Optimization Rogers et al collected data on the photosynthetic rate of soybean l

Question

Optimization

Rogers et al collected data on the photosynthetic rate of soybean leaves over the course of one day. On day 215 in the soybean plant life cycle, the photosynthetic rate of the upper leaves of the plant can be approximated with the equation:

P(t) = -0.374t2 + 9.44t - 41.3

where t is measured in SECONDS and P(t) has units micromol m^-2 s^-1 . Find the maximum photosynthetic rate over the course of one day (in micromol m^-2 s^-1).

Note* Im assuming you have to convert to seconds per day but other than that im not sure

Explanation / Answer

P(t) = -0.374t2 + 9.44t - 41.3

for maximum rate dp/dt=0

dP/dt = -(0.374*2)t + 9.44 - 0 =0

==>t=9.44/(0.374*2)

==>t=12.62 sec

P(12.62) = -0.374(12.62)2 + 9.44*12.62 - 41.3

=18.27

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