Upper bounds for $n^{1/p} p$

Question

I'm trying to find, for given natural $n$, the positive real value for $p$ that minimizes

$$f(n,p) = n^{1/p} p$$

So there is a minimum value, for each natural number $n$, that $f(n,p)$ takes on as $p$ ranges over all positive reals, and we can call this value $g(n)$, since it is really only a function of $n$, because $p$ is pre-determined, given $n$. I'm looking for to find the best bounds $O(g(n))$.

Answer 1

Partially differentiating $f$ wrt to $p$ gives;

$$\frac{\delta f}{\delta p}=n^{\frac{1}{p}}(1-\frac{1}{p}\ln{n})$$

This is equal to zero for stationary points. This gives either $n=0$ or $n=e^p$.

Similarly Partially differentiating $f$ wrt to $n$ gives;

$$\frac{\delta f}{\delta n}=n^{\frac{1}{p}-1}$$

Which is equal to zero for stationary points. This gives $n=0$.

Now $f(0,p)=0$, hence the stationary points are on the line $n=0$ have value of $0$.

Now $f(e^p,p)=ep$, hence the stationary points are on the line $n=e^p$ have value of $ep$.

Further work using the Hessian is needed to show if points on $n=0$ and $n=e^p$ are minima or maxima.

I am not able to advise on an asymptotic approximation.