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Let x and y be nonnegative real numbers. if x^2+5y^2=30, what is the max value of x + y?

 Jun 14, 2024
 #1
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x^2 + 5y^2  = 30

 

Take the derivative with respect to x and set to 0

2x = 0

Take the derivative with respect ot y and  set to 0

10y = 0

 

This implies that

2x =10y

x = 5y

 

So

 

(5y)^2 + 5y^2   =30

 

30y^2  = 30

 

y = 1

 

x = 5y  =  5

 

Mav  x + y   =  6

 

cool cool cool

 Jun 15, 2024

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