The rate (In mg carbon/m3/h) at which photosynthesis takes place for a species of phytoplankton is modeled by the function 110I 12 +1+ 9 whe

Question

The rate (In mg carbon/m3/h) at which photosynthesis takes place for a species of phytoplankton is modeled by the function 110I 12 +1+ 9 where I is the light intensity (measured in thousands of foot-candles). For what light intensity is P a maximum?

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Thanh Hà 3 years 2021-08-08T12:36:08+00:00 1 Answers 789 views 0

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    2021-08-08T12:37:41+00:00

    Answer:

    P is maximum at I = 2

    Step-by-step explanation:

    Here is the complete question

    The rate (in mg carbon/m³/h) at which photosynthesis takes place for a species of phytoplankton is modeled by the function P = 100I/(I² + I + 4) where I is the light intensity (measured in thousands of foot candles). For what light intensity P is a maximum?

    To find the value of I at which P is maximum, we differentiate P with respect to I and equate it to zero.

    So, dP/dI =  d[100I/(I² + I + 4)]/dI

    = [(I² + I + 4)d(100I)/dI – 100Id(I² + I + 4)/dI]/(I² + I + 4)²

    = [(I² + I + 4)100 – 100I(2I + 1)]/(I² + I + 4)²

    = [100I² + 100I + 400 – 200I² – 100I]/(I² + I + 4)²

    = [-100I² + 400]/(I² + I + 4)²

    =  -100[I² – 4]/(I² + I + 4)²

    Since dP/dI = 0,  -100[I² – 4]/(I² + I + 4)² = 0 ⇒ I² – 4 = 0 ⇒ I² = 4 ⇒ I = ±√4

    I = ±2

    Since I cannot be negative, we ignore the minus sign

    To determine if this is a maximum point, we differentiate dP/dI. So,

    d(dP/dI)/dI = d²P/dI² = d[-100[I² – 4]/(I² + I + 4)²]/dI

    = [(I² + I + 4)²d(-100[I² – 4])/dI – (-100[I² – 4])d(I² + I + 4)²/dt]/[(I² + I + 4)²]²

    = [(I² + I + 4)²(-200I) + 100[I² – 4]) × (2I + 1) × 2(I² + I + 4)]/(I² + I + 4)⁴

    = [-200I(I² + I + 4)² + 200[I² – 4])(2I + 1)(I² + I + 4)]/(I² + I + 4)⁴

    = [-200(I² + I + 4)[I(I² + I + 4) – [I² – 4])(2I + 1)]]/(I² + I + 4)⁴

    = [-200(I² + I + 4)[I³ + I² + 4I – I² + 4])(2I + 1)]]/(I² + I + 4)⁴

    = [-200(I² + I + 4)[I³ + 4I + 8])(2I + 1)]]/(I² + I + 4)⁴

    Substituting I = 2 into d²P/dI², we have

    = [-200(2² + 2 + 4)[2³ + 4(2) + 8])(2(2) + 1)]]/(2² + 2 + 4)⁴

    = [-200(4 + 2 + 4)[8 + 8 + 8])(4 + 1)]]/(4 + 2 + 4)⁴

    = [-200(10)[24](5)]]/(10)⁴

    = -240000/10⁴

    = -24

    Since d²P/dI² = -24 < 0 at I = 2,  this shows that it I = 2 is a maximum point.

    So, P is maximum at I = 2

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