MHB S8. 3.7.10 at which photosynthesis takes place (min/max)

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The discussion focuses on determining the maximum rate of photosynthesis for a phytoplankton species based on the light intensity function P. The derivative P' is calculated, revealing critical points at I = ±2. However, the relevance of I = -2 is questioned since light intensity cannot be negative. The conversation emphasizes the need to confirm whether I = 2 indeed represents a maximum for the function. Ultimately, the viability of negative intensity values is a recurring concern in the analysis.
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10. The rate (in mg carbon$/m^3/h$) at which photosynthesis takes place for a species of phytoplankton is modeled by the function

$P=\dfrac{100I}{I^2 + I + 4} $
where I is the light intensity (measured in thousands of foot-candles). For what light intensity is $P$ a maximum?

OK I presume first we derive P'

$P'=\dfrac{ -100 (I^2 - 4)}{(I^2 + I + 4)}^2$

$P'(I)=0 $ at $x=\pm 2$

so far !
 
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$P’=\dfrac{100(4-I^2)}{(I^2+I+4)^2}$

does $I =-2$ make sense given what $I$ represents?

So, how do you determine if $I=2$ determines a maximum?
 
Can Intensity be negative?

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Can Intensity be negative?

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Can Intensity be negative?
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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