Peak drug concentration and the dosing window
A pharmacology class is modelling the plasma concentration of a drug after an oral dose. For t hours after the dose, t >= 0, the model is C(t) = 20t/(t^2 + 4) mg/L. A table of values gives C(1)=4.0, C(2)=5.0 and C(3)≈4.62, and one student claims the peak is at t=2 because that is the largest table entry. Another student differentiates, finds a critical point, but is unsure whether it is a maximum or whether the endpoints matter. The class must decide whether a dose is effective: the concentration has to be at least 3 mg/L for 4 hours. A modified formulation is proposed with C2(t) = 20t/(t^2 + 3). The safety rule for the modified formulation is that concentration must never exceed 5 mg/L. The class has to advise the formulator using the models, not just the table.
1. Determine the maximum concentration and when it occurs. Justify that it is a global maximum on t >= 0.
Read the worked answer
C(t)=20t/(t^2+4). Differentiate with the quotient rule:
C'(t)=[20(t^2+4)-20t(2t)]/(t^2+4)^2 = [80-20t^2]/(t^2+4)^2 = 20(4-t^2)/(t^2+4)^2.
For t>=0 the denominator is positive. C'(t)=0 when t=2. C'(t)>0 for 0<=t<2 and C'(t)<0 for t>2, so C increases then decreases. C(2)=40/8=5 mg/L. The endpoints give C(0)=0 and lim as t tends to infinity C(t)=0. Therefore the global maximum on t>=0 is 5 mg/L at t=2 h. The table happened to agree at t=2, but the derivative and sign analysis are what justify a global maximum; a table alone could miss the peak or mistake a local value for the global one.
2. Solve C(t) >= 3 and determine whether the concentration stays at least 3 mg/L for 4 hours.
Read the worked answer
Solve 20t/(t^2+4) >= 3. Since t^2+4>0, multiply:
20t >= 3t^2+12 0 >= 3t^2-20t+12 3t^2-20t+12 <= 0.
The roots are t=(20±√(400-144))/6=(20±16)/6, so t=2/3 and t=6. The quadratic opens upward, so it is <=0 between the roots: 2/3 <= t <= 6. The duration is 6-2/3=16/3≈5.33 h, greater than 4 h. Therefore the concentration is at least 3 mg/L for about 5.33 h, so the dose meets the 4-hour effectiveness requirement. The denominator check is essential; multiplying by t^2+4 is safe because it is always positive.
3. For the modified model C2(t)=20t/(t^2+3), a colleague says it has the same peak of 5 mg/L because the numerator is unchanged. Check this and decide whether C2 meets the safety rule that concentration must never exceed 5 mg/L.
Read the worked answer
For C2(t)=20t/(t^2+3),
C2'(t)=[20(t^2+3)-20t(2t)]/(t^2+3)^2 = [60-20t^2]/(t^2+3)^2 = 20(3-t^2)/(t^2+3)^2.
The maximum occurs at t=√3. Then C2(√3)=20√3/(3+3)=20√3/6=10√3/3≈5.77 mg/L. Since 5.77>5, C2 violates the rule 'never exceed 5 mg/L'. Keeping the same numerator does not keep the same peak, because the denominator changes both the time and the value of the maximum. The modified formulation is not safe under the stated rule.