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# help

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Find the largest value of $$\frac{y}{x}$$ for pairs of real numbers $$(x,y)$$ that satisfy $$(x - 3)^2 + (y - 3)^2 = 6.$$

Apr 9, 2019

#1
+4809
+3

$$y = \pm \sqrt{6-(x-3)^2}+ 3\\ \dfrac{y}{x} = \dfrac{ \pm \sqrt{6-(x-3)^2}+ 3}{x}\\ \text{The maximum will be when the numerator has a positive sign}\\ \dfrac{y}{x} = \dfrac{ \sqrt{6-(x-3)^2}+ 3}{x}\\ \dfrac{d}{dx} \dfrac{y}{x} = -\dfrac{3 (x-1)}{x^2 \sqrt{-x^2+6 x-3}}\\ \text{and this clearly equals 0 at }x=1$$

$$\text{The second derivative evaluated at 1 is less than zero so this is a maximum}$$

$$\dfrac{y}{x} = \sqrt{6-(1-3)^2}+3 = 3+\sqrt{2}$$

.
Apr 9, 2019
#2
+21978
+5

Find the largest value of
$$\dfrac{y}{x}$$
for pairs of real numbers
$$(x,y)$$
that satisfy
$$(x - 3)^2 + (y - 3)^2 = 6$$.

$$\text{Let OC^2 = x_c^2+y_c^2 = 3^2+3^2 = 2\cdot 3^2 }$$

$$\begin{array}{|rcll|} \hline OT^2 +r^2 &=& OC^2 \\ OT^2 + 6 &=& 2\cdot 3^2 \\ OT^2 &=& 18 - 6 \\ &=& 12 \\ &=& 3\cdot 4 \\ \mathbf{OT} & \mathbf{=} & \mathbf{2\sqrt{3}} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \tan(C) &=& \dfrac{y_c}{x_c} \\ &=& \dfrac{3}{3} \\ \mathbf{\tan(C) } &\mathbf{=}&\mathbf{1} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \tan(B) &=& \dfrac{r}{OT} \\ &=& \dfrac{\sqrt{6}}{2\sqrt{3}} \\ &=& \dfrac{\sqrt{2}\sqrt{3}}{2\sqrt{3}} \\ \mathbf{\tan(B) } &\mathbf{=}&\mathbf{\dfrac{\sqrt{2} }{2 }} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \tan(A) &=& \tan(B+C) \\\\ &=& \dfrac{\tan(B)+\tan(C)} {1-\tan(B)\tan(C)} \\\\ &=& \dfrac{1+\dfrac{\sqrt{2} }{2 }} {1-\dfrac{\sqrt{2} }{2 }} \\\\ &=& \dfrac{2+ \sqrt{2} } {2- \sqrt{2} } \\\\ &=& \dfrac{2+ \sqrt{2} } {2- \sqrt{2} }\cdot \left( \dfrac{2+ \sqrt{2} } {2+ \sqrt{2} } \right) \\\\ &=& \dfrac{\left(2+ \sqrt{2}\right)^2 } {4-2 } \\\\ &=& \dfrac{6+4\sqrt{2}} {2 } \\\\ \mathbf{\tan(A) } &\mathbf{=}&\mathbf{3+2\sqrt{2} } \\ \hline \end{array}$$

The largest value of $$\dfrac{y}{x} = \tan(A) = 3+2\sqrt{2}$$

Apr 9, 2019
#3
+99580
+2

That's a pretty unique method, Heureka.....where did you find it  ????

CPhill  Apr 9, 2019
#4
+21978
+2

Hallo CPhill,

Thank you,
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