Chapter 5.EA, Problem 1EA

To determine

To find:

The maximum and minimum steady-state concentration for the given conditions.

Answer to Problem 1EA

Solution:

The maximum and minimum steady-state concentration for the given conditions are $775\text{mcg}/\text{mL}$ and $185\text{mcg}/\text{mL}$.

Explanation of Solution

Given:

The given conditions are:

$C_M\left(t\right)=1.99D{e}^{-0.14t}-1.62D{e}^{-2.08t}$

and $500-\text{mg}$ dose is given every 12 hours.

Approach:

First derivative test:

If c is a critical number for f on the open interval $\left(a,b\right)$, f is continuous on $\left(a,b\right)$, differentiable on $\left(a,b\right)$ (except possibly at c), then

1. $f\left(c\right)$ is a relative maximum if $f^{\prime }\left(x\right)>0$ on $\left(a,c\right)$ and $f^{\prime }\left(x\right)<0$ on $\left(c,b\right)$.

2. $f\left(c\right)$ is a relative minimum if $f^{\prime }\left(x\right)<0$ on $\left(a,c\right)$ and $f^{\prime }\left(x\right)>0$ on $\left(c,b\right)$.

Calculation:

Consider the given function:

$C_M\left(t\right)=1.99D{e}^{-0.14t}-1.62D{e}^{-2.08t}$

Substitute the value $500-\text{mg}$ for $D$, we get

$\begin{array}{rll}{C}_M\left(t\right)& =& 1.99\left(500\right)e^{-0.14t}-1.62\left(500\right)e^{-2.08t}\\ & =& 995e^{-0.14t}-810e^{-2.08t}\end{array}$

Differentiate the function with respect to t:

$\begin{array}{rll}{C}_M{}^{\prime }\left(t\right)& =& 995\left(-0.14\right)e^{-0.14t}-810\left(-2.08\right)e^{-2.08t}\\ & =& -139.3e^{-0.14t}+1684.8e^{-2.08t}\end{array}$

Find the critical point:

$\begin{array}{rll}-139.3e^{-0.14t}+1684.8e^{-2.08t}& =& 0\\ e^{-0.14t}\left(-139.3+1684.8e^{-1.94t}\right)& =& 0\\ -139.3+1684.8e^{-1.94t}& =& 0\\ e^{-1.94t}& =& \frac{139.3}{1684.8}\end{array}$

Take log both side:

$\begin{array}{rll}-1.94t& =& \ln \left(\frac{139.3}{1684.8}\right)\\ t& =& 1.285\end{array}$

Substitute in the function:

$\begin{array}{rll}{C}_M\left(1.285\right)& =& 995e^{-0.14\left(1.285\right)}-810e^{-2.08\left(1.285\right)}\\ & =& 775\text{mcg}/\text{mL}\end{array}$

and minimum is:

$\begin{array}{rll}{C}_M\left(0\right)& =& 995e^{-0.14\left(0\right)}-810e^{-2.08\left(0\right)}\\ & =& 185\text{mcg}/\text{mL}\end{array}$