Answer:
the concentration is maximum at 2 hours after the drug is given, and the maximum concentration is \( 0.02 \) mg/cm³.
Step-by-step explanation:
To find the maximum concentration of the drug and the time at which it occurs, we need to find the critical points of the function [tex]\( C(t) \)[/tex] and then determine whether these points correspond to a maximum or minimum.
First, let's find the derivative of [tex]\( C(t) \)[/tex] concerning [tex]\( t \)[/tex]:
[tex]\[ C'(t) = \frac{d}{dt} \left( \frac{0.16t}{t^2 + 4t + 4} \right) \][/tex]
We can use the quotient rule:
[tex]\[ C'(t) = \frac{(0.16(t^2 + 4t + 4) - 0.16t(2t + 4))}{(t^2 + 4t + 4)^2} \][/tex]
Simplify this expression:
[tex]\[ C'(t) = \frac{0.16t^2 + 0.64t + 0.64 - 0.32t^2 - 0.64t}{(t^2 + 4t + 4)^2} \][/tex]
[tex]\[ C'(t) = \frac{-0.16t^2 + 0.64}{(t^2 + 4t + 4)^2} \][/tex]
To find the critical points, we set the derivative equal to zero and solve for [tex]\( t \)[/tex]:
[tex]\[ -0.16t^2 + 0.64 = 0 \][/tex]
[tex]\[ -0.16t^2 = -0.64 \][/tex]
[tex]\[ t^2 = 4 \][/tex]
[tex]\[ t = \pm 2 \][/tex]
Now, we need to determine whether[tex]\( t = 2 \) or \( t = -2 \)[/tex] corresponds to a maximum or minimum. We can use the second derivative test or analyze the behavior of the function around these points.
Since [tex]\( t \)[/tex] represents the number of hours after the drug is taken, we discard the negative solution [tex]\( t = -2 \).[/tex]
To determine if t = 2 corresponds to a maximum or minimum, we can analyze the behavior of the derivative around this point or use the second derivative test.
Taking the second derivative of C(t):
[tex]\[ C''(t) = \frac{d}{dt} \left( \frac{-0.16t^2 + 0.64}{(t^2 + 4t + 4)^2} \right) \][/tex]
After computing the second derivative, you'll find that [tex]\( C''(2) < 0 \)[/tex]. This means that the function has a local maximum at [tex]\( t = 2 \)[/tex], indicating that the concentration is maximum 2 hours after the drug is given.
To find the maximum concentration, plug [tex]\( t = 2 \)[/tex] into the original function:
[tex]\[ C(2) = \frac{0.16 \times 2}{2^2 + 4 \times 2 + 4} \][/tex]
[tex]\[ C(2) = \frac{0.32}{4 + 8 + 4} \][/tex]
[tex]\[ C(2) = \frac{0.32}{16} \][/tex]
[tex]\[ C(2) = 0.02 \text{ mg/cm}^3 \][/tex]
So, the concentration is maximum at 2 hours after the drug is given, and the maximum concentration is \( 0.02 \) mg/cm³.
