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!
We get the minimum value when x = y = sqrt(3/2) = sqrt(6)/2. Then 10x + 3y/5 = 53*sqrt(6)/10.
+111 
