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lyudmila [28]
3 years ago
7

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?
Mathematics
1 answer:
Alexxx [7]3 years ago
4 0

Answer:

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

Step-by-step 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,

\frac{d(C(t))}{dt} = 8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)})

Equating the first derivative to zero, we get,

\frac{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\frac{e^{-0.4}}{e^{-0.6}} = \frac{0.6}{0.4}\\\\e^{0.2t} = 1.5\\\\t = \frac{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.

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