Drug Dosing Dilemma

Calculus Level 4

A doctor has prescribed an ingestible painkiller for a patient after intensive surgery. The hospital brings you an information sheet before they work on another patient who needs intensive care. However, the printer was having problems and only printed one sheet containing the composition, statistics, and kinetics of the drug.

The drug is an opaque, liquid-filled capsule that ruptures open once the outer layer is dissolved. For the sake of this problem, we assume the drug starts working immediately after ingestion.

Inside the capsule, there is 10 mL 10\text{ mL} of the solution. Assuming the solution has the same density 10 mg/mL , 10\text{ mg/mL}, the actual drug content is 60% by mass.

In the pharmacokinetics section, it lists that the first-order half-life of this drug in the body is 2 hours.

In the statistics section are the following:

  • The drug loses its effectiveness once the amount of the drug in the body is equal to or below 10 mg 10\text{ mg} on average.
  • At any given time, if there is more than 100 mg 100\text{ mg} of the drug in the body, there is an increased chance of side effects.
  • At around 80 mg 80\text{ mg} in the body, it has maximum effectiveness.

Assume you take 1 capsule per dose. How many (integer) hours in between each dose should you wait in order for the drug to achieve maximum effectiveness?

( ( Hint: The maximum amount after each dose should converge to 80 mg . ) 80\text{ mg}.)

2 4 6 8

David Qiu by David Qiu , 24, USA

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1 solution

Hosam Hajjir
Oct 15, 2017

Let T be the waiting time between doses. And let t be measured from the time of

the first dose. We're interested in discrete values of t which are multiples of T.

From the decay formula, y ( N T ) y(N T) is given by,

y ( N T ) = ∑ n = 0 N A ( 1 2 ) ( N T − n T ) / 2 y(NT) = \sum_{n=0}^{N} A (\frac{1}{2})^{(NT - nT)/2}

where A is the content of the drug in one dose which is, A = 0.6 ∗ 100 = 60 A = 0.6 *100 = 60 mg.

Hence,

y ( N T ) = A ∑ n = 0 N ( 1 2 ) ( N − n ) T / 2 y(NT) = A \sum_{n=0}^{N} (\frac{1}{2})^{(N-n)T/2}

let k = N − n k = N - n , and the summation becomes,

y ( N T ) = A ∑ k = 0 N ( 1 2 ) k T / 2 y(NT) = A \sum_{k=0}^{N} (\frac{1}{2})^ {kT/2}

The sum is the sum of a geometric series, and is given by,

y ( N T ) = A ( 1 − ( 1 2 ) ( N + 1 ) T / 2 ) ( 1 − ( 1 / 2 ) T / 2 ) y(NT) = A \dfrac{(1 - (\frac{1}{2})^{(N+1)T/2} )}{(1 - (1/2)^{T/2} ) }

Therefore, for large N, the above becomes,

y ( N T ) = A ( 1 − ( 1 2 ) ( T / 2 ) ) y(NT) = \dfrac{A}{(1 - (\frac{1}{2})^{(T/2)} )}

Since y ( N T ) y(NT) must be 80, then

80 = 60 ( 1 − ( 1 2 ) ( T / 2 ) ) 80 = \dfrac{60}{(1 - (\frac{1}{2})^{(T/2)} )}

from which,

1 − ( 1 2 ) ( T / 2 ) = 60 / 80 = 0.75 1 - (\frac{1}{2})^{(T/2)} = 60/80 = 0.75

and,

( 1 2 ) ( T / 2 ) = 0.25 (\frac{1}{2})^{(T/2)} = 0.25

Hence T / 2 = 2 T/2 = 2 , and T = 4 T = 4

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