The rate (in mg carbon/m3/h) at which photosynthesis takes place for a species of phytoplankton is modeled by the function
P = 110I / ( I^2 + I + 4 )
where I is the light intensity (measured in thousands of foot-candles). For what light intensity is P a maximum?
I'm not sure how to go about this problem.
The steps we learned in class are:
Make a diagram (if needed)
determine objective function
determine constraints
substitute constraints into objective function
differentiate objective function
find critical points
check with Second derivative test
but I'm not sure how to apply these steps to this problem. Could someone please show me how to do this problem? I'd really appreciate it! Thanks!
P = 110I / ( I^2 + I + 4 )
where I is the light intensity (measured in thousands of foot-candles). For what light intensity is P a maximum?
I'm not sure how to go about this problem.
The steps we learned in class are:
Make a diagram (if needed)
determine objective function
determine constraints
substitute constraints into objective function
differentiate objective function
find critical points
check with Second derivative test
but I'm not sure how to apply these steps to this problem. Could someone please show me how to do this problem? I'd really appreciate it! Thanks!
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derive the first equation with respect to x to get
dP/dx = (-110(x^2-4))/(x^2+x+4)^2
then we find where dP/dx = 0, which is at -2 and 2
then we find where the graph of the derivative peaks
at any number less than -2, you get a negative sign
between -2 and 2, you get a positive sign
at any number above 2, you get a negative sign
the graph peaks when x = 2, and therefore when P = 22
I substituted the I for x since its easier to work with
so, we now know that the light intensity must be 2 in order to obtain a maximum
dP/dx = (-110(x^2-4))/(x^2+x+4)^2
then we find where dP/dx = 0, which is at -2 and 2
then we find where the graph of the derivative peaks
at any number less than -2, you get a negative sign
between -2 and 2, you get a positive sign
at any number above 2, you get a negative sign
the graph peaks when x = 2, and therefore when P = 22
I substituted the I for x since its easier to work with
so, we now know that the light intensity must be 2 in order to obtain a maximum