Answer
The maximum average BAC during the first three hours is $0.177~mg/mL$
This occurs after 21.4 minutes.
Work Step by Step
We can find the value of $t$ when $C'(t) = 0$:
$C(t) = 1.35t~e^{-2.802t}$
$C'(t) = 1.35~e^{-2.802t}+(1.35t)(-2.802e^{-2.802t})=0$
$1.35~e^{-2.802t}=(1.35t)(2.802e^{-2.802t})$
$1=2.802~t$
$t = \frac{1}{2.802}$
$t = 0.357~h$
$t = 21.4~min$
We can find the value of $C(\frac{1}{2.802})$:
$C(t) = 1.35t~e^{-2.802t}$
$C(\frac{1}{2.802}) = 1.35(\frac{1}{2.802})~e^{(-2.802\cdot \frac{1}{2.802})}$
$C(\frac{1}{2.802}) = \frac{1.35}{2.802}~e^{-1}$
$C(\frac{1}{2.802}) = 0.177~mg/mL$
The maximum average BAC during the first three hours is $0.177~mg/mL$
This occurs after 21.4 minutes.