Respuesta :
Answer:
the maximum concentration of the antibiotic during the first 12 hours is 1.185 [tex]\mu g/mL[/tex] 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 [tex]\mu g/mL[/tex]
[tex]C(t) = 8(e^{(-0.4t)}-e^{(-0.6t)})[/tex]
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,
[tex]\frac{d(C(t))}{dt} = 8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)})[/tex]
Equating the first derivative to zero, we get,
[tex]\frac{d(C(t))}{dt} = 0\\\\8(-0.4e^{(-0.4t)}+ 0.6e^{(-0.6t)}) = 0[/tex]
Solving, we get,
[tex]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[/tex]
At t = 0
[tex]C(0) = 8(e^{(0)}-e^{(0)}) = 0[/tex]
At t = 2
[tex]C(2) = 8(e^{(-0.8)}-e^{(-1.2)}) = 1.185[/tex]
At t = 12
[tex]C(12) = 8(e^{(-4.8)}-e^{(-7.2)}) = 0.059[/tex]
Thus, the maximum concentration of the antibiotic during the first 12 hours is 1.185 [tex]\mu g/mL[/tex] at t= 2 hours.
