mathrm{34}−\mathrm{15}=\mathrm{19} \\ $$$$\Rightarrow\boldsymbol{\mathrm{AB}}=\sqrt{\mathrm{19}}\: \\ $$$$\boldsymbol{\mathrm{AB}}^{\mathrm{2}} =\boldsymbol{\mathrm{AC}}^{\mathrm{2}} +\boldsymbol{\mathrm{CB}}^{\mathrm{2}} \:\:\: \\ $$$$\left(\measuredangle\boldsymbol{\mathrm{ACB}}=\mathrm{90}\right)\:\:\left[\mathrm{Courte}\:\mathrm{distance}\:\mathrm{entre}\:\boldsymbol{\mathrm{A}}\:\boldsymbol{\mathrm{B}}\right] \\ $$$$\Delta\boldsymbol{\mathrm{OAC}}\:\:\:\:\:\:\:\:\:\measuredangle\mathrm{DOC}=\boldsymbol{\mathrm{x}} \\ $$$$\:\boldsymbol{\mathrm{AC}}^{\mathrm{2}} =\boldsymbol{\mathrm{OC}}^{\mathrm{2}} +\boldsymbol{\mathrm{AD}}^{\mathrm{2}} −\mathrm{2}.\boldsymbol{\mathrm{OC}}.\boldsymbol{\mathrm{AD}}\mathrm{cos}\:\boldsymbol{\mathrm{x}} \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{26}−\mathrm{10cos}\:\boldsymbol{\mathrm{x}} \\ $$$$\Rightarrow\:\:\boldsymbol{\mathrm{AC}}=\sqrt{\mathrm{26}−\mathrm{10cos}\:\boldsymbol{\mathrm{x}}} \\ $$$$ \\ $$$$\Delta\mathrm{z}\boldsymbol{\mathrm{OBC}}\:\: \\ $$$$\:\boldsymbol{\mathrm{BC}}^{\mathrm{2}} =\boldsymbol{\mathrm{OC}}^{\mathrm{2}} +\boldsymbol{\mathrm{OB}}^{\mathrm{2}} −\mathrm{2}.\boldsymbol{\mathrm{OC}}.\boldsymbol{\mathrm{OB}}\mathrm{cos}\:\left(\mathrm{60}−\boldsymbol{\mathrm{x}}\right) \\ $$$$=\mathrm{10}−\mathrm{6}\left(\mathrm{cos}\:\mathrm{60cos}\:\boldsymbol{\mathrm{x}}+\mathrm{sin}\:\mathrm{60sin}\:\boldsymbol{\mathrm{x}}\right) \\ $$$$=\mathrm{10}−\mathrm{3}\left(\mathrm{cos}\:\boldsymbol{\mathrm{x}}+\sqrt{\mathrm{3}}\:\mathrm{sin}\:\boldsymbol{\mathrm{x}}\right) \\ $$$$ \\ $$$$\:\boldsymbol{\mathrm{BC}}=\sqrt{\mathrm{10}−\mathrm{3}\left(\mathrm{cos}\:\boldsymbol{\mathrm{x}}+\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{x}\right)} \\ $$$$\:\:\: \\ $$$$\boldsymbol{\mathrm{AC}}^{\mathrm{2}} +\boldsymbol{\mathrm{BC}}^{\mathrm{2}} =\boldsymbol{\mathrm{AB}}^{\mathrm{2}} \\ $$$$\mathrm{19}=\:\mathrm{26}−\mathrm{10cos}\:\boldsymbol{\mathrm{x}}+\mathrm{10}−\mathrm{3cos}\:\boldsymbol{\mathrm{x}}−\mathrm{3}\sqrt{\mathrm{3}}\:\mathrm{sin}\:\boldsymbol{\mathrm{x}} \\ $$$$\mathrm{13cos}\:\boldsymbol{\mathrm{x}}+\mathrm{3}\sqrt{\mathrm{3}}\:\mathrm{sin}\:\boldsymbol{\mathrm{x}}−\mathrm{36}=−\mathrm{19} \\ $$$$ \\ $$$$\mathrm{13cos}\:\boldsymbol{\mathrm{x}}+\mathrm{3}\sqrt{\mathrm{3}}\:\mathrm{sin}\:\boldsymbol{\mathrm{x}}−\mathrm{17}=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{posonz}}\:\:\boldsymbol{\mathrm{z}}=\mathrm{cos}\:\boldsymbol{\mathrm{x}} \\ $$$$\mathrm{13}\boldsymbol{\mathrm{z}}+\mathrm{3}\sqrt{\mathrm{3}−\mathrm{3}\boldsymbol{\mathrm{z}}^{\mathrm{2}} }−\mathrm{17}=\mathrm{0} \\ $$$$\sqrt{\mathrm{3}−\mathrm{3z}^{\mathrm{2}} \:}\:=\left(\frac{\mathrm{17}−\mathrm{13}\:\mathrm{z}}{\mathrm{9}}\right)^{\mathrm{2}} \\ $$$$\mathrm{27}−\mathrm{27}\boldsymbol{\mathrm{z}}^{\mathrm{2}} =\mathrm{17}^{\mathrm{2}} +\mathrm{169}\boldsymbol{\mathrm{z}}^{\mathrm{2}} −\mathrm{442}\boldsymbol{\mathrm{z}} \\ $$$$\mathrm{169}\boldsymbol{\mathrm{z}}^{\mathrm{2}} −\mathrm{442}\boldsymbol{\mathrm{z}}+\mathrm{262}=\mathrm{0} \\ $$$$ \\ $$$$\:\:\boldsymbol{\mathrm{z}}=\mathrm{cos}\:\boldsymbol{\mathrm{x}}=\mathrm{0},\mathrm{90}\:\:\:\Rightarrow\:\:\:\:\boldsymbol{\mathrm{x}}=\mathrm{24},\mathrm{77}° \\ $$$$ \\ $$$$\:\boldsymbol{\mathrm{Longeur}}\:\:\boldsymbol{\mathrm{minimale}}\:\boldsymbol{\mathrm{est}}\:\: \\ $$$$\boldsymbol{\mathrm{L}}=\boldsymbol{\mathrm{AC}}+\boldsymbol{\mathrm{BC}}= \\ $$$$\:\boldsymbol{\mathrm{AC}}=\left(\:\sqrt{\mathrm{26}−\mathrm{10cos}\:\mathrm{24},\mathrm{77}}\:\right)\:\:=\mathrm{4},\mathrm{113396} \\ $$$$\:\:\: \\ $$$$\boldsymbol{\mathrm{BC}}=\left(\sqrt{\mathrm{10}−\mathrm{3}\left(\mathrm{cos}\:\mathrm{24},\mathrm{77}+\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{24},\mathrm{77}\right.}\:\right) \\ $$$$\:\:\:\:\:\:\:=\mathrm{2},\mathrm{258083} \\ $$$$\boldsymbol{\mathrm{L}}=\mathrm{4},\mathrm{113396}+\mathrm{2},\mathrm{258083} \\ $$$$ \\ $$$$\:\:\:\boldsymbol{\mathrm{L}}=\mathrm{6},\mathrm{371}\:\boldsymbol{\mathrm{Km}} \\ $$$$\:\: \\ $$$$ \\ $$
Commented by mr W last updated on 19/Dec/23
wrong! there is no reason why ∠ACB should be 90°.
