If-x-and-y-satisfy-the-equation-x-2-2-y-2-3-what-is-the-maximum-value-of-y-x-

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

If x and y satisfy the equation

{(x - 2)}^{2} + y^{2} = 3, what is the maximum

value of \frac{y}{x} ?

Answered by liberty last updated on 15/Jan/21

f (x) = \frac{\pm \sqrt{3 - {(x - 2)}^{2}}}{x} = \frac{\pm \sqrt{4 x - x^{2} - 1}}{x}

f^{'} (x) = \pm [\frac{x (\frac{2 - x}{\sqrt{4 x - x^{2} - 1}}) - \sqrt{4 x - x^{2} - 1}}{x^{2}}] = 0

\Rightarrow \pm (2 x - x^{2} - (4 x - x^{2} - 1)) = 0

\Rightarrow \pm (1 - 2 x) = 0 \Rightarrow x = \frac{1}{2}

f (\frac{1}{2}) = \pm \frac{\sqrt{3 - {(\frac{1}{2} - 2)}^{2}}}{\frac{1}{2}} = \pm 2 \sqrt{\frac{3}{4}}

{(\frac{y}{x})}_{\max} = \sqrt{3}; {(\frac{y}{x})}_{\min} = - \sqrt{3}

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

y^{2} = 3 - {(x - 2)}^{2} = - 1 + 4 x - x^{2}

{(\frac{y}{x})}^{2} = - \frac{1}{x^{2}} + \frac{4}{x} - 1

{(\frac{y}{x})}^{2} = - {(\frac{1}{x} - 2)}^{2} + 3 ⩽ 3

{(\frac{y}{x})}_{m a x} = \sqrt{3}

Commented by liberty last updated on 15/Jan/21

your method is nice

{(\frac{y}{x})}^{2} ⩽ 3 \Rightarrow (\frac{y}{x} - \sqrt{3}) (\frac{y}{x} + \sqrt{3}) ⩽ 0

we get - \sqrt{3} ⩽ \frac{y}{x} ⩽ \sqrt{3}

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

y o u a r e f a s t!

Commented by liberty last updated on 15/Jan/21

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

Commented by bramlexs22 last updated on 15/Jan/21

yes . . thanks

Answered by ajfour last updated on 15/Jan/21

Commented by ajfour last updated on 15/Jan/21

{(\frac{y}{x})}_{m a x} f o r {(x - h)}^{2} + y^{2} = r^{2} i s

{(\frac{y}{x})}_{m a x} = \frac{r}{\sqrt{h^{2} - r^{2}}}

h e r e = \frac{\sqrt{3}}{1} .

Commented by liberty last updated on 15/Jan/21

\frac{y}{x} = slope = \frac{r}{\tan θ} = \frac{\sqrt{3}}{{(\sqrt{3})}^{2} - {(\sqrt{2})}^{2}} = \sqrt{3}

Answered by Olaf last updated on 15/Jan/21

{(x - 2)}^{2} + y^{2} = 3

{(\frac{y}{x})}^{2} = \frac{3}{x^{2}} - {(1 - \frac{2}{x})}^{2}

2 \frac{d}{d x} (\frac{y}{x}) \frac{y}{x} = - \frac{6}{x^{3}} - \frac{4}{x^{2}} (1 - \frac{2}{x})

\frac{d}{d x} (\frac{y}{x}) = 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 {(\frac{y}{x})}^{2} = \frac{3}{{(1 / 2)}^{2}} - {(1 - \frac{2}{1 / 2})}^{2} = 3

\frac{y}{x} = \pm \sqrt{3}

\sqrt{3} is a maximum .

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