Question Number 181619 by mr W last updated on 27/Nov/22
Commented by mr W last updated on 27/Nov/22
$${a}\:{goat}\:{is}\:{connected}\:{through}\:{a}\:{ring}\:{with} \\ $$$${a}\:{rope}\:{of}\:\mathrm{30}\:{m}\:{length}\:{whose}\:{ends}\:{are} \\ $$$${fixed}\:{at}\:{two}\:{points}\:{with}\:{a}\:{distance}\:{of} \\ $$$$\mathrm{20}\:{m}\:{as}\:{shown}.\:{a}\:{gate}\:{at}\:{point}\:{C}\:{on} \\ $$$${the}\:{fence}\:{is}\:\mathrm{20}\:{m}\:{apart}\:{from}\:\:{point}\:{A}. \\ $$$$ \\ $$$${find}\:{the}\:{maximum}\:{area}\:{where}\:{the}\: \\ $$$${goat}\:{can}\:{graze}.\:{what}\:{is}\:{the}\:{smallest}\: \\ $$$${distance}\:{the}\:{goat}\:{can}\:{get}\:{to}\:{the}\:{gate}. \\ $$
Answered by Acem last updated on 27/Nov/22
Commented by Acem last updated on 27/Nov/22
$${Farthest}\:{distance}\:{is}\:{x}_{\mathrm{1}} '\:{not}\:{the}\:{point}\:{c}\:_{\delta=\:\mathrm{32}.\mathrm{01}\:{m}} \\ $$$$\:{Far}._{{dis}} =\:\mathrm{34}.\mathrm{07}\:{m}\:;\:{x}_{\mathrm{1}} '=\:\mathrm{15}.\mathrm{24}\:{m} \\ $$$$ \\ $$
Commented by Acem last updated on 27/Nov/22
$$\:{c}\:\bullet\:\boldsymbol{{Additional}}\:\boldsymbol{{question}}: \\ $$$$\:{Assume}\:{that}\:{the}\:{gate}\:{wide}\:{is}\:\mathrm{6}{m},\:{and}\:{we}\:{want} \\ $$$$\:{to}\:{make}\:{that}\:{goat}\:{can}\:{get}\:{to}\:{middle}\:{of}\:{it}, \\ $$$$\:{what}'{s}\:{the}\:{length}\:{of}\:{the}\:{rope}\:{should}\:{be}?\:{with}\:{the} \\ $$$$\:{same}\:{two}\:{previous}\:{anchor}\:{points}. \\ $$$$\:{The}\:{begining}\:{of}\:{the}\:{gate}\:{is}\:\mathrm{20}\:{m}\:{away}\:{from}\:{the}\:{pointA} \\ $$
Commented by mr W last updated on 27/Nov/22
$${thanks}\:{sir}! \\ $$
Commented by Acem last updated on 28/Nov/22
$${Thank}\:{you}\:{too}\:{Sir}! \\ $$
Answered by mr W last updated on 27/Nov/22
Commented by mr W last updated on 27/Nov/22
![the maximum area the goat can reach is an ellipse with semi axes: a=15 b=(√(15^2 −10^2 ))=5(√5) ((x/(15)))^2 +((y/(5(√5))))^2 =1 ⇒y=5(√5)(√(1−((x/(15)))^2 )) A=2×5(√5)∫_(−10) ^(15) (√(1−((x/(15)))^2 ))dx =75(√5)[sin^(−1) (x/(15))+(x/(15))(√(1−((x/(15)))^2 ))]_(−10) ^(15) =75(√5)((π/2)+sin^(−1) (2/3)+((2(√5))/9))≈469.14 m^2 say P(−a cos θ, b sin θ) ((10−15 cos θ)/(20−5(√5) sin θ))=((5(√5) cos θ)/(15 sin θ)) ⇒(3/(cos θ))−((2(√5))/(sin θ))=2 ⇒θ≈1.1249 d_(min) =(√((10−15 cos θ)^2 +(20−5(√5) sin θ)^2 )) ≈10.29 m](Q181635.png)
$${the}\:{maximum}\:{area}\:{the}\:{goat}\:{can}\:{reach} \\ $$$${is}\:{an}\:{ellipse}\:{with}\:{semi}\:{axes}: \\ $$$${a}=\mathrm{15} \\ $$$${b}=\sqrt{\mathrm{15}^{\mathrm{2}} −\mathrm{10}^{\mathrm{2}} }=\mathrm{5}\sqrt{\mathrm{5}} \\ $$$$\left(\frac{{x}}{\mathrm{15}}\right)^{\mathrm{2}} +\left(\frac{{y}}{\mathrm{5}\sqrt{\mathrm{5}}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow{y}=\mathrm{5}\sqrt{\mathrm{5}}\sqrt{\mathrm{1}−\left(\frac{{x}}{\mathrm{15}}\right)^{\mathrm{2}} } \\ $$$${A}=\mathrm{2}×\mathrm{5}\sqrt{\mathrm{5}}\int_{−\mathrm{10}} ^{\mathrm{15}} \sqrt{\mathrm{1}−\left(\frac{{x}}{\mathrm{15}}\right)^{\mathrm{2}} }{dx} \\ $$$$\:\:\:=\mathrm{75}\sqrt{\mathrm{5}}\left[\mathrm{sin}^{−\mathrm{1}} \frac{{x}}{\mathrm{15}}+\frac{{x}}{\mathrm{15}}\sqrt{\mathrm{1}−\left(\frac{{x}}{\mathrm{15}}\right)^{\mathrm{2}} }\right]_{−\mathrm{10}} ^{\mathrm{15}} \\ $$$$\:\:\:=\mathrm{75}\sqrt{\mathrm{5}}\left(\frac{\pi}{\mathrm{2}}+\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{9}}\right)\approx\mathrm{469}.\mathrm{14}\:{m}^{\mathrm{2}} \\ $$$${say}\:{P}\left(−{a}\:\mathrm{cos}\:\theta,\:{b}\:\mathrm{sin}\:\theta\right) \\ $$$$\frac{\mathrm{10}−\mathrm{15}\:\mathrm{cos}\:\theta}{\mathrm{20}−\mathrm{5}\sqrt{\mathrm{5}}\:\mathrm{sin}\:\theta}=\frac{\mathrm{5}\sqrt{\mathrm{5}}\:\mathrm{cos}\:\theta}{\mathrm{15}\:\mathrm{sin}\:\theta} \\ $$$$\Rightarrow\frac{\mathrm{3}}{\mathrm{cos}\:\theta}−\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{sin}\:\theta}=\mathrm{2} \\ $$$$\Rightarrow\theta\approx\mathrm{1}.\mathrm{1249} \\ $$$${d}_{{min}} =\sqrt{\left(\mathrm{10}−\mathrm{15}\:\mathrm{cos}\:\theta\right)^{\mathrm{2}} +\left(\mathrm{20}−\mathrm{5}\sqrt{\mathrm{5}}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\approx\mathrm{10}.\mathrm{29}\:{m} \\ $$
Commented by mr W last updated on 27/Nov/22
