Question

After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modeled by the function C(t)=3(e^(−0.4t)−e^(−0.6t)) where the time t is measured in hours and C is measured in mu g/mL. What is the maximum concentration of the antibiotic during the first 12 hours?

Answer 1

0.444 µg/mL

Explanation:

You want the maximum value of C(t) = 3(e^(−0.4t)−e^(−0.6t)) in the period 0 < t < 12. C is measured in µg/mL, and t is measured in hours.

Derivative

The maximum concentration will be at the point where the derivative of C(t) with respect to t is zero.

C' = 3(-0.4e^(-0.4t) +0.6e^(-0.6t)) = 0

3e^(-0.6t) -2e^(-0.4t) = 0 . . . . . divide by 0.6

3e^(-0.2t) -2 = 0 . . . . . . . . . divide by e^(-0.4t)

e^(-0.2t) = 2/3 . . . . . . . . . add 2, divide by 3

Max C(t)

Now, we can find the values of the exponential terms:

e^(-0.4t) = (e^(-0.2t))^2 = (2/3)^2 = 4/9

e^(-0.6t) = (e^(-0.2t))^3 = (2/3)^3 = 8/27

So, ...

C(t) = 3(4/9 -8/27) = 3(4/27) = 4/9 . . . . . maximum concentration

The maximum concentration of antibiotic is 0.444 µg/mL.

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Additional comment

The time at which the maximum occurs will be approximately ...

t = ln(2/3)/-0.2 ≈ 2.027 . . . . hours after the tablet is taken

You might normally expect the value of t to be found first, then C(t) evaluated for that value of t. By skipping those steps, we eliminate some rounding errors and arrive at a more accurate value for the maximum concentration.