Antibiotic Dosing: One Dose Fits, Two Doses Do Not
A hospital pharmacy is reviewing the dosing schedule for a new oral antibiotic. A single 400 mg dose produces a plasma concentration modelled by (C(t) = 12t,e^{-0.35t}) mg/L, where t is hours after the dose, and the model is stated to be valid for (0 \le t \le 8). The therapeutic window is 2 mg/L to 15 mg/L: below 2 the drug is ineffective, above 15 there is a risk of hearing damage. The current protocol gives a dose every 6 hours. A pharmacist is worried because concentrations are assumed to add, so after a second dose the total is the sum of the two curves. Nursing staff report that most doses are given on time, but the night shift is often 90 minutes late. The committee wants to know whether the protocol is safe, whether the total drug exposure is acceptable, and how far the model can be pushed before it stops being trustworthy.
1. Find the maximum concentration after a single dose and the time at which it occurs.
Read the worked answer
Differentiate using the product rule:
(C'(t) = 12e^{-0.35t} + 12t(-0.35)e^{-0.35t} = 12e^{-0.35t}(1 - 0.35t)).
Since (e^{-0.35t} > 0), (C'(t) = 0) requires (1 - 0.35t = 0), so t = 1/0.35 = 2.857 h.
The derivative is positive for t < 2.857 and negative afterwards, so this is a maximum.
(C(2.857) = 12(2.857)e^{-0.35 \times 2.857} = 34.29,e^{-1} = 34.29 \times 0.36788 = 12.6) mg/L.
So a single dose peaks at about 12.6 mg/L at 2.86 h, which is inside the therapeutic window and below the 15 mg/L toxicity threshold, with a margin of about 2.4 mg/L.
Also note (C(0) = 0): the model describes a rise from zero as the dose is absorbed, not an immediate jump, which is why the peak is delayed rather than instantaneous.
2. Calculate the total drug exposure \(\int_0^8 C(t)\,dt\) and interpret it.
Read the worked answer
Use (\int t,e^{-at},dt = -\left(\dfrac{t}{a} + \dfrac{1}{a^2}\right)e^{-at}), with a = 0.35, so (1/a^2 = 8.1633):
At t = 8: (\left(\dfrac{8}{0.35} + 8.1633\right)e^{-2.8} = (22.857 + 8.1633)(0.060810) = 31.020 \times 0.060810 = 1.8863)
At t = 0: ((0 + 8.1633)(1) = 8.1633)
(\int_0^8 t,e^{-0.35t},dt = 8.1633 - 1.8863 = 6.2770)
(\int_0^8 C(t),dt = 12 \times 6.2770 = 75.3) mg·h/L.
Interpretation: about 75 mg·h/L of total exposure over the 8-hour window. This is the standard quantity for comparing formulations or doses - twice the dose gives twice the area if elimination stays linear - and it is not the same as the peak. A drug can have a safe peak but excessive total exposure, and here the area continues to accumulate well past the peak, since (C(t)) is still 5.8 mg/L at t = 8. The upper limit of the integral is also the limit of the model's stated validity.
3. Evaluate the 6-hour protocol using superposition, and advise the committee.
Read the worked answer
After a second dose given at t = 6, the total for (6 \le t \le 8) is
(C_{tot}(t) = 12t,e^{-0.35t} + 12(t-6)e^{-0.35(t-6)}).
At t = 6 the residual from the first dose alone is (12(6)e^{-2.1} = 8.82) mg/L - already most of the way to the limit before the second dose has been absorbed.
Sample values: t = 7 gives 7.25 + 8.46 = 15.7 mg/L; t = 8 gives 5.84 + 11.92 = 17.76 mg/L. The sum peaks near t ≈ 8.3 h at about 17.8 mg/L, above the 15 mg/L threshold.
So even with perfect timing the 6-hour schedule breaches the toxicity limit for roughly an hour around each second dose. Late night doses make it marginally worse in timing, not better: shifting the second dose later moves the peak but does not lower it, because the residual curve is already falling slowly. Repeating the calculation for other spacings, the peak stays under 15 mg/L only once the interval reaches about 9 hours, and even then the margin is under 0.2 mg/L - too thin to rely on.
Advice: do not continue 6-hourly dosing unmonitored. Options are extending the interval to 9–10 h, reducing the dose, or measuring a second-dose concentration to test the superposition assumption. That assumption is the weak point: the model is fitted to single-dose data over 0–8 h and assumes linear elimination, so accumulation predictions beyond that are extrapolation and should be verified before the protocol is changed.