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find-maximum-value-2x-2-y-2-with-constraint-x-2-y-2-4x-2y-1-0-

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Question Number 87989 by john santu last updated on 07/Apr/20
find maximum value 2x^2 +y^2 with constraint x^2 +y^2 −4x+2y+1=0
$$\mathrm{find}\:\mathrm{maximum}\:\mathrm{value} \\ $$$$\mathrm{2x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \:\mathrm{with}\:\mathrm{constraint} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{4x}+\mathrm{2y}+\mathrm{1}=\mathrm{0}\: \\ $$$$ \\ $$
Answered by mr W last updated on 07/Apr/20
x^2 +y^2 −4x+2y+1=0 ⇒(x−2)^2 +(y+1)^2 =4 let x=2+2 cos θ let y=−1+2 sin θ k=2x^2 +y^2 =8(1+cos θ)^2 +(−1+2 sin θ)^2 (dk/dθ)=16(1+cos θ)(−sin θ)+4(−1+2 sin θ)cos θ=0 4sin θ+2cos θsin θ+cos θ=0 ⇒(4/(cos θ))+2+(1/(sin θ))=0 ⇒θ≈−0.1659, 2.7118 k_(min) ≈0.094 with θ=2.7118 k_(max) ≈33.33 with θ=−0.1659
$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{4x}+\mathrm{2y}+\mathrm{1}=\mathrm{0}\: \\ $$$$\Rightarrow\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\left({y}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4} \\ $$$${let}\:{x}=\mathrm{2}+\mathrm{2}\:\mathrm{cos}\:\theta \\ $$$${let}\:{y}=−\mathrm{1}+\mathrm{2}\:\mathrm{sin}\:\theta \\ $$$${k}=\mathrm{2}{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{8}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)^{\mathrm{2}} +\left(−\mathrm{1}+\mathrm{2}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} \\ $$$$\frac{{dk}}{{d}\theta}=\mathrm{16}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)\left(−\mathrm{sin}\:\theta\right)+\mathrm{4}\left(−\mathrm{1}+\mathrm{2}\:\mathrm{sin}\:\theta\right)\mathrm{cos}\:\theta=\mathrm{0} \\ $$$$\mathrm{4sin}\:\theta+\mathrm{2cos}\:\theta\mathrm{sin}\:\theta+\mathrm{cos}\:\theta=\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{4}}{\mathrm{cos}\:\theta}+\mathrm{2}+\frac{\mathrm{1}}{\mathrm{sin}\:\theta}=\mathrm{0} \\ $$$$\Rightarrow\theta\approx−\mathrm{0}.\mathrm{1659},\:\mathrm{2}.\mathrm{7118} \\ $$$$ \\ $$$${k}_{{min}} \approx\mathrm{0}.\mathrm{094}\:{with}\:\theta=\mathrm{2}.\mathrm{7118} \\ $$$${k}_{{max}} \approx\mathrm{33}.\mathrm{33}\:{with}\:\theta=−\mathrm{0}.\mathrm{1659} \\ $$
Commented by mr W last updated on 07/Apr/20
the question is to find the smallest and the largest ellipse which touches a given circle. “exact” solution maybe not possible.
$${the}\:{question}\:{is}\:{to}\:{find}\:{the}\:{smallest} \\ $$$${and}\:{the}\:{largest}\:{ellipse}\:{which}\:{touches} \\ $$$${a}\:{given}\:{circle}.\:“{exact}''\:{solution}\:{maybe} \\ $$$${not}\:{possible}. \\ $$
Answered by john santu last updated on 08/Apr/20
Commented by mr W last updated on 08/Apr/20
when i use t=tan (θ/2), i get also a quadratic equation about t: t^4 −4t^3 −12t+1=0.
$${when}\:{i}\:{use}\:{t}=\mathrm{tan}\:\frac{\theta}{\mathrm{2}},\:{i}\:{get}\:{also}\:{a} \\ $$$${quadratic}\:{equation}\:{about}\:{t}: \\ $$$${t}^{\mathrm{4}} −\mathrm{4}{t}^{\mathrm{3}} −\mathrm{12}{t}+\mathrm{1}=\mathrm{0}. \\ $$
Commented by john santu last updated on 08/Apr/20
yes. we cannot get the exact value
$${yes}.\:{we}\:{cannot}\:{get}\:{the}\:{exact}\:{value} \\ $$

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