Accumulating Drug Doses
A hospital prescribes a drug with a 6-hour dosing interval. Each dose is 40 mg, and after 6 hours 25% of the drug present just after the previous dose remains in the bloodstream. Let a_n be the amount in mg immediately after the nth dose, so a_1=40 and a_n=0.25a_{n−1}+40 for n≥2. The clinical safety cutoff is 60 mg; a nurse also wants to keep the peak below 50 mg if possible. A pharmacist claims that because each new 40 mg dose is below 60, the amount can never exceed 60; another claims the amount will eventually stabilize at 160/3 mg. The ward needs to know whether the 50 mg target is ever crossed, whether 60 mg is ever reached, and what constant dose would give a steady-state peak of exactly 60 mg. The drug is not cleared completely between doses, and the 25% retention rate is assumed constant.
1. Solve the recurrence for a_n, confirm or refute the pharmacist's steady-state value, and decide whether the 60 mg safety cutoff is ever reached.
Read the worked answer
At a fixed point L, L=0.25L+40, so 0.75L=40 and L=160/3≈53.33 mg. The closed form is
a_n=L+(a_1−L)(1/4)^{n−1}=160/3−(40/3)(1/4)^{n−1}.
Since 0<1/4<1, a_n increases to 160/3. Because 160/3<60, the amount never reaches 60 mg. The pharmacist's steady-state value is correct. The reasoning 'each dose is below 60' is not by itself a proof, because repeated doses can accumulate above a single dose; the correct reason is that the limiting value is below 60.
2. Determine whether the peak ever exceeds the nurse's 50 mg target; if so, after which dose does it first happen?
Read the worked answer
Solve a_n>50:
160/3−(40/3)(1/4)^{n−1}>50 ⇒ 160−40(1/4)^{n−1}>150 ⇒ (1/4)^{n−1}<1/4 ⇒ n−1>1 ⇒ n>2.
At n=2, a_2=0.25·40+40=50 exactly; at n=3, a_3=0.25·50+40=52.5. Thus the level is exactly 50 mg after the second dose and exceeds 50 mg from the third dose onward. If the target means strictly below 50 mg, it is not satisfied from the third dose.
3. What constant dose d would make the steady-state peak exactly 60 mg? Comment on whether a finite number of doses reaches exactly 60 mg.
Read the worked answer
The steady-state limit is
L=d/(1−0.25)=4d/3.
Set 4d/3=60, giving d=45 mg. With d=45 and a_1=45,
a_n=60+(45−60)(1/4)^{n−1}=60−15(1/4)^{n−1}.
This approaches 60 from below but never equals 60 at any finite n. Thus 'steady-state peak exactly 60' means the asymptotic limit is 60, while all finite peaks remain below 60.