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Question Number 172253 by cortano1 last updated on 25/Jun/22

$$MaxandminP=\sqrt{2}x+\sqrt{3}y$$$$subjecttoconstraint$$$$\frac{x^2}{9}+\frac{y^2}{25}⩽1⩽x^2+y^2$$

Answered by mr W last updated on 25/Jun/22

$$MethodI$$$$y=\frac{P-\sqrt{2}x}{\sqrt{3}}$$$$\frac{x^2}{9}+\frac{1}{25}{\left(\frac{P-\sqrt{2}x}{\sqrt{3}}\right)}^2=1$$$$25x^2+3(P^2-2\sqrt{2}Px+2x^2)=9\times 25$$$$31x^2-6\sqrt{2}Px+3P^2-225=0$$$$\mathrm{\Delta }=36\times 2P^2-4\times 31(3P^2-225)=0$$$$P^2=93$$$$P=\pm \sqrt{93}$$$$\Rightarrow P_{max}=\sqrt{93}$$$$\Rightarrow P_{min}=-\sqrt{93}$$

Commented by mr W last updated on 25/Jun/22

Commented by Tawa11 last updated on 25/Jun/22

$$\mathrm{Great}\mathrm{sir}$$

Answered by mr W last updated on 25/Jun/22

$$MethodII$$$$\frac{x^2}{3^2}+\frac{y^2}{5^2}=1$$$$\Rightarrow x=3\cos \theta ,y=5\sin \theta$$$$P=\sqrt{2}x+\sqrt{3}y=3\sqrt{2}\cos \theta +5\sqrt{3}\sin \theta$$$$P=\sqrt{{\left(3\sqrt{2}\right)}^2+{\left(5\sqrt{3}\right)}^2}(\frac{3\sqrt{2}}{\sqrt{{\left(3\sqrt{2}\right)}^2+{\left(5\sqrt{3}\right)}^2}}\cos \theta +\frac{5\sqrt{3}}{\sqrt{{\left(3\sqrt{2}\right)}^2+{\left(5\sqrt{3}\right)}^2}}\sin \theta )$$$$P=\sqrt{93}(\frac{3\sqrt{2}}{\sqrt{93}}\cos \theta +\frac{5\sqrt{3}}{\sqrt{93}}\sin \theta )$$$$P=\sqrt{93}\sin (\theta +{\tan }^{-1}\frac{3\sqrt{2}}{5\sqrt{3}})$$$$\Rightarrow P_{max}=\sqrt{93}at\theta =\frac{\pi }{2}-{\tan }^{-1}\frac{3\sqrt{2}}{5\sqrt{3}}$$$$\Rightarrow P_{min}=-\sqrt{93}at\theta =-\frac{\pi }{2}-{\tan }^{-1}\frac{3\sqrt{2}}{5\sqrt{3}}$$

Commented by mr W last updated on 25/Jun/22

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