Question

when a certain drug is taken orally, the concentration of the drug in the patients bloodstream after t minutes is given by c(t)=0.06t - 0.0002t², where 0 ≤ t ≤ 240. (a) when is the maximum serum - concentration reached? (b) what is the maximum concentration?

Explanation:

Step1: Find the derivative of C(t)

The derivative of \(C(t)=0.06t - 0.0002t^{2}\) using the power - rule \((x^n)^\prime=nx^{n - 1}\) is \(C^\prime(t)=0.06-0.0004t\).

Step2: Set the derivative equal to zero to find critical points

Set \(C^\prime(t) = 0\), so \(0.06-0.0004t=0\). Solving for \(t\) gives \(0.0004t=0.06\), then \(t=\frac{0.06}{0.0004}=150\) minutes.

Step3: Check the second - derivative to confirm it's a maximum

The second - derivative \(C^{\prime\prime}(t)=- 0.0004<0\), so \(t = 150\) is a point of maximum.

Step4: Find the maximum concentration

Substitute \(t = 150\) into \(C(t)\): \(C(150)=0.06\times150-0.0002\times(150)^{2}=9 - 4.5 = 4.5\) mg/L.

Answer:

The maximum serum concentration is reached at \(t = 150\) minutes and the maximum concentration is \(4.5\) mg/L.