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The real numbers $x$ and $y$ satisfy $x^2 + y^2 = 4x + 4y.$ Find the largest possible value of $x.$

 Mar 19, 2020
 #1
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x^2  + y^2  =  4x  +  4y      rearrange  as

 

x^2 - 4x  + y^2  - 4y  =   0        complete the square  on x and y

 

x^2 - 4x + 4  + y^2 - 4y + 4   =  4 + 4      simplify

 

(x - 2)^2  + (y - 2)^2  =  8

 

x  will be maximzed when  y  = 2   ....so we  have that

 

(x - 2)^2  + (2 - 2)^2  =  8

 

(x - 2)^2  =  8       take the positive root

 

x  - 2  = √8

 

x = √8 + 2

 

x  = 2√2 + 2

 

x = 2 (√2 + 1)

 

 

 

cool cool cool

 Mar 19, 2020
 #2
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0

stop cheating on your aops homework bruh

 Mar 19, 2020

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