The rate (in mg carbon/m3/h) at which photosynthesis takes place for a species of phytoplankton is modeled by the function below, where I is the light intensity (measured in thousands of foot-candles).

P = (120 I) / (I^2 + I + 1)
For what light intensity is P a maximum?

To find the light intensity at which P is a maximum, we need to first find the derivative of the function P with respect to I and then set it equal to zero. Let's start by finding the derivative.

P = (120 I) / (I^2 + I + 1)

To find the derivative, we can use the quotient rule:

dP/dI = [(120)(I^2 + I + 1) - (120 I)(2I + 1)] / (I^2 + I + 1)^2

Now, let's simplify the numerator:

dP/dI = (120I^2 + 120I + 120 - 240I^2 - 120I) / (I^2 + I + 1)^2

Combine like terms:

dP/dI = (-120I^2 - 120I + 120) / (I^2 + I + 1)^2

Now, let's set the derivative equal to zero and solve for I:

(-120I^2 - 120I + 120) / (I^2 + I + 1)^2 = 0

Since the numerator is equal to zero, we can ignore the denominator:

-120I^2 - 120I + 120 = 0

Divide by -120 to simplify the equation:

I^2 + I - 1 = 0

Now, we can solve this quadratic equation using the quadratic formula:

I = (-1 ± sqrt(1^2 - 4(1)(-1))) / (2(1))

I = (-1 ± sqrt(1 + 4)) / 2

I = (-1 ± sqrt(5)) / 2

Therefore, the light intensity at which the rate of photosynthesis is a maximum is given by:

I = (-1 + sqrt(5)) / 2 or I = (-1 - sqrt(5)) / 2

To find the light intensity at which P is a maximum, we need to find the critical points of the function. The critical points occur when the derivative of the function is equal to zero or undefined.

First, let's find the derivative of P with respect to I.

To simplify the calculations, let's rewrite P = (120 I) / (I^2 + I + 1) as P = 120 I (I^2 + I + 1)^-1.

Now, we need to use the quotient rule to find the derivative of P.

Let Q = I^2 + I + 1.

By applying the quotient rule:
P' = (120Q - 120I(Q')) / Q^2, where Q' represents the derivative of Q.

Calculating the derivative of Q:
Q' = (2I + 1) + 1 = 2I + 2.

Substituting Q' back into the expression for P':
P' = (120Q - 120I(2I + 2)) / Q^2
P' = (120Q - 240I^2 - 240I) / Q^2.

Now, let's set P' equal to zero and solve for I:
0 = (120Q - 240I^2 - 240I) / Q^2.

To simplify the expression, let's multiply both sides by Q^2:
0 = 120Q - 240I^2 - 240I.

Next, rearrange the equation:
240I^2 + 240I - 120Q = 0.

Divide the entire equation by 120 to simplify it:
2I^2 + 2I - Q = 0.

Now, we can use the quadratic formula to solve for I:
I = (-b ± √(b^2 - 4ac)) / (2a),
where a = 2, b = 2, and c = -Q.

Plugging in the values:
I = (-2 ± √(2^2 - 4(2)(-Q))) / (2(2))
I = (-2 ± √(4 + 8Q)) / 4
I = (-1 ± √(1 + 2Q)) / 2.

Therefore, the light intensity at which P is a maximum occurs when I = (-1 ± √(1 + 2Q)) / 2.

Note: To find the value of Q (I^2 + I + 1) at this light intensity, substitute the expression for I into Q.

I=3

p = (120i)/(i^2 + i + 1)

dp/di = 120[(i^2+i+1)-i(2i+1)]/(i^2+i+1)^2
where is that 0?
where the numerator is zero (assuming the denominator is no 0 there)
(i^2+i+1) = i(2i+1)
0 = i^2 -1
i^2 = 1
i = +1 or -1
if i = +1
p = 120/3 = 40
if i = -1, p = -120 (meaningless for negative intensity anyway)
check the arithmetic, that is the method.