Integer Optimization for a rational function

Question

Let $f(m,n)=\frac{1+(2^{n-1}-1)(2^{m-1}-1)}{2^{mn}}$ where $m,n \in \mathbb{Z}^+ - \{0\}$. Suppose $mn=10000$. How should we choose $m$ and $n$ so that $f(m,n)$ is minimized.

Answer 1

If you already know, that $mn=10\,000=10^4$, then your function is $f(m,n)=1+10^{-4}(2^{m-1}-1)(2^{n-1}-1)$. Since $10^{-4}>0$, it suffices to minimize $(2^{m-1}-1)(2^{n-1}-1)$. For $m=10^4/n$, we get $$g(n)=(2^{10^4/n-1}-1)(2^{n-1}-1).$$ Since both factors are positive for $1<n<10^4$, we get the minimal value 0 at $n\in \{1,10^4\}$. Equivalently, the minimum of $f$ is attained at $(m,n)=(1,10^4)$ or $(m,n)=(10^4,1)$.