Single Variable Calculus, Chapter 4, 4.7, Section 4.7, Problem 8

Suppose that the equation $\displaystyle P = \frac{100 I}{I^2 + I + 4}$ represents the rate $\displaystyle \left( \frac{\text{in mg carbon}}{\frac{m^3}{h}} \right)$ at which photosyntehsis takes place for a species of Phytoplankton where $I$ is the light intensity (measured in thousands of foot candles). For what light intensity is $P$ a maximum?

$
\begin{equation}
\begin{aligned}
\text{if } P &= \frac{100 I}{I^2 + I +4} \quad \text{, then by using Quotient Rule}\\
\\
P' &= \frac{(I^2 + I 4) (100) - (100I)(2I +1)}{(I^2 + I +4)^2}
\end{aligned}
\end{equation}
$

when $P'=0$

$
\begin{equation}
\begin{aligned}
0 &= 100I^2 + \cancel{100I} + 400 - 200I^2 - \cancel{100I}\\
\\
0 &= -100I^2 + 400\\
\\
100I^2 &= 400\\
\\
I &= \sqrt{\frac{400}{100}}\\
\\
I &= 2 \quad \text{and} \quad I = -2
\end{aligned}
\end{equation}
$

If we evaluate $P$ with these values,


$
\begin{equation}
\begin{aligned}
\text{when } I & = 2 &&& \text{when } I & = -2\\
\\
P & = \frac{100(2)}{2^2+2+4} &&& P &= \frac{100(-2)}{(-2)^2-2+4}\\
\\
P &= 20 &&& P &= -5.20
\end{aligned}
\end{equation}
$

$P(2) > P(-2)$, therefore the light intensity that will make $P$ maximum is $I=2$ thousands of foot candles.