Answer:
Both the numbers are 8.
Step-by-step explanation:
Let the two numbers are p and 64/p.
The sum is given by
[tex]S = p +\frac{64}{p}\\\\\frac{dS}{dp}= 1 - \frac{64}{p^2}\\\\\frac{dS}{dp}=0\\\\\frac{64}{p^2}=1\\\\p= \pm 8[/tex]
So, the sum is minimum for p = 8 0r - 8, so the two numbers 8.
