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Question Number 129344 by bramlexs22 last updated on 15/Jan/21

$$\mathrm{If}\mathrm{x}\mathrm{and}\mathrm{y}\mathrm{satisfy}\mathrm{the}\mathrm{equation}$$$${(\mathrm{x}-2)}^2+{\mathrm{y}}^2=3,\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{maximum}$$$$\mathrm{value}\mathrm{of}\frac{\mathrm{y}}{\mathrm{x}}?$$

Answered by liberty last updated on 15/Jan/21

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\pm \sqrt{3-{(\mathrm{x}-2)}^2}}{\mathrm{x}}=\frac{\pm \sqrt{4\mathrm{x}-{\mathrm{x}}^2-1}}{\mathrm{x}}$$$$\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=\pm \left[\frac{\mathrm{x}\left(\frac{2-\mathrm{x}}{\sqrt{4\mathrm{x}-{\mathrm{x}}^2-1}}\right)-\sqrt{4\mathrm{x}-{\mathrm{x}}^2-1}}{{\mathrm{x}}^2}\right]=0$$$$\Rightarrow \pm (2\mathrm{x}-{\mathrm{x}}^2-(4\mathrm{x}-{\mathrm{x}}^2-1))=0$$$$\Rightarrow \pm (1-2\mathrm{x})=0\Rightarrow \mathrm{x}=\frac{1}{2}$$$$\mathrm{f}\left(\frac{1}{2}\right)=\pm \frac{\sqrt{3-{(\frac{1}{2}-2)}^2}}{\frac{1}{2}}=\pm 2\sqrt{\frac{3}{4}}$$$${\left(\frac{\mathrm{y}}{\mathrm{x}}\right)}_{\max }=\sqrt{3};{\left(\frac{\mathrm{y}}{\mathrm{x}}\right)}_{\min }=-\sqrt{3}$$

Answered by mr W last updated on 15/Jan/21

$$y^2=3-{(x-2)}^2=-1+4x-x^2$$$${\left(\frac{y}{x}\right)}^2=-\frac{1}{x^2}+\frac{4}{x}-1$$$${\left(\frac{y}{x}\right)}^2=-{(\frac{1}{x}-2)}^2+3⩽3$$$${\left(\frac{y}{x}\right)}_{max}=\sqrt{3}$$

Commented by liberty last updated on 15/Jan/21

$$\mathrm{your}\mathrm{method}\mathrm{is}\mathrm{nice}$$$${\left(\frac{\mathrm{y}}{\mathrm{x}}\right)}^2⩽3\Rightarrow (\frac{\mathrm{y}}{\mathrm{x}}-\sqrt{3})(\frac{\mathrm{y}}{\mathrm{x}}+\sqrt{3})⩽0$$$$\mathrm{we}\mathrm{get}-\sqrt{3}⩽\frac{\mathrm{y}}{\mathrm{x}}⩽\sqrt{3}$$

Commented by mr W last updated on 15/Jan/21

$$youarefast!$$

Commented by liberty last updated on 15/Jan/21

$$3-{(\mathrm{x}-2)}^2=3-({\mathrm{x}}^2-4\mathrm{x}+4)=4\mathrm{x}-{\mathrm{x}}^2-1$$

Commented by bramlexs22 last updated on 15/Jan/21

$$\mathrm{yes}..\mathrm{thanks}$$

Answered by ajfour last updated on 15/Jan/21

Commented by ajfour last updated on 15/Jan/21

$${\left(\frac{y}{x}\right)}_{max}for{(x-h)}^2+y^2=r^{2}is$$$${\left(\frac{y}{x}\right)}_{max}=\frac{r}{\sqrt{h^2-r^2}}$$$$here=\frac{\sqrt{3}}{1}.$$

Commented by liberty last updated on 15/Jan/21

$$\frac{\mathrm{y}}{\mathrm{x}}=\mathrm{slope}=\frac{\mathrm{r}}{\tan \theta }=\frac{\sqrt{3}}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2}=\sqrt{3}$$

Answered by Olaf last updated on 15/Jan/21

$$$$$${(x-2)}^2+y^2=3$$$${\left(\frac{y}{x}\right)}^2=\frac{3}{x^2}-{(1-\frac{2}{x})}^2$$$$2\frac{d}{dx}\left(\frac{y}{x}\right)\frac{y}{x}=-\frac{6}{x^3}-\frac{4}{x^2}(1-\frac{2}{x})$$$$\frac{d}{dx}\left(\frac{y}{x}\right)=0\Rightarrow \frac{4}{x^2}(1-\frac{2}{x})=-\frac{6}{x^3}$$$$x(1-\frac{2}{x})=-\frac{3}{2}$$$$x=\frac{1}{2}\Rightarrow {\left(\frac{y}{x}\right)}^2=\frac{3}{{\left(1/2\right)}^2}-{(1-\frac{2}{1/2})}^2=3$$$$\frac{y}{x}=\pm \sqrt{3}$$$$\sqrt{3}\mathrm{is}\mathrm{a}\mathrm{maximum}.$$

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