Calculus: Maximum and Minimum Values After the consumption of an alcoholic beverage, the concentration of alcohol in the bloodstream (blood alcohol concentration, or BAC) surges as the alcohol is absorbed, followed by a gradual decline as the alcohol is metabolized. The function
$$C(t) = 1.35te^{−2.802t}$$
models the average BAC, measured in mg/mL, of a group of eight male subjects t hours after rapid consumption of $15 mL$ of ethanol (corresponding to one alcoholic drink).
What is the maximum average BAC during the first 6 hours?
When does it occur?
 A: I would do this problem as follows because the product rule is messy.
First, hide the constants because it's just going to waste time writing:
$$f(t) = \alpha t e^{-\beta t}$$
Let's rewrite this in a convenient form:
$$f(t) = \alpha e^{\ln t} e^{-\beta t} = \alpha e^{\ln t - \beta t}$$
Since $e^x$ is strictly increasing (make sure you understand what I mean here, at least intuitively), it will be maximized exactly when its argument is maximized, so we can maximize instead:
$$g(t) = \ln t - \beta t$$
This has the simple first order condition:
$$g'(t) = \frac{1}{t} - \beta = 0$$
Which clearly has solution $t^{*} = \frac{1}{\beta}$. Going back to the problem formulation we see $\beta = 2.802$.
Technically to be sure it's maximized we need to check the second order condition, and that of $g$ will suffice (again, check your understanding of why this is true, intuitively and/or through math):
$$g''(t^{*}) = -\frac{1}{(t^{*})^2}<0$$
And we're done.
A: You have to compute the derivative of $C(t)$:
$$C'(t)=1.35t\cdot(-2.802)e^{-2.802t}+1.35e^{-2.802t}$$
It has an optimum $t_{0}$ if $C'(t_{0})=0$ and $C''(t_{0})\neq 0$. Here, we have $$\begin{align}
C'(t_{0}) &=0\\
1.35t_{0}\cdot(-2.802)e^{-2.802t_{0}}+1.35e^{-2.802t_{0}}&=0\\
e^{-2.802t_{0}}\left(-1.35\cdot 2.802t_{0}+1.35 \right) &=0
\end{align}$$
It is clear that $e^{-2.802t_{0}}\neq 0$ for all $t_{0}\in\mathbb{R}$, so that 
$$-1.35\cdot 2.802t_{0}+1.35=0$$
and you get
$$t_{0}=\frac{1}{2.802}\approx0.36$$
I assume $t$ is in hours, so that it makes $t_{0}=0.36\,\text{h}\approx 21.41\min$. This is the only optimum and one should verify it is indeed a maximum, i.e. $C''(t_{0})<0$.
You can easily compute the maximum average BAC, which is $C(t_{0})=C\left(2.802^{-1}\right)$. WolframAlpha can be an useful tool to check your answers.
A: Just as MichaelChirico suggested to do, you want to maximize $$f = \alpha\, t \,e^{-\beta t}$$ That is to say to find the zero of its derivative $f'$.
When you have this kind of expressions, logarithmic differentiation makes life easier. $$\log(f)=\log(\alpha)+\log(t)-\beta t$$ Differentiation both sides $$\frac {f'}f=\frac 1t-\beta$$ and, since $f'=0$, then $\frac {f'}f=0$ and finally $t=\frac 1 \beta$.
