Calculus application problem

Question

The current $C$ is given by a battery of $N$ similar voltaic cells is $C=nE/[R+(rn^2/N)]$, where $E,R, r$ are constants and $n$ is the number of cells coupled in series. Find the proportion of $n$ to $N$ for which the current is greatest. I am learning calculus from 'calculus made easy, by Thompson' . I got this problem there for practice. I don't know how to start with!

Answer 1

The problem is asking for the proportion $n/N$ so try to make some $n/N$'s appear in the problem. For instance, divide the top and bottom of your fraction by $N$ to get

$$C = \frac{ \frac{nE}{N}}{\frac{R}{N} + \frac{rn^2}{N^2}} $$

Let $x = n/N$ and you have

$$C=\frac{Ex}{R/N +rx^2}.$$

So maximize $C$ with respect to $x$.

Answer 2

Similar to B. Goddard's answer , you could do it directly.

For a given $N$, since $$C=\frac{n E}{R+\frac{n^2 r}{N}}= \frac{E n N}{n^2 r+N R}$$ differentiate with respect to $n$ to get $$C'=\frac{E N \left(N R-n^2 r\right)}{\left(n^2 r+N R\right)^2}=0 \implies n=\sqrt{\frac {N R}r}\implies \color{blue}{\frac n N=\sqrt{\frac R{N r}}}$$

Perform a sanity check to verify that this leads to a maximum value for $C$. We have $$C''=\frac{2 E n N r \left(n^2 r-3 N R\right)}{\left(n^2 r+N R\right)^3}$$ which, for $n=\sqrt{\frac {N R}r}$ gives $$C''= -\frac{E \sqrt{r}}{2 \sqrt{N} R^{3/2}} <0$$