Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.3* The Natural Exponential Function - 6.3* Exercises - Page 454: 79


$0.172$ mg/mL

Work Step by Step

Given: The function $C(t) =1.35te^{-2.802t}$ Calculate the maximum 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. For this, we will have to take differentiae of the function $C(t) =1.35te^{-2.802t}$ $C'(t)= 1.35(e^{-2.802t}-t(2.802)e^{-2.802t})$ Put $C'(t)=0$ $1.35(e^{-2.802t}-t(2.802)e^{-2.802t})=0$ $t=0.3569$ hours Therefore, the maximum average BAC will be at $t=0.3569$ hours This implies $C(0.3569)=1.35(0.3569)e^{-2.802(0.3569)}=0.172$ mg/mL
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