Can we minimize $3^m 2^{n/m}$, given $n$?

Question

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 $m$.

Answer 1

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 $0$.

Answer 2

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 $m=\sqrt{n\log_32}$.

Answer 3

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.

Answer 4

Hint: $am+\frac{b}{m}$ is minimized at $m=\sqrt{\frac{b}{a}}$.

Answer 5

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 $m = \sqrt{n \log_3 2}$