MATH SOLVE

4 months ago

Q:
# 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?

Accepted Solution

A:

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.