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Find the maximum value of $y/x$ over all real numbers $x$ and $y$ that satisfy \[(x - 3)^2 + (y - 3)^2 = 6.\]

michaelcai  Oct 30, 2017
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 #1
avatar+78756 
+2

 (x - 3)^2 + (y - 3)^2 = 6

 

This is a circle with a center of  (3, 3)    and a radius  = √ 6

 

I don't know how to prove this but WolframAlpha gives the answer as

 

{ x , y }  =  ( 2 - √ 2 , 2 + √ 2 )

 

And   y / x   ≈   [ 2 + √ 2 ]  / [ 2  - √ 2 ]  ≈  5.828

 

 

cool cool cool

CPhill  Oct 31, 2017
 #2
avatar+78756 
+2

Here's the   Calculus solution for this one...it's a little difficult  !!!

 

 (x - 3)^2 + (y - 3)^2 = 6

 

(y - 3)^2  =  6 -  [ x^2 - 6x + 9]

 

( y - 3)^2  =  6x- x^2 - 3

 

y - 3  = √ [ 6x- x^2 - 3] 

 

y = √ [ 6x- x^2 - 3]  + 3

 

So......call y/x  =  z  ....so we have

 

z =  y/x  =   [ √ [ 6x- x^2 - 3]  + 3]  / x

 

The derivative  of this is messy, but we have

 

z'  =   - 3 ( x  + √ [ 6x- x^2 - 3] - 1 )  /   [ x^2 * √ [ 6x- x^2 - 3] ]

 

We  can find a solution that minimizes  x   by  solving   this :

 

( x  + √ [ 6x- x^2 - 3] - 1 )  =  0

 

x - 1  =   -√ [ 6x- x^2 - 3]        square both sides

 

x^2 - 2x + 1  =  6x- x^2 - 3

 

2x^2 - 8x + 4  =  0

 

x^2 - 4x + 2  = 0

 

The solution that minimizes x  is      2 - √ 2

 

And using    (x - 3)^2 + (y - 3)^2 = 6    

 

When  x =  2 - √ 2     we have

 

( 2 - √ 2 - 3 )^2 +  (y - 3)^2 = 6

 

( 1 + √ 2)^2 +  (y - 3)^2 = 6

 

1 + 2 + 2√ 2 + (y - 3)^2 = 6

 

 (y - 3)^2 = 3 -  2√ 2

 

y =  √ [ 3 -  2√ 2] + 3    ⇒ √  [ ( 1 -√2  ) ^2 ]  + 3   or √  [ ( √2 - 1 ) ^2 ]  + 3  ⇒   

 

y =  4 - √ 2     or   y =   2 + √ 2

 

And   

 

[  2 + √ 2 ] / [  2 - √ 2 ]   maximizes  y / x

 

 

cool cool cool

CPhill  Nov 1, 2017

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