Can we minimize $3^m 2^{n/m}$, given $n$? If we are given $n$, a positive real, can we find a the positive real $m$ that minimizes the function:
$$3^m 2^{n/m}$$
I'd prefer to find the function that gives a value for $m$, but I'm also interested in asymptotic bounds for $m$.
 A: Minimizing $3^m2^{n/m}$ is the same as minimizing
$$\log 3^m2^{n/m} = m\log 3 + \frac nm\log2,$$
which we can solve by setting the derivative in $m$ equal to $0$.
A: If you take the logarithm, your expression turns into
$$m\ln 3+\frac{n\ln 2}{m}.$$
By the arithmetic-geometric-mean inequality
$$ m\ln 3+\frac{n\ln 2}{m}\ge 2\sqrt{n\ln 3\ln2}$$
with equality iff $m\ln 3=\frac{n\ln 2}{m}$, that is iff $m=\sqrt{n\log_32}$.
A: Take the logarithm of the expression,
$$\log \left(3^m 2^{n/m}\right) = m\log 3 + \frac{n}{m}\log 2,$$
differentiate with respect to $m$,
$$\log 3 - \frac{n}{m^2}\log 2$$
to find the critical point
$$m = \sqrt{n\frac{\log 2}{\log 3}}.$$
Since the derivative is negative for small $m$ and positive for large, that is the global minimum.
A: Hint: $am+\frac{b}{m}$ is minimized at $m=\sqrt{\frac{b}{a}}$.
A: Taking logs we have $m \log 3 + \dfrac{n}{m} \log 2$ to minimise.  As this is a sum whose product is a constant, minimum occurs when the terms are equal.  
So solve $m \log 3 = \dfrac{n}{m} \log 2$ to get $m = \sqrt{n \log_3 2}$
