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

 Nov 10, 2017

Best Answer 

 #1
avatar+7352 
+1

(x + 3)2  +  (y - 3)2  =  6          Let's solve this for  y .

(y - 3)2   =   6 - (x + 3)2

 y - 3   =   ±√[ 6 - (x + 3)2 ]

 y   =   ±√[ 6 - (x + 3)2 ]  +  3           So....using this value for  y....

 

\(\frac{y}{x}\,=\,\frac{\pm\sqrt{6-(x+3)^2}+3}{x}\)

 

We can say

 

\(Y\,=\,\frac{\pm\sqrt{6-(x+3)^2}+3}{x}\)      and we want to know the maximum Y value here.

 

We can get an approximation by looking at a graph.  (about  5.828)

 

If you take the derivative and set it = 0, you will get  x = √2 - 2

Then plug this in for  x  and we can find that the exact maximum value  =  3 + 2√2

 Nov 11, 2017
 #1
avatar+7352 
+1
Best Answer

(x + 3)2  +  (y - 3)2  =  6          Let's solve this for  y .

(y - 3)2   =   6 - (x + 3)2

 y - 3   =   ±√[ 6 - (x + 3)2 ]

 y   =   ±√[ 6 - (x + 3)2 ]  +  3           So....using this value for  y....

 

\(\frac{y}{x}\,=\,\frac{\pm\sqrt{6-(x+3)^2}+3}{x}\)

 

We can say

 

\(Y\,=\,\frac{\pm\sqrt{6-(x+3)^2}+3}{x}\)      and we want to know the maximum Y value here.

 

We can get an approximation by looking at a graph.  (about  5.828)

 

If you take the derivative and set it = 0, you will get  x = √2 - 2

Then plug this in for  x  and we can find that the exact maximum value  =  3 + 2√2

hectictar Nov 11, 2017

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