Question

The rate (in mg carbon/m3/h) at which photosynthesis takes place for a species of phytoplankton is modeled by the function P = 120I I2 + I + 1 where I is the light intensity (measured in thousands of foot-candles). For what light intensity is P a maximum?

Answer 1

Answer:

Light intensity = L =1

Step-by-step explanation:

The given function is:

$P = \frac{120L}{L^{2}+L+1 }$                    (Equation.1)

Taking derivative of the whole equation:

$\frac{dP}{dL}$ =$\frac{d}{dL}$ $(\frac{120L}{L^{2}+L+1 } )$

$\frac{dP}{dL}$ =120 $\frac{1 (L^{2}+L+1) - (\frac{d}{dL}L^{2} +\frac{d}{dL} L+\frac{d}{dL}1)L}{(L^{2}+L+1)^{2} }$

$\frac{dP}{dL}$ =$\frac{120 (L^{2}-1)}{L^{2}+L+1 }$

$\frac{dP}{dL}$ =$\frac{120 (L+1)(L-1)}{L^{2}+L+1 }$

For first derivative test, there is a maxima. for that , we need to take:

L+1 =0 and L-1 =0

we get:

L= -1

or L=1

Since light intensity cannot be negative so, L=1

put this in equation 1:

$P(1) = \frac{120}{1^{2}+1+1 }$

P = $\frac{120}{3}$

P= 40  which is the maximum value of P (At L=1)