The rate (in mg carbon/m^3/h) at which photosynthesis takes place for a species of phytoplankton is modeled by the function
P = (120I)/(I^2 + I + 4)
where I is the light intensity (measured in thousands of foot-candles). For what light intensity is P a maximum?
p = 120 i / (i^2 + i + 4)
dp/di = 0 at max or min
0 = [(i^2 + i + 4)120 - 120 i(2i+1)]/(i^2 + i + 4)2
look for zeros of numerator
0 = 120 i^2 +120 i + 480 -240 i^2 -120 i
0 = -120 i^2 +480
i^2 = 4
i = 2
To find the light intensity at which the rate of photosynthesis is a maximum (P is a maximum), we need to find the critical points of the function P(I).
To do this, we will find the derivative of the function P(I) with respect to I and set it equal to zero.
First, let's determine the derivative of P(I):
P'(I) = dP(I)/dI
Using the quotient rule of differentiation, the derivative becomes:
P'(I) = [(I^2 + I + 4)(120) - (120I)(2I + 1)] / (I^2 + I + 4)^2
Simplifying the expression, we have:
P'(I) = (120I^2 + 120I + 480 - 240I^2 - 120I) / (I^2 + I + 4)^2
P'(I) = (-120I^2 + 360) / (I^2 + I + 4)^2
Setting P'(I) equal to zero and solving for I:
-120I^2 + 360 = 0
-120I^2 = -360
I^2 = 360/120
I^2 = 3
I = √3 or I = -√3
Since light intensity cannot be negative, we discard I = -√3 and conclude that the light intensity at which the rate of photosynthesis is a maximum is I = √3.