In the first part of this lab, we described an exponential decay model for the concentration of a drug in the blood of a patient. Recall that if the first dose of the drug produces a maximum concentration of , then the concentration of the drug is given by

We also said that a drug would have a minimum effective concentration,
*m* and a maximum save concentration *M*. The problem of devising a
safe and effective treatment program is to come up with a dose of the
drug and an interval between doses, so that the drug concentration is
always above the minimum effective level and below the maximum safe
level. For simplicity, we will assume that the doses all supply the
same amount of the drug and that the time between doses is fixed.

The first problem is to figure out a reasonable time between
doses. One way to get a handle on this is to calculate the maximum
time between doses. That is, suppose a dose is given such that the
concentration immediately after the dose is given is *M*, the maximum
safe dose. If we calculate the time at which the concentration has
decayed to *m*, then this gives the maximum time interval between
doses. For example, consider the drug from the exercises in part 1 of
this lab. To help us later on, we'll refer to this drug as drug B. The
experimental data given said that the initial
concentration was 3.3 mg/ml and the concentration 6 hours later was
1.55 mg/ml. From this data we can determine the value of *k* with the
following command.

> k2 := solve(1.55=3.3*exp(-6*k),k);

Then we can calculate the maximum time between doses with the command shown below to be about 11.8 hours.

> solve( 8*exp(-k2*t)=1.8,t);

This calculation tells us that we can try to use any time interval less than 11.8 hours between doses. Many factors could be important in determining the value to use, including practical concerns like hospital schedules and shift changes.

Fri Feb 7 13:14:44 EST 1997