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The rate (in $ mg $ $ carbon/m^3/h $) at which photosynthesis takes place for a species of phytoplankton is modeled by the function

$$ P = \dfrac{100I}{I^2 + I + 4} $$

where $ I $ is the light intensity (measured in thousands of foot-candles). For what light intensity is $ P $ a maximum?

$\mathrm{P}$ is maximum when $I=2$

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we're told the rate at which photosynthesis takes place for a species of phytoplankton is model model. By the function, P equals 100. I over I squared plus I plus four. You listen to our where I is The light intensity measured in thousands of foot candles were asked for what light intensity pea is a maximum. To answer this question, we want to differentiate p with respect to I Then find critical values upside down. Some bitch had a year on mhm. Right, Right. You're what you read those? So we have the derivative of p with respect to I promise I this is I squared. Plus I plus four bottom times that are the top 100. Yeah. Last the top 100 I comes the derivative, the bottom to I plus one all over the bottom squared. Plus I plus four squared. Now I can factor at 100 from the top. So and so this is equal to 100 times And then we have I squared minus two. I squared is negative. Eyes squared. We have positive. I Yeah, minus I zero I and then plus four. No, no, no. All over I squared plus I plus four squared. No clothes. Yeah. Question. We can factor this as negative 100 times. I minus two times I plus two over I squared. Plus I plus four squared. We set this equal to zero. This gives us the critical values I equals. Plus or minus two. You're scared. Comes being. Of course, you can't really have negative light intensity. So I is just positive to This is our only critical value. You grew up dreamy. She's dead on now. Right, Blue Ranger. Billy? Yes. Notice that. Come. Both Jamaica The rate p this approaches zero as the intensity I approaches infinity. So it follows that our function P has an absolute Yeah, maximum on the interval. 02 infinity. So we simply have to test the value of P at zero. This is zero and the value of P at two. Now the value P two is 200 over four plus two is six plus four is 10, which is 20 here. Yeah, so clearly this is greater than zero. And therefore it follows that P has an absolute maximum value at I equals two here. And this is in thousands of foot candles. Vast Ribery firebug. How would you