Question

The rate (in mg carbon/m3/h) at which photosynthesis takes place for a species of phytoplankton is modeled by the function P = 90I I2 + I + 9 where I is the light intensity (measured in thousands of foot-candles). For what light intensity is P a maximum?

Answer 1

Answer:

At light intensity I = 3, is P a maximum

Explanation:

Given:

$P=\frac{90I}{I^2+I+9}$

now differentiating the above equation with respect to Intensity 'I' we get

$\frac{dp}{dI}=\frac{(I^2+I+9).\frac{d(90I)}{dI}-90I.\frac{d((I^2+I+9)}{dI}}{(I^2+I+9)^2}$

or

$\frac{dp}{dI}=\frac{(I^2+I+9).90-90I.(2I+1)}{(I^2+I+9)^2}$

or

$\frac{dp}{dI}=\frac{90I^2+90I+810)-(180I^2+90I)}{(I^2+I+9)^2}$

or

$\frac{dp}{dI}=\frac{-90I^2+810)}{(I^2+I+9)^2}$

Now for the maxima $\frac{dP}{dI}=0$

thus,

$0=\frac{-90I^2+810)}{(I^2+I+9)^2}$

or

$-90I^2+810=0$

or

$I^2=\frac{810}{90}$

or

$I^2=9$

or

I = 3

thus, for the value of intensity I = 3, the P is maximum

at I = 3

$P=\frac{90\times3}{3^2+3+9}$

or

$P=\frac{270}{21}$

or

$P=12.85$