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Question Number 87989 by john santu last updated on 07/Apr/20

$$\mathrm{find}\mathrm{maximum}\mathrm{value}$$$${2\mathrm{x}}^2+{\mathrm{y}}^2\mathrm{with}\mathrm{constraint}$$$${\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}+2\mathrm{y}+1=0$$$$$$

Answered by mr W last updated on 07/Apr/20

$${\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}+2\mathrm{y}+1=0$$$$\Rightarrow {(x-2)}^2+{(y+1)}^2=4$$$$letx=2+2\cos \theta$$$$lety=-1+2\sin \theta$$$$k=2x^2+y^2=8{(1+\cos \theta )}^2+{(-1+2\sin \theta )}^2$$$$\frac{dk}{d\theta }=16(1+\cos \theta )(-\sin \theta )+4(-1+2\sin \theta )\cos \theta =0$$$$4\sin \theta +2\cos \theta \sin \theta +\cos \theta =0$$$$\Rightarrow \frac{4}{\cos \theta }+2+\frac{1}{\sin \theta }=0$$$$\Rightarrow \theta \approx -0.1659,2.7118$$$$$$$$k_{min}\approx 0.094with\theta =2.7118$$$$k_{max}\approx 33.33with\theta =-0.1659$$

Commented by mr W last updated on 07/Apr/20

$$thequestionistofindthesmallest$$$$andthelargestellipsewhichtouches$$$$agivencircle.‘‘{exact}^{″}solutionmaybe$$$$notpossible.$$

Answered by john santu last updated on 08/Apr/20

Commented by mr W last updated on 08/Apr/20

$$wheniuset=\tan \frac{\theta }{2},igetalsoa$$$$quadraticequationaboutt:$$$$t^4-4t^3-12t+1=0.$$

Commented by john santu last updated on 08/Apr/20

$$yes.wecannotgettheexactvalue$$

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