Menu Close

Question-181619




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 30 m length whose ends are  fixed at two points with a distance of  20 m as shown. a gate at point C on  the fence is 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.
$${a}\:{goat}\:{is}\:{connected}\:{through}\:{a}\:{ring}\:{with} \\ $$$${a}\:{rope}\:{of}\:\mathrm{30}\:{m}\:{length}\:{whose}\:{ends}\:{are} \\ $$$${fixed}\:{at}\:{two}\:{points}\:{with}\:{a}\:{distance}\:{of} \\ $$$$\mathrm{20}\:{m}\:{as}\:{shown}.\:{a}\:{gate}\:{at}\:{point}\:{C}\:{on} \\ $$$${the}\:{fence}\:{is}\:\mathrm{20}\:{m}\:{apart}\:{from}\:\:{point}\:{A}. \\ $$$$ \\ $$$${find}\:{the}\:{maximum}\:{area}\:{where}\:{the}\: \\ $$$${goat}\:{can}\:{graze}.\:{what}\:{is}\:{the}\:{smallest}\: \\ $$$${distance}\:{the}\:{goat}\:{can}\:{get}\:{to}\:{the}\:{gate}. \\ $$
Answered by Acem last updated on 27/Nov/22
Commented by Acem last updated on 27/Nov/22
Farthest distance is x_1 ′ not the point c _(δ= 32.01 m)    Far._(dis) = 34.07 m ; x_1 ′= 15.24 m
$${Farthest}\:{distance}\:{is}\:{x}_{\mathrm{1}} '\:{not}\:{the}\:{point}\:{c}\:_{\delta=\:\mathrm{32}.\mathrm{01}\:{m}} \\ $$$$\:{Far}._{{dis}} =\:\mathrm{34}.\mathrm{07}\:{m}\:;\:{x}_{\mathrm{1}} '=\:\mathrm{15}.\mathrm{24}\:{m} \\ $$$$ \\ $$
Commented by Acem last updated on 27/Nov/22
 c • Additional question:   Assume that the gate wide is 6m, 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 20 m away from the pointA
$$\:{c}\:\bullet\:\boldsymbol{{Additional}}\:\boldsymbol{{question}}: \\ $$$$\:{Assume}\:{that}\:{the}\:{gate}\:{wide}\:{is}\:\mathrm{6}{m},\:{and}\:{we}\:{want} \\ $$$$\:{to}\:{make}\:{that}\:{goat}\:{can}\:{get}\:{to}\:{middle}\:{of}\:{it}, \\ $$$$\:{what}'{s}\:{the}\:{length}\:{of}\:{the}\:{rope}\:{should}\:{be}?\:{with}\:{the} \\ $$$$\:{same}\:{two}\:{previous}\:{anchor}\:{points}. \\ $$$$\:{The}\:{begining}\:{of}\:{the}\:{gate}\:{is}\:\mathrm{20}\:{m}\:{away}\:{from}\:{the}\:{pointA} \\ $$
Commented by mr W last updated on 27/Nov/22
thanks sir!
$${thanks}\:{sir}! \\ $$
Commented by Acem last updated on 28/Nov/22
Thank you too Sir!
$${Thank}\:{you}\:{too}\:{Sir}! \\ $$
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
$${the}\:{maximum}\:{area}\:{the}\:{goat}\:{can}\:{reach} \\ $$$${is}\:{an}\:{ellipse}\:{with}\:{semi}\:{axes}: \\ $$$${a}=\mathrm{15} \\ $$$${b}=\sqrt{\mathrm{15}^{\mathrm{2}} −\mathrm{10}^{\mathrm{2}} }=\mathrm{5}\sqrt{\mathrm{5}} \\ $$$$\left(\frac{{x}}{\mathrm{15}}\right)^{\mathrm{2}} +\left(\frac{{y}}{\mathrm{5}\sqrt{\mathrm{5}}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow{y}=\mathrm{5}\sqrt{\mathrm{5}}\sqrt{\mathrm{1}−\left(\frac{{x}}{\mathrm{15}}\right)^{\mathrm{2}} } \\ $$$${A}=\mathrm{2}×\mathrm{5}\sqrt{\mathrm{5}}\int_{−\mathrm{10}} ^{\mathrm{15}} \sqrt{\mathrm{1}−\left(\frac{{x}}{\mathrm{15}}\right)^{\mathrm{2}} }{dx} \\ $$$$\:\:\:=\mathrm{75}\sqrt{\mathrm{5}}\left[\mathrm{sin}^{−\mathrm{1}} \frac{{x}}{\mathrm{15}}+\frac{{x}}{\mathrm{15}}\sqrt{\mathrm{1}−\left(\frac{{x}}{\mathrm{15}}\right)^{\mathrm{2}} }\right]_{−\mathrm{10}} ^{\mathrm{15}} \\ $$$$\:\:\:=\mathrm{75}\sqrt{\mathrm{5}}\left(\frac{\pi}{\mathrm{2}}+\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{9}}\right)\approx\mathrm{469}.\mathrm{14}\:{m}^{\mathrm{2}} \\ $$$${say}\:{P}\left(−{a}\:\mathrm{cos}\:\theta,\:{b}\:\mathrm{sin}\:\theta\right) \\ $$$$\frac{\mathrm{10}−\mathrm{15}\:\mathrm{cos}\:\theta}{\mathrm{20}−\mathrm{5}\sqrt{\mathrm{5}}\:\mathrm{sin}\:\theta}=\frac{\mathrm{5}\sqrt{\mathrm{5}}\:\mathrm{cos}\:\theta}{\mathrm{15}\:\mathrm{sin}\:\theta} \\ $$$$\Rightarrow\frac{\mathrm{3}}{\mathrm{cos}\:\theta}−\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{sin}\:\theta}=\mathrm{2} \\ $$$$\Rightarrow\theta\approx\mathrm{1}.\mathrm{1249} \\ $$$${d}_{{min}} =\sqrt{\left(\mathrm{10}−\mathrm{15}\:\mathrm{cos}\:\theta\right)^{\mathrm{2}} +\left(\mathrm{20}−\mathrm{5}\sqrt{\mathrm{5}}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\approx\mathrm{10}.\mathrm{29}\:{m} \\ $$
Commented by mr W last updated on 27/Nov/22

Leave a Reply

Your email address will not be published. Required fields are marked *