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 + 4 where I is the light intensity (measured in thousands of foot-candles). For what light intensity is P a maximum? I = thousand foot-candle

P is maximum at I = 2

Step-by-step explanation:

We are given that;

P = 120I/(I² + I + 4)

Now, using differential coefficient, we have;

dP/dI = [(I² + I + 4)•(120I)' - (120I)•(I² + I + 4)']/(I² + I + 4)²

dP/dI = [(I² + I + 4)•(120) - (120I)•(2I + 1)]/(I² + I + 4)²

dP/dI = [(120I² + 120I + 480 - 240I² - 120I)]/(I² + I + 4)²

dP/dI = (-120I² + 480)/(I² + I + 4)²

dP/dI = -120(I² - 4)/(I² + I + 4)²

dP/dI = (-120(I + 2)(I - 2))/(I² + I + 4)²

At dP/dI = 0, I = 2 or - 2

Now,light intensity cannot be negative, thus we pick 2.

So there is local maxima at I = 2 because the sign of the derivative changes from positive to negative at that point.

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