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
