Question

After an alcoholic beverage is consumed, 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

\(C\left( t \right) = {\bf{0}}{\bf{.135}}t{e^{ - {\bf{2}}{\bf{.802}}t}}\)

models the average BAC, measured in g/dL, 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 3 hours? When does it occur?

Short answer

The maximum average BAC is 0.0177 g/dL, and it will occur at \(t = 0.36\;{\rm{h}}\) or 21.4 min.

Step-by-step solution

Differentiate the equation of \(C\left( t \right)\)

Differentiate the equation \(C\left( t \right) = 0.135t{e^{ - 2.802t}}\) with respect to t.

\(\begin{aligned}{c}C'\left( t \right) &= \frac{{\rm{d}}}{{{\rm{d}}t}}\left( {0.135t{e^{ - 2.802t}}} \right)\\ &= 0.135\left( {t{e^{ - 2.802t}}\left( {2.802} \right) + {e^{ - 2.802t}}\left( 1 \right)} \right)\\ &= 0.135{e^{ - 2.802t}}\left( {2.802t + 1} \right)\end{aligned}\)

Find the root of the equation \(C'\left( t \right) = 0\).

\(\begin{aligned}{c}0.135{e^{ - 2.802t}}\left( {2.802t + 1} \right) &= 0\\t &= - \frac{1}{{2.802}}\\ &= 0.36\end{aligned}\)

Find the value of \(C\left( t \right)\) at critical points

The value of \(C\left( t \right)\) at \(t = 0\);

\(\begin{aligned}{c}C\left( 0 \right) &= 0.135\left( 0 \right){e^{ - 0}}\\ &= 0\end{aligned}\)

At \(t = 0.36\),

\(\begin{aligned}{c}C\left( {0.36} \right) &= 0.135\left( {0.36} \right){e^{ - 2.802\left( {0.36} \right)}}\\ &\approx 0.0177\end{aligned}\)

At \(t = 3\),

\(\begin{aligned}{c}C\left( 3 \right) &= 0.135\left( 3 \right){e^{ - 2.802\left( 3 \right)}}\\ &\approx 0.00009\end{aligned}\)

Thus, the maximum average BAC during the first three hours is about 0.0177 g/dL and it will occur at \(t = 0.36\;{\rm{h}}\) or 21.4 min.