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Question Number 152478 by imjagoll last updated on 28/Aug/21

Answered by mr W last updated on 28/Aug/21

$$x=5\cos \theta$$$$y=5\sin \theta$$$$k=2\times 25{\cos }^2\theta +6\times 25\cos \theta \sin \theta -4\times 25{\sin }^2\theta$$$$=25(6{\cos }^2\theta +3\sin 2\theta -4)$$$$=25(6{\cos }^2\theta +3\sin 2\theta -3-1)$$$$=25(3\cos 2\theta +3\sin 2\theta -1)$$$$=75\sqrt{2}\cos (2\theta -\frac{\pi }{4})-25$$$$max=75\sqrt{2}-25$$$$min=-75\sqrt{2}-25$$

Commented by mr W last updated on 28/Aug/21

Answered by bramlexs22 last updated on 29/Aug/21

$$\mathrm{f}(\mathrm{x},\mathrm{y})=2({\mathrm{x}}^2+{\mathrm{y}}^2)+6\mathrm{xy}-{6\mathrm{y}}^2=6\mathrm{xy}-{6\mathrm{y}}^2+50$$$$\mathrm{where}\mathrm{y}=\sqrt{25-{\mathrm{x}}^2}$$$$\mathrm{f}\left(\mathrm{x}\right)=6\mathrm{x}\sqrt{25-{\mathrm{x}}^2}-6(25-{\mathrm{x}}^2)+50$$$$\mathrm{f}\left(\mathrm{x}\right)=6\mathrm{x}\sqrt{25-{\mathrm{x}}^2}+{6\mathrm{x}}^2-100$$$$\frac{\mathrm{df}\left(\mathrm{x}\right)}{\mathrm{dx}}=6\sqrt{25-{\mathrm{x}}^2}-\frac{{6\mathrm{x}}^2}{\sqrt{25-{\mathrm{x}}^2}}+12\mathrm{x}=0$$$$\Rightarrow \sqrt{25-{\mathrm{x}}^2}+2\mathrm{x}=\frac{{\mathrm{x}}^2}{\sqrt{25-{\mathrm{x}}^2}}$$$$\Rightarrow 25-{\mathrm{x}}^2+2\mathrm{x}\sqrt{25-{\mathrm{x}}^2}={\mathrm{x}}^2$$$$\Rightarrow {2\mathrm{x}}^2-25=2\mathrm{x}\sqrt{25-{\mathrm{x}}^2}$$$$\Rightarrow {4\mathrm{x}}^4-{100\mathrm{x}}^2+625={4\mathrm{x}}^2(25-{\mathrm{x}}^2)$$$$\Rightarrow {8\mathrm{x}}^4-{200\mathrm{x}}^2+625=0$$$$\Rightarrow {\mathrm{x}}^2=\frac{200+\sqrt{{200}^2-32\times 625}}{16}=\frac{200+100\sqrt{2}}{16}$$$$\Rightarrow {\mathrm{x}}^2=\frac{50+25\sqrt{2}}{4}\mathrm{then}{\mathrm{y}}^2=25-\left(\frac{50+25\sqrt{2}}{4}\right)=\frac{50-25\sqrt{2}}{4}$$$$\Rightarrow {\mathrm{x}}^2{\mathrm{y}}^2=\left(\frac{50+25\sqrt{2}}{4}\right)\left(\frac{50-25\sqrt{2}}{4}\right)=\frac{1250}{16}$$$$\Rightarrow \mathrm{xy}=\sqrt{\frac{1250}{16}}=\pm \frac{25\sqrt{2}}{4}$$$$\max ({2\mathrm{x}}^2+6\mathrm{xy}-{4\mathrm{y}}^2)=6\left(\frac{25\sqrt{2}}{4}\right)-6\left(\frac{50-25\sqrt{2}}{4}\right)+50$$$$=\frac{150\sqrt{2}-300+150\sqrt{2}+200}{4}$$$$=75\sqrt{2}-25$$$$\min ({2\mathrm{x}}^2+6\mathrm{xy}-{4\mathrm{y}}^2)=-6\left(\frac{25\sqrt{2}}{4}\right)-6\left(\frac{50-25\sqrt{2}}{4}\right)+50$$$$=\frac{-150\sqrt{2}-300+150\sqrt{2}+200}{4}$$$$$$

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