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 + 1 where I is the light intensity (measured in thousands of foot-candles). For what light intensity is P a maximum?
Accepted Solution
A:
Answer:Light intensity = L =1 Step-by-step explanation:The given function is:[tex]P = \frac{120L}{L^{2}+L+1 }[/tex] (Equation.1)Taking derivative of the whole equation:[tex]\frac{dP}{dL}[/tex] =[tex]\frac{d}{dL}[/tex] [tex](\frac{120L}{L^{2}+L+1 } )[/tex][tex]\frac{dP}{dL}[/tex] =120 [tex]\frac{1 (L^{2}+L+1) - (\frac{d}{dL}L^{2} +\frac{d}{dL} L+\frac{d}{dL}1)L}{(L^{2}+L+1)^{2} }[/tex][tex]\frac{dP}{dL}[/tex] =[tex]\frac{120 (L^{2}-1)}{L^{2}+L+1 }[/tex][tex]\frac{dP}{dL}[/tex] =[tex]\frac{120 (L+1)(L-1)}{L^{2}+L+1 }[/tex]For first derivative test, there is a maxima. for that , we need to take: L+1 =0 and L-1 =0we get:L= -1or L=1Since light intensity cannot be negative so, L=1put this in equation 1:[tex]P(1) = \frac{120}{1^{2}+1+1 }[/tex]P = [tex]\frac{120}{3}[/tex]P= 40 which is the maximum value of P (At L=1)