Calculate Function global maximum

Question

$f_n: (0, ∞) →ℝ$, $f_n(x) = x/n^2 \cdot e^{-x/n}$

Show that the function $f_n$ has a global maximum with value $1/(ne)$.

Answer 1

To show this, just proceed in the usual way to show a maximum, assuming $n$ is a constant. That is, check the point $x^*$ at which the derivative is zero, verify it is a maximum, and then evaluate that point.

Let's begin by taking derivatives,

$$ \frac{\partial}{\partial x}(x/n^2) e^{-x/n} = (1/n^2) e^{-x/n}-(x/n^3) e^{-x/n} =0$$

So, the condition that needs to be verified is, $$ x^* = n $$ Which is maximum given the positive sign on the left and the negative sign on the right of $f^{'}(x^*)$. That is, the function grows (positive derivative) up to $x^*$ and then in decreases (negative derivative) after $x^*$. Now, we verify the functional value of $x^*$, as follows, $$ f(x^*)=f(n)=(n/n^2) e^{-n/n}=1/ne^{-1}=\frac{1}{en} $$ Completing the proof.