Diluted Dose with a Hidden Domain Restriction
A pharmacist models the concentration of a drug in a patient's bloodstream using C(t) = (12t)/(t^2 + 4) for t ≥ 0, where C is measured in mg/L and t in hours after administration. A colleague proposes a simplified model C2(t) = 3t/(t + 1) after arguing that for large t the two formulas behave similarly. A safety threshold is set at C = 2 mg/L: the drug is considered unsafe whenever concentration reaches or exceeds this value. The pharmacist must determine whether the simplified model would ever flag the drug as unsafe, and whether the two models agree on the maximum concentration and the time it occurs.
1. Determine the maximum value of C(t) on its stated domain and the time at which it occurs, giving exact values.
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Differentiate using the quotient rule. C(t) = 12t/(t^2+4), so C'(t) = [12(t^2+4) - 12t(2t)]/(t^2+4)^2 = [12t^2 + 48 - 24t^2]/(t^2+4)^2 = (48 - 12t^2)/(t^2+4)^2.
Setting C'(t)=0 gives 48 = 12t^2, so t^2 = 4, t = 2 (reject t = -2 since t ≥ 0).
At t = 2: C(2) = 24/(4+4) = 24/8 = 3 mg/L.
Check endpoints: C(0) = 0 and as t → ∞, C(t) → 0. Since C'(t) > 0 on (0,2) and C'(t) < 0 on (2,∞), t = 2 is a global maximum on the domain.
Maximum is 3 mg/L at t = 2 hours.
2. Analyse the simplified model C2(t) = 3t/(t+1) for t ≥ 0: state its range, whether it ever reaches the 2 mg/L threshold, and compare its peak behaviour with the original model.
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For C2(t) = 3t/(t+1) = 3 - 3/(t+1), the derivative is C2'(t) = 3/(t+1)^2 > 0 for all t ≥ 0, so C2 is strictly increasing. C2(0) = 0 and lim(t→∞) C2(t) = 3, so the range is [0, 3).
Solving C2(t) = 2: 3t = 2(t+1), so 3t = 2t + 2, t = 2. Thus C2 reaches 2 mg/L at t = 2 hours and exceeds it for all t > 2, approaching 3 but never reaching it.
Comparison: both models share the same limiting value 3, but the original model attains 3 as a finite maximum at t = 2 and then decays, whereas the simplified model rises monotonically toward 3 as an asymptote. The simplified model would flag the drug as unsafe for all t > 2, while the original model flags it only briefly (C ≥ 2 between roughly t ≈ 0.76 and t ≈ 5.24, symmetric about t = 2). The simplification is therefore unsound for safety decisions despite matching long-run limits.
3. Without solving a full equation, explain how the symmetry of C(t) affects the interval during which the drug is unsafe, and state how many times C(t) = 2.
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The function C(t) = 12t/(t^2+4) satisfies C(1/t)·? - more directly, note C(t) = C(4/t) for t > 0: substituting 4/t gives 12(4/t)/((16/t^2)+4) = (48/t)/((16+4t^2)/t^2) = 48t/(16+4t^2) = 12t/(4+t^2) = C(t). So the graph is symmetric under t ↦ 4/t, which fixes t = 2 (since t = 4/t gives t = 2).
Because the maximum is at t = 2 and the curve is symmetric about that reciprocal pairing, any horizontal line C = k with 0 < k < 3 meets the curve at two points whose product is 4. So C(t) = 2 has exactly two positive solutions, and the unsafe interval is centred (in the multiplicative sense) on t = 2. This confirms the drug is unsafe only for a bounded window rather than indefinitely, unlike the monotonic simplified model.