## Calculus: Early Transcendentals 8th Edition

The numbers are $x$ and $\frac{100}{x}$. We want to minimize $x+\frac{100}{x}$, which we will call $f(x)$. $f'(x)=1-\frac{100}{x^{2}}$, or $\frac{x^{2}-100}{x^{2}}$. Therefore there is a minimum value at $x=10$.