Answer
The maximum concentration of the antibiotic during the first 12 hours is $1.18~\mu g/mL$
Work Step by Step
We can find the value of $t$ when $C'(t) = 0$:
$C(t) =8(e^{-0.4t}-e^{-0.6t})$
$C'(t) =8(-0.4e^{-0.4t}+0.6e^{-0.6t}) = 0$
$-0.4e^{-0.4t}+0.6e^{-0.6t} = 0$
$0.6e^{-0.6t} = 0.4e^{-0.4t}$
$\frac{e^{-0.4t}}{e^{-0.6t}}=\frac{0.6}{0.4}$
$e^{0.2t}=1.5$
$0.2t=ln(1.5)$
$t = \frac{ln(1.5)}{0.2}$
$t = 2.027$
We can find the value of $C(2.027)$:
$C(t) =8(e^{-0.4t}-e^{-0.6t})$
$C(2.027) =8[e^{(-0.4)(2.027)}-e^{(-0.6)(2.027)}]$
$C(2.027) =8(e^{-0.8108}-e^{-1.2162})$
$C(2.027) =8(0.148)$
$C(2.027) = 1.18$
Note that $C(0) = 0$
Also, $C(12) =8[e^{(-0.4)(12)}-e^{(-0.6)(12)}] = 0.06$
The maximum concentration of the antibiotic during the first 12 hours is $1.18~\mu g/mL$
