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Consider all the points in the plane that solve the equation \(x^2 + 2y^2 = 16.\) Find the maximum value of the product xy on this graph.

 Jul 29, 2019
 #1
avatar+111389 
+2

x^2 + 2y^2  = 16

 

2y^2  =  16 - x^2

 

y^2  =  (16 - x^2)/2      take the positive root on both sides

 

y  =  [ (1/2)(16 - x^2)]^(1/2)    =   [ 8 - x^2/2]^(1/2)

 

xy  =  x [ 8 - x^2/2]^(1/2)       take the derivative of this with respect to x   and set to 0

 

y '  = [ 8 - x^2/2]^(1/2)  - (x)(1/2) [ 8 - x^2/2] ^(-1/2) (x)  =

 

[8 - x^2/2]^(1/2) - (x^2/2) [ 8 - x^2/2]^(-1/2)  = 0      factor

 

[8 -x^2/2]^(-1/2)  [ [8-x^2/2  -x^2/2]  = 0     

 

[8 -x^2/2]^(-1/2)  [  8 - x^2]  = 0       divide both sides by the first factor and we have that

 

8 - x^2  = 0

 

x^2  =  8       take the positive root

 

x = √8  = 2√2

 

And y  =  √ [ 8 - (√8)^2 / 2 ]  =  √[ 8 -8/2] = √ 8 -4]  = √4  = 2

 

So.....the max value of xy on the graph is  (2√2) (2)  =  4√2

 

 

 

cool cool cool

 Jul 29, 2019
 #2
avatar+1198 
+1

Thank you CPhill!!!

 Jul 30, 2019

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