The rate (in mg carbon/m3/h) at which photosynthesis takes place for a species of phytoplankton is modeled by the function
P =
110I
I2 + I + 4
where I is the light intensity (measured in thousands of foot-candles). For what light intensity is P a maximum?
To find the light intensity at which the rate of photosynthesis is a maximum, we need to determine the critical points of the function. In other words, we need to find the values of I for which the derivative of P with respect to I is equal to zero.
Let's calculate the derivative of P with respect to I using the quotient rule. The quotient rule states that if you have a function of the form f(x) = u(x) / v(x), then the derivative of f(x) with respect to x is (u'(x)v(x) - v'(x)u(x)) / v^2(x).
In this case, u(x) is 110I and v(x) is I^2 + I + 4. So, we have:
P = (110I) / (I^2 + I + 4)
To find P', the derivative of P with respect to I, we can use the quotient rule:
P' = [(110)(I^2 + I + 4) - (110I)(2I + 1)] / (I^2 + I + 4)^2
Now, we set P' equal to zero and solve for I:
[(110)(I^2 + I + 4) - (110I)(2I + 1)] / (I^2 + I + 4)^2 = 0
Simplifying the equation:
[(110)(I^2 + I + 4) - (220I^2 + 110I)] / (I^2 + I + 4)^2 = 0
[110I^2 + 110I + 440 - 220I^2 - 110I] / (I^2 + I + 4)^2 = 0
[-110I^2] / (I^2 + I + 4)^2 = 0
Since the numerator is zero, we must have -110I^2 = 0. This means that I = 0 or I cannot be determined.
Therefore, the maximum light intensity at which photosynthesis occurs cannot be determined from the given function.