Organism A grows on substrate $S$ and produces product $P$, which is the only substrate that organism $B$ can utilize. The batch kinetics are
$$
\begin{aligned}
\frac{d X_{A}}{d t} &=\frac{\mu_{A} S X_{A}}{K_{s}+S} \\
\frac{d X_{B}}{d t} &=\frac{\mu_{B} P X_{B}}{K_{p}+P} \\
\frac{d P}{d t} &=Y_{P / A} \frac{\mu_{A} S X_{A}}{K_{S}+S}-\frac{\mu_{B} P X_{B}}{Y_{X_{B} / P}\left(K_{P}+P\right)} \\
\frac{d S}{d t} &=-\frac{\mu_{A} S X_{A}}{Y_{X_{A} / S}\left(K_{S}+S\right)}-\frac{Y_{P / A}}{Y_{P / S}} \frac{\mu_{A} S X_{A}}{\left(K_{S}+S\right)}
\end{aligned}
$$
Assume the following parameter values:
$$
\begin{aligned}
\mu_{A} &=0.18 \mathrm{hr}^{-1}, & K_{s} &=0.42 \mathrm{~g} / \mathrm{l}, & \mu_{B} &=0.29 \mathrm{hr}^{-1} \\
K_{P} &=0.30 \mathrm{~g} / \mathrm{l}, & Y_{X_{A} / s} &=0.3 \mathrm{~g} / \mathrm{g}, & Y_{X_{B} / P} &=0.5 \mathrm{~g} / \mathrm{g} \\
Y_{P / S} &=1.0 \mathrm{~g} / \mathrm{g}, & Y_{P / A} &=4.0 \mathrm{~g} / \mathrm{g}, & S_{0} &=10 \mathrm{~g} / \mathrm{l}
\end{aligned}
$$
Determine the behavior of these two organisms in a chemostat. Plot $S, P, X_{A}$, and $X_{B}$ versus dilution rate. Discuss what happens to organism $B$ as the dilution rate approaches the washout dilution rate for organism $A$. (Courtesy of $\mathrm{L}$. Erickson, from "Collected Coursework Problems in Biochemical Engineering," compiled by H. W. Blanch for $1977 \mathrm{Am}$. Soc. Eng. Educ.