Problem+8+Sophia

Excellent work!!(ms sweeney)

Problem 8 The rate (in mg carbon/m^3/h) at which photosynthesis takes place for a species of phytoplankton is modeled by the function P = 100 I / (I^2+I+4) where I is the light intensity (measured in thousands of foot-candles). For what light intensity is P a maximum?

Because the function given is already in terms of just one variable, I, we can just take the derivative and set it equal to 0 to find the critical values.

P' = (I^2 + I + 4)(100)-(100I)(2I + 1) / (I^2 + I + 4)^2 P' = 100I^2 + 100I + 400 - (200I^2 + 100I)/ (I^2 + I + 4)^2 0 = -100I^2 + 400/ (I^2+I+4)^2 0 = -100I^2+400 100I^2 = 400 I^2=4 I = + - 2 I = 2 (cant have negative light intensity) P is at a maximum when I is at a light intensity of 2

Another thing is that if you take the limit of the function as x goes to infinite, the limit is zero. If you take the limit of the function as x approaches 0 (light intensity cannot be negative) then the answer is zero. Both of these limits confirm that at I=2 there is a maximum, because they are both smaller than 20 (the value of P at zero).