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Let $x$ and $y$ be real numbers. If $x^2 + 3y^2 = 18$, then find the maximum value of $x + y$.

 Jun 9, 2024
 #1
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x^2 + 3y^2   =18

 

Take the derivative with respect to x and  set to 0

2x = 0

 

Take the derivative with respect to y and set to 0

6y = 0

 

This implies that   2x  = 6y

y = (1/3)x

 

So

 

x^2  + 3 [(1/3) x]^2  = 18

 

(4/3]x^2 =18

 

x^2  = 54/4 

 

x =  sqrt [54/4]    =  sqrt [27/2]  =  3sqrt (3)  /sqrt (2)  = (3/2)sqrt 6

 

y = (1/2)sqrt 6

 

Max is

 x + y =    (3/2)sqrt (6)  + (1/2)sqrt (6)  =    2sqrt 6

 

cool cool cool

 Jun 10, 2024

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