Example of using derivatives to find optimal drug dosage

June 16, 2024September 7, 2024by Kurious Fox

Dosage Optimization: Pharmacologists use derivatives to find the optimal drug dosage that maximizes therapeutic effects while minimizing side effects. The concentration of the drug in the bloodstream is modeled as a function of time, and derivatives help identify peak concentration times.

Example: If the drug concentration function is $C(t) = 50t e^{-0.5t}$ , the derivative $C'(t) = 50e^{-0.5t}(1 - 0.5t)$ . Setting $C'(t) = 0$ finds the time $t$ at which concentration peaks, allowing us to find the optimal dosage. More specifically,

Problem: Find the optimal drug dosage time that maximizes the drug concentration $C(t) = 50t e^{-0.5t}$ .

Solution:

First, find the first derivative of $C(t)$ :
$C'(t) = \frac{d}{dt}(50t e^{-0.5t})$
Use the product rule: $(uv)' = u'v + uv'$ , where $u = 50t$ and $v = e^{-0.5t}$ :
$u' = 50, \quad v' = -0.5 e^{-0.5t}$
$C'(t) = 50 e^{-0.5t} + 50t (-0.5 e^{-0.5t}) = 50 e^{-0.5t} - 25t e^{-0.5t}$
$C'(t) = e^{-0.5t} (50 - 25t)$
Set the first derivative to zero to find critical points:
$e^{-0.5t} (50 - 25t) = 0$
Since $e^{-0.5t} \neq 0$ ,
$50 - 25t = 0 \implies t = 2$
To confirm if $t = 2$ is a maximum, find the second derivative $C''(t)$ :
$C''(t) = \frac{d}{dt} \left[ e^{-0.5t} (50 - 25t) \right]$
Use the product rule again:
$C''(t) = e^{-0.5t} \left( -0.5(50 - 25t) \right) + (50 - 25t) \left( -0.5 e^{-0.5t} \right)$
$C''(t) = e^{-0.5t} \left( -25 + 12.5t \right) - 0.5 e^{-0.5t} (50 - 25t)$
$C''(t) = e^{-0.5t} \left( -25 + 12.5t - 25 + 12.5t \right)$
$C''(t) = e^{-0.5t} \left( -50 + 25t \right)$
Evaluate $C''(t)$ at $t = 2$ :
$C''(2) = e^{-1} \left( -50 + 50 \right) = 0$
Since $C''(2) \neq 0$ , check if the function $C(t)$ has a maximum using the first derivative test or by analyzing the function’s behavior around $t = 2$ . For this function, since the concentration increases and then decreases, $t = 2$ corresponds to a maximum point.