Question

Given that xy = 3/2 and both x and y are nonnegative real numbers, find the minimum value of 10x + 3y/5

Answer 1

y = 3/(2x)
so we want the minimum of

10x + 3/5 * 3/(2x)
= 10x + 9/(10x)

take the derivative and you have

10 - 9/(10x^2)

set that to zero and you get

100x^2 - 9 = 0
(10x-3)(10x+3) = 0
x = ±3/10

So, at x = 3/10, y = 5
and 10x+3y/5 = 3+3 = 6

Answer 2

thanks for the help!