Question #31214

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Answer 1 · 2017-01-18T17:44:08

See the explanation section, below.

A Surge Function has the form $f(x) = axe^(-bx)$ for positive $a,b$ .

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

$f'(x) = ae^(-bx) - abxe^(-bx) = ae^(-bx)(1-bx)$ .

$f'(x) = 0$ when $1-bx =0$ . Which happens at $x=1/b$ .

We know that $a$ and $e^(-bx)$ are both positive, so the sign of $f'(x)$ agrees with that of $(1-bx)$ .

$f'(x) < 0$ for $x < 1/b$ (test $1/(2b)$ )
and
$f'(x) > 0$ for $x > 1/b$ (test $2/b$ ).

Therefore, $f(1/b) = a/(be)$ is the maximum.