Find two positive numbers that satisfy the given requirements

Question

Question: The product is 147 and the sum of the first number plus three times the second number is a minimum.

Answer 1

XY = 147

x +3y = x + 3(147/x) = (x^2 + 441)/x = f(x)

0= f'(x) = (x ( 2x) - x^2 - 441)/x^2

= 2x^2 - x^2 -441)/x^2

= (x^2 - 441)/x^2

So then x^2 - 441 = 0

x = 21 , as x>0

x=21 and y=7,

the max sum is 21 +3(7) = 42