Let the two numbers be x & y
Given xy = 64 ⇒ y = \(\frac{64}{x}\)
Let s = x + y = x + \(\frac{64}{x}\)
\(\frac{ds}{dx}\) = 1 – \(\frac{64}{x^2}\). \(\frac{ds}{dx}\) = 0 ⇒ x2 = 64 ⇒ x = ±8
\(\frac{d^2s}{dx^2}\) = \(\frac{128}{x^3}\)
When x = 8, \(\frac{d^2s}{dx^2}\)= \(\frac{128}{8^3}\) > 0 ⇒ s is minimum
When x = -8, \(\frac{d^2s}{dx^2}\) = \(\frac{128}{8^3}\) < 0 ⇒ s is maximum
The two numbers are 8 & 8.
