+0  
 
+1
102
1
avatar+102 

 Given that \(xy = \dfrac32 \) and both \(x\) and \(y\) are nonnegative real numbers, find the minimum value of \(10x + \dfrac{3y}5\).

 

Any help would be appriciated. Thank you!

 Feb 21, 2020
 #1
avatar
0

We get the minimum value when x = y = sqrt(3/2) = sqrt(6)/2.  Then 10x + 3y/5 = 53*sqrt(6)/10.

 Feb 21, 2020

12 Online Users

avatar
avatar
avatar
avatar