\( 3 / 4 \) Han is prescribed a course of antibiotics for an infection. He is told to take a 150 mg dose of the antibiotic regularly every 12 hours for 15 days. Han is curious about the antibiotic and learns that at the end of the 12 hours, only \( 5 \% \) of the dose is still in his body. When will the total amount of the antibiotic in Han be highest over the course of the 15 day treatment? After dose 5 After dose 30 After dose 1 After dose 15

Let \( A_n \) denote the amount of antibiotic in Han’s body immediately after the \( n^\text{th} \) dose. Each dose is 150 mg and every 12 hours the previous amount is reduced to \( 5\% \) (i.e. multiplied by 0.05). Thus, the amount just after the next dose is given by the recurrence \[ A_{n+1} = 150 + 0.05\, A_n, \] with the initial condition \[ A_1 = 150. \] This is a linear recurrence with a constant coefficient. Its steady‐state (maximum) value, as \( n \) becomes large, can be found by solving \[ A = 150 + 0.05\, A. \] Subtract \( 0.05\, A \) from both sides: \[ A - 0.05\, A = 150, \] \[ 0.95\, A = 150, \] \[ A = \frac{150}{0.95} \approx 157.89. \] The solution of the recurrence is \[ A_n = \frac{150}{0.95}\Bigl(1 - (0.05)^n\Bigr). \] This formula tells us that as \( n \) increases, \( (0.05)^n \) becomes very small and \( A_n \) approaches \( \frac{150}{0.95} \). Since Han takes a dose every 12 hours for 15 days, he receives a total of \[ 15 \text{ days} \times 2 \text{ doses per day} = 30 \text{ doses}. \] Because \( A_n \) is strictly increasing with \( n \) (each dose increases the total amount, approaching the steady state), the highest amount in his body occurs after the last (30th) dose. Thus, the total amount of the antibiotic in Han will be highest after dose 30.