The population , in thousands, of a certain bacterial culture is modeled by the logistic equation where is the time, in hours, since observations began and and are positive constants. At , the population is bacteria, and at the population is bacteria. According to this model, at what time will the population be increasing at its greatest rate?

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SAT Nonlinear Functions Question 27 | Free SAT Prep | Aniko

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Question 27·Hard·Nonlinear Functions

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The population $P(t)$ , in thousands, of a certain bacterial culture is modeled by the logistic equation

P(t)=\frac{120}{1+Ae^{-kt}},

where $t$ is the time, in hours, since observations began and $A$ and $k$ are positive constants. At $t=0$ , the population is $20{,}000$ bacteria, and at $t=1$ the population is $24{,}000$ bacteria. According to this model, at what time will the population be increasing at its greatest rate?

For logistic growth questions that ask when the population is increasing fastest, first identify the limiting population (the horizontal asymptote) from the formula; the maximum growth rate occurs when the population is at half this value. Use the given data points to solve for any unknown parameters in the model, then set the function equal to half the limiting value and solve for the time. This avoids differentiating directly and turns the problem into straightforward algebra with exponentials and logarithms.

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Use the initial condition

Remember that $P(t)$ is in thousands. Substitute $t=0$ and $P(0)=20$ into the formula to solve for $A$ first.

Use the second data point to find k

After you find $A$ , plug in $t=1$ and $P(1)=24$ into $P(t) = \dfrac{120}{1 + A e^{-kt}}$ and solve the resulting equation for $k$ .

Think about what “increasing at the greatest rate” means for a logistic curve

The rate of change is greatest where the graph is steepest, which for a logistic (S-shaped) curve happens at its inflection point. For logistic models of the form in this problem, that inflection point occurs when the population is half the limiting value (here, half of 120).

Turn the condition on P into an equation for t

Once you know the population value at which the growth rate is greatest, set $P(t)$ equal to that value and solve the resulting equation for $t$ using your value of $k$ .

Desmos Guide

Enter the constants and the logistic function

In Desmos, define the constant k = ln(5/4) (from solving the two given data points) and then enter P(t) = 120/(1+5*e^(-k*t)) as your function.

Use the half-capacity condition

Because the growth rate of a logistic function is greatest when the population is half of its limiting value, enter the horizontal line y = 60 to represent this population level.

Find the corresponding time

Zoom or adjust the viewing window so you can see where the curve y = P(t) crosses the line y = 60, tap the intersection point, and read off its $t$ -coordinate. That $t$ -value is the time when the population is increasing at its greatest rate.

Step-by-step Explanation

Use the initial population to find A

Because $P(t)$ is measured in thousands, $20{,}000$ bacteria means $P(0)=20$ .

Substitute $t=0$ into the model:

P(0) = \frac{120}{1 + A e^{-k\cdot 0}} = \frac{120}{1 + A e^{0}} = \frac{120}{1 + A}.

We are told $P(0) = 20$ , so

20 = \frac{120}{1 + A} \Rightarrow 1 + A = \frac{120}{20} = 6 \Rightarrow A = 5.

So the model becomes

P(t) = \frac{120}{1 + 5 e^{-kt}}.

Use the population at t = 1 to find k

Now use $P(1)=24$ (since $24{,}000$ bacteria is 24 thousand):

24 = \frac{120}{1 + 5 e^{-k}}.

Solve for $k$ :

1 + 5 e^{-k} = \frac{120}{24} = 5 \Rightarrow 5 e^{-k} = 4 \Rightarrow e^{-k} = \frac{4}{5}.

Take natural logs:

-k = \ln\!\left(\frac{4}{5}\right) \Rightarrow k = -\ln\!\left(\frac{4}{5}\right) = \ln\!\left(\frac{5}{4}\right).

So the model is now completely specified:

P(t) = \frac{120}{1 + 5 e^{-\ln(5/4)\, t}}.

Find when the growth rate is greatest (use the logistic property)

The question asks when the population is increasing at its greatest rate. That means we want the time when the rate of change $P'(t)$ is largest.

For this logistic form

P(t) = \frac{120}{1 + 5 e^{-kt}},

its derivative can be rewritten (by differentiating and simplifying) as

P'(t) = k\, P(t)\left(1 - \frac{P(t)}{120}\right).

Here $k$ is a positive constant, so maximizing $P'(t)$ is the same as maximizing the expression

P(t)\left(1 - \frac{P(t)}{120}\right).

Treat that as a function of $P$ :

f(P) = P\left(1 - \frac{P}{120}\right).

This is a quadratic in $P$ with zeros at $P=0$ and $P=120$ , so its graph is a downward opening parabola whose maximum occurs exactly halfway between the zeros, at

P = \frac{0 + 120}{2} = 60.

So the population is increasing at its greatest rate when $P(t) = 60$ thousand.

Solve for the time when P(t) = 60

Set $P(t)$ equal to 60 and solve for $t$ using $A=5$ and $k = \ln(5/4)$ :

60 = \frac{120}{1 + 5 e^{-kt}}.

Solve step by step:

\begin{aligned} 1 + 5 e^{-kt} &= \frac{120}{60} = 2 \\ 5 e^{-kt} &= 1 \\ e^{-kt} &= \frac{1}{5}. \end{aligned}

Take natural logs:

-kt = \ln\!\left(\frac{1}{5}\right) = -\ln 5 \Rightarrow t = \frac{\ln 5}{k}.

Substitute $k = \ln(5/4)$ :

t = \frac{\ln 5}{\ln(5/4)} \approx 7.2 \, \text{hours}.

From the answer choices, this corresponds to $\boxed{7.2}$ .

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Step-by-step Explanation

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