Step 1:

Let the two positive numbers be x and $\large\frac{100}{ x}$ (since their product is 100)

Step 2:

The sum of two numbers $s= x +\large\frac{100}{x}$

$\large\frac{ds}{dx}=1-\large\frac{100}{x^2}$

At extreme values of $s, \large\frac{ds}{dx}$$=0=>1-\large\frac{100}{x^2}$$=0$

$=>x^2=100\;or\; x=10$

Now $\large\frac{d^2s}{dx^2}=\frac{200}{x^3}$

$\large\frac{d^2s}{dx^2(x=10)}=\frac{200}{1000}=\frac{1}{5} > 0$

$\therefore x=10$ corresponds to a minimum value of S.

The two numbers are 10 and 10