Question

After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modelled by the function C(t)=8(e^(-0.4t)-e^(-0.6t)) where the time t is measured in hours and C is measured in mewg/mL.

What is the maximum concentration of the antibiotic during the first 12 hours?

Answer 1

the maximum concentration of the antibiotic during the first 12 hours is 1.185
`\mu g/mL` at t= 2 hours.

Explanation:

We are given the following information:

After an antibiotic tablet is taken, the concentration of the antibiotic in the bloodstream is modeled by the function where the time t is measured in hours and C is measured in
`\mu g/mL`

`C(t) = 8(e^((-0.4t))-e^((-0.6t)))`

Thus, we are given the time interval [0,12] for t.

• We can apply the first derivative test, to know the absolute maximum value because we have a closed interval for t.
• The first derivative test focusing on a particular point. If the function switches or changes from increasing to decreasing at the point, then the function will achieve a highest value at that point.

First, we differentiate C(t) with respect to t, to get,

`(d(C(t)))/(dt) = 8(-0.4e^((-0.4t))+ 0.6e^((-0.6t)))`

Equating the first derivative to zero, we get,

`(d(C(t)))/(dt) = 0\\\\8(-0.4e^((-0.4t))+ 0.6e^((-0.6t))) = 0`

Solving, we get,

`8(-0.4e^((-0.4t))+ 0.6e^((-0.6t))) = 0\\\displaystyle(e^(-0.4))/(e^(-0.6)) = (0.6)/(0.4)\\\\e^(0.2t) = 1.5\\\\t = (ln(1.5))/(0.2)\\\\t \approx 2`

At t = 0

`C(0) = 8(e^((0))-e^((0))) = 0`

At t = 2

`C(2) = 8(e^((-0.8))-e^((-1.2))) = 1.185`

At t = 12

`C(12) = 8(e^((-4.8))-e^((-7.2))) = 0.059`

Thus, the maximum concentration of the antibiotic during the first 12 hours is 1.185
`\mu g/mL` at t= 2 hours.