# Question #31214

Jan 18, 2017

See the explanation section, below.

#### Explanation:

A Surge Function has the form $f \left(x\right) = a x {e}^{- b x}$ for positive $a , b$.

In order to find the maximum, we must find the derivative and the critical numbers for $f$.

$f ' \left(x\right) = a {e}^{- b x} - a b x {e}^{- b x} = a {e}^{- b x} \left(1 - b x\right)$.

$f ' \left(x\right) = 0$ when $1 - b x = 0$. Which happens at $x = \frac{1}{b}$.

We know that $a$ and ${e}^{- b x}$ are both positive, so the sign of $f ' \left(x\right)$ agrees with that of $\left(1 - b x\right)$.

$f ' \left(x\right) < 0$ for $x < \frac{1}{b}$ (test $\frac{1}{2 b}$)
and
$f ' \left(x\right) > 0$ for $x > \frac{1}{b}$ (test $\frac{2}{b}$).

Therefore, $f \left(\frac{1}{b}\right) = \frac{a}{b e}$ is the maximum.