$${wrong}! \\ $$$${there}\:{is}\:{no}\:{reason}\:{why}\:\angle{ACB}\: \\ $$$${should}\:{be}\:\mathrm{90}°. \\ $$
Commented by Frix last updated on 19/Dec/23
I don′t think we can get an exact solution. I get ≈6.354km with ∡COB=32.485°
$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{we}\:\mathrm{can}\:\mathrm{get}\:\mathrm{an}\:\mathrm{exact}\:\mathrm{solution}. \\ $$$$\mathrm{I}\:\mathrm{get}\:\approx\mathrm{6}.\mathrm{354km}\:\mathrm{with}\:\measuredangle{COB}=\mathrm{32}.\mathrm{485}° \\ $$
Commented by mr W last updated on 19/Dec/23
agree! l_(min) ≈6.35405 km
$${agree}! \\ $$$${l}_{{min}} \approx\mathrm{6}.\mathrm{35405}\:{km} \\ $$
Commented by mr W last updated on 19/Dec/23
Commented by a.lgnaoui last updated on 19/Dec/23
((d(AB+BC))/dx)=0 d((√(26−10cos x)) )/dx+ d((√(10−3(cos x+(√3) sin x)) )/dx =0 soit ((10sin x)/(2(√(36−10cos x))))+((3((√3)cos x−sin x))/(2(√(10−3(cos x+(√3) sin x)))))=0 resolutiin de l equatiin Soit c valeur de x cherche { ((sin c=h)),((cos c=k)) :} Alors (AB+BC)= (√(26−10k)) +(√(10−3(k+(√3) h)) (I think that this methode can resolve the probleme)
$$\frac{\mathrm{d}\left(\mathrm{AB}+\mathrm{BC}\right)}{\mathrm{dx}}=\mathrm{0} \\ $$$$\mathrm{d}\left(\sqrt{\mathrm{26}−\mathrm{10cos}\:\boldsymbol{\mathrm{x}}}\:\right)/\boldsymbol{\mathrm{dx}}+ \\ $$$$\boldsymbol{\mathrm{d}}\left(\sqrt{\mathrm{10}−\mathrm{3}\left(\mathrm{cos}\:\boldsymbol{\mathrm{x}}+\sqrt{\mathrm{3}}\:\mathrm{sin}\:\boldsymbol{\mathrm{x}}\right.}\:\right)/\boldsymbol{\mathrm{dx}}\:\:\:=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{soit}} \\ $$$$\frac{\mathrm{10sin}\:\boldsymbol{\mathrm{x}}}{\mathrm{2}\sqrt{\mathrm{36}−\mathrm{10cos}\:\boldsymbol{\mathrm{x}}}}+\frac{\mathrm{3}\left(\sqrt{\mathrm{3}}\mathrm{cos}\:\boldsymbol{\mathrm{x}}−\mathrm{sin}\:\boldsymbol{\mathrm{x}}\right)}{\mathrm{2}\sqrt{\mathrm{10}−\mathrm{3}\left(\mathrm{cos}\:\boldsymbol{\mathrm{x}}+\sqrt{\mathrm{3}}\:\mathrm{sin}\:\boldsymbol{\mathrm{x}}\right)}}=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{resolutiin}}\:\boldsymbol{\mathrm{de}}\:\boldsymbol{\mathrm{l}}\:\boldsymbol{\mathrm{equatiin}} \\ $$$$\:\boldsymbol{\mathrm{Soit}}\:\boldsymbol{\mathrm{c}}\:\:\boldsymbol{\mathrm{valeur}}\:\boldsymbol{\mathrm{de}}\:\boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{cherche}} \\ $$$$\begin{cases}{\mathrm{sin}\:\boldsymbol{\mathrm{c}}=\boldsymbol{\mathrm{h}}}\\{\mathrm{cos}\:\boldsymbol{\mathrm{c}}=\boldsymbol{\mathrm{k}}}\end{cases} \\ $$$$\boldsymbol{\mathrm{Alors}}\: \\ $$$$\left(\boldsymbol{\mathrm{AB}}+\boldsymbol{\mathrm{BC}}\right)= \\ $$$$\sqrt{\mathrm{26}−\mathrm{10}\boldsymbol{\mathrm{k}}}\:+\sqrt{\mathrm{10}−\mathrm{3}\left(\boldsymbol{\mathrm{k}}+\sqrt{\mathrm{3}}\:\boldsymbol{\mathrm{h}}\right.} \\ $$$$\:\left(\boldsymbol{{I}}\:\boldsymbol{{think}}\:\boldsymbol{{th}\mathrm{a}{t}}\:\boldsymbol{\mathrm{this}}\:\boldsymbol{{methode}}\:\boldsymbol{{c}\mathrm{a}{n}}\:\boldsymbol{{resolve}}\right. \\ $$$$\left.\boldsymbol{{the}}\:\boldsymbol{{probleme}}\right) \\ $$
Commented by mr W last updated on 19/Dec/23
yes, this is a right approach.
$${yes},\:{this}\:{is}\:{a}\:{right}\:{approach}. \\ $$

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