- Topic
- Optimization
- Difficulty
- Introductory
- Suggested time
- 3 min
- Marks
- 2
Optimization practice question 4
The concentration of a drug in a patient’s blood hours after a dose is modelled by the function below, in mg/L. When does the concentration peak?
Give the time in hours.
At the peak the concentration stops rising and starts to fall. What is at that instant?
Differentiate with the product rule, then factor out . An exponential is never zero, so only the other factor can vanish.
An exact expression, or a decimal to two decimal places.
Theory UBC Math 100 §8CLP-1 §3.5Key results
Worked solution
Strategy
The peak is where C′ changes from positive to negative. Differentiate with the product rule, factor out the exponential, and check the sign of C′ on either side.
Working
Product rule, with the chain rule giving the factor −0.3.
The exponential is never zero, so only the bracket can vanish.
The level rises and then falls, so this is the global maximum for t ≥ 0.
The peak level itself, for reference.
Answer
The answer is 10/3 hours, that is 3 hours 20 minutes after the dose.
Check
C(3) ≈ 18.30 and C(3.5) ≈ 18.37 mg/L both fall short of C(10/3) = 50/e ≈ 18.39 mg/L.
Where this goes wrong
Solving 1 − 0.3t = 0 as t = 0.3, or answering with the peak concentration 50/e instead of the time. For any model C(t) = At e−kt the peak comes at t = 1/k, whatever A is.
Leans on
The product rule with an exponential, from The chain rule and logarithms
Try a similar question
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