The product is 27 and the sum of the first plus three times the second is a minimum.

Answer 1

xy = 27

y=27/x

minimize x+ 3y

x+3y = x + 3(27/x)

take the derivative and set = 0

f'(x) = 1 -81/x^2 = 0

81/x^2 = 1

x^2 = 81

x = 9

y = 27/9 = 3

the numbers are 3 and 9

9 + 3(3) = 18 = minimum possible x+3y