Chapter 4.6, Problem 8E

The rate (in mg carbon/m³/h) at which photosynthesis takes place for a species of phytoplankton is modeled by the function

$P=\frac{100I}{I^2+I+4}$

where I is the light intensity (measured in thousands of foot-candles). For what light intensity is P a maximum?

To determine

To find: The light intensity for which P is maximum.

Answer to Problem 8E

The light intensity is maximum at $I=2$.

Explanation of Solution

Given:

Given model: $P=\frac{100I}{I^2+I+4}$.

Calculation:

Differentiate with respect to I,

$\begin{array}{c}\frac{dP}{dI}=\frac{\left(I^2+I+4\right)\times 100-100I\times \left(2I+1\right)}{{\left(I^2+I+4\right)}^2}\\ =\frac{-100I^2+400}{{\left(I^2+I+4\right)}^2}\end{array}$

For critical points, set $\frac{dP}{dI}=0$

$\begin{array}{c}\frac{-100I^2+400}{{\left(I^2+I+4\right)}^2}=0\\ -100\cdot I^2+400=0\\ I^2-4=0\\ I^2=4\end{array}$

Then, $I=\pm 2$.

Since the light intensity can never be negative, reject $I=-2$.

Thus, a possible light intensity for the maximum P occurrence is at $I=2$.

Again differentiate $\frac{dP}{dI}$ with respect to I,

$\begin{array}{c}\frac{d^{2}P}{d{I}^2}=\frac{{\left(I^2+I+4\right)}^2\times \left(-200I\right)-\left(-100\cdot I^2+400\right)\times 2\times \left(I^2+I+4\right)\times \left(2I+1\right)}{{\left(I^2+I+4\right)}^4}\\ =\frac{\left(I^2+I+4\right)\times (-200I)-(-100I^2+400)\left(4I+2\right)}{{\left(I^2+I+4\right)}^3}\\ =\frac{200I^3-2400\cdot I-800}{{\left(I^2+I+4\right)}^3}\end{array}$

When $I=2$, the value of $\frac{d^{2}p}{d{I}^2}$ becomes,

$\begin{array}{c}\frac{d^{2}p}{d{I}^2}=\frac{200\cdot 2^3-2400\cdot 2-800}{{\left(2^2+2+4\right)}^3}\\ =\frac{1600-4800-800}{{10}^3}\\ =\frac{-4000}{1000}\end{array}$

Therefore, $\frac{d^{2}p}{d{I}^2}=-4<0$.

Thus, by second derivative test we have a maximum at $I=2$.