We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
81
2
avatar+1040 

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+103148 
+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+1040 
+1

Thank you CPhill!!!

 Jul 30, 2019

6 Online Users

avatar