Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 5 - Graphs and the Derivative - Extended Application - Exercises - Page 302: 1

Answer

Maximum steady state concentration is $775.24 mcg/mL$. Minimum steady state concentration is approximately $185 mcg/mL$.

Work Step by Step

The steady-state concentration function is given by, $C_{ss}(t)=1.99De^{-0.14t}-1.62De^{-2.08t} mcg/mL$ $D$ is the size of the dose in milligrams administered every $12 hrs$. When $D=500 mg$, $C_{ss}(t)=995e^{-0.14t}-810e^{-2.08t} mcg/mL$ At critical points, $C'_{ss}(t)=0$. $C'_{ss}(t)=-139.3e^{-0.14t}+1684.8e^{-2.08t}$ $C'_{ss}(t)=0 \implies t=1.285 hrs$. $C_{ss}(1.285)=775.24 mcg/mL$ $C_{ss}(1)=763.817mcg/mL$ Therefore we can conclude that at $t=1,285 hrs$ the steady-state function is maximum. At $t=0 $ and $t=12$, $C_{ss}(t)\approx185 mcg/mL$ which is the minimum steady- state concentration.
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