Science:Math Exam Resources/Courses/MATH102/December 2016/Question B 07 (b)/Solution 1

To maximize the net revenue $I(N)$ on the domain $\mathbb {N} \cup \{0\}$, we first find its derivative.

Recall that $I(N)=2000(1-e^{-kN})F-1000N$ from part (a).

Then, we have $I'(N)=(2000(1-e^{-kN})F-1000N)'=2000F(1-e^{-kN})'-1000=2000Fke^{-kN}-1000$.

Since $2000Fke^{-kN}-1000=0$ at $N=500\ln \left({\frac {F}{250}}\right)=N_{0}$, we have $I'(N)>0$ for $N<N_{0}$ and $I'(N)<0$ for $N>N_{0}$. Therefore, $I(N)$ achieves its maximum at $N=N_{0}$, provided that $N_{0}\geq 0\iff F\geq 250$.

Answer: $\color {blue}N=500\ln \left({\frac {F}{250}}\right),F\geq 250$