Say we have function:
$ f(r) = \frac{b}{r} (n + 2^r), r > 0 $
where $b$ and $n$ are some constants large than $0$.
How can we determine the minimal value of this function?
Compute the derivative:
$f'(r) = \frac{-b}{r^2} (n + 2^r) + \frac{b}{r}(2^r \ln 2)$
How can I solve the equation $f'(r) = 0$?
Since draks ... gave you the answer for the solution of $f'(r)=0$, for $$r_0=\frac{W\left(\frac{n}{e}\right)+1}{\log (2)}$$ you could find, after some simplifications using Lambert function properties, that $$f(r_0)=\frac{b n \log (2)}{W\left(\frac{n}{e}\right)}=\frac{b n \log (2)}{r_0 \log (2)-1}$$
Set $f'(r) = \frac{-b}{r^2} (n + 2^r) + \frac{b}{r}(2^r \ln 2)=0$. You get: $$ \frac{1}{r} (n + 2^r) = (2^r \ln 2)\\ n= r2^r\ln2-2^r=2^r(r\ln2 -1) $$ Substitute $x=r\ln2-1$, then $$ n=2^{(x+1)/\ln2}x=2^{x/\ln2}2^{1/\ln2}x=e^xex\\ n/e=xe^x $$ This can be solved using Lambert's W function: $Y=Xe^X \to X=W(Y)$. After resubstitution you get $$r=\frac{W(n/e)+1}{\ln2}$$
