(a)
At t = 0 we have:
\displaystyle P = \frac {100 e^0} {1 + 3e^0} + 40 = \frac {100} 4 + 40 = 25 + 40 = 65
(b)
We have using the quotient rule:
\begin{align*}\frac {\mathrm dP} {\mathrm dt} &= \frac {100e^{-0.1 t} \times (-0.1) (1 + 3e^{-0.9t}) - 100e^{-0.1t} \times 3e^{-0.9t} \times (-0.9)} {(1 + 3e^{-0.9t})^2} \\ & = \frac {-10e^{-0.1t} - 30e^{-0.1t-0.9t} + 90 \times 3 e^{-0.9t - 0.1t}} {(1 + 3e^{-0.9t})^2} \\ & = \frac {240 e^{-t} - 10e^{-0.1t}} {(1 + 3e^{-0.9t})^2}\end{align*}
(ci) At the maximum we have:
\displaystyle \frac {\mathrm dP} {\mathrm dt} = 0
So:
\displaystyle \frac {240 e^{-T} - 10e^{-0.1T}} {(1 + 3e^{-0.9T})^2} = 0
So:
240e^{-T} = 10e^{-0.1T}
giving:
24 = e^{0.9T}
So:
0.9T = \ln 24
Hence:
\displaystyle T = \frac {10} 9 \ln 24 = 3.53 \text { (2dp)}
(cii)
Substituting T = \frac {10} 9 \ln 24 gives P_T = 102.444\ldots = 102 \text { (0dp)}
(d)
As t \to \infty, 100e^{-0.1t} \to 0 and (1 + 3e^{-0.9t}) \to 1 so:
\displaystyle \frac {100 e^{-0.1t}} {1 + 3e^{-0.9t}} \to \frac 0 1 = 0
So P = \displaystyle \frac {100 e^{-0.1t}} {1 + 3e^{-0.9t}} + 40 \to 40, so k = 40.
