The rate of change $\frac{d P}{d t}$ of the number of algae in a tank is modeled by the following differential equation: $\newline$ $\frac{d P}{d t}=\frac{2317}{10614} P\left(1-\frac{P}{662}\right)$ $\newline$ At $t=0$ , the number of algae in the tank is $174$ and is increasing at a rate of $28$ algae per minute. At what value of $P$ is $P(t)$ growing the fastest? $\newline$ Answer:

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Q. The rate of change

\frac{d P}{d t}

of the number of algae in a tank is modeled by the following differential equation:

\newline

\frac{d P}{d t}=\frac{2317}{10614} P\left(1-\frac{P}{662}\right)

\newline

t=0

, the number of algae in the tank is

174

and is increasing at a rate of

28

algae per minute. At what value of

P

P(t)

growing the fastest?

\newline

Answer:

Simplify Equation: The growth rate of $P(t)$ is given by the differential equation $\frac{dP}{dt} = \frac{2317}{10614}P\left(1 - \frac{P}{662}\right)$ . To find the value of $P$ where $P(t)$ is growing the fastest, we need to find the maximum point of the growth rate function. This occurs where the derivative of the growth rate with respect to $P$ is zero. Let's first simplify the growth rate function.
Find Derivative: Simplify the differential equation by dividing the constant $2317$ by $10614$ to get a simplified constant factor. $\newline$ $\frac{2317}{10614} \approx 0.2183$ (rounded to four decimal places for simplicity). $\newline$ So, the simplified differential equation is $\frac{dP}{dt} \approx 0.2183P(1 - \frac{P}{662})$ .
Set Derivative Equal: Now, let's find the derivative of the growth rate function with respect to $P$ , which is $\frac{d}{dP}\left[\frac{dP}{dt}\right]$ . This will give us the rate of change of the growth rate with respect to $P$ . The derivative of $0.2183P(1 - \frac{P}{662})$ with respect to $P$ is $0.2183(1 - \frac{P}{662}) - 0.2183P\left(\frac{1}{662}\right) = 0.2183 - \frac{0.2183\times 2P}{662}$ .
Eliminate Denominator: Set the derivative equal to zero to find the critical points: $\newline$ $0 = 0.2183 - \frac{0.2183\times 2P}{662}$ . $\newline$ This simplifies to $0 = 2183 - \frac{2183\times 2P}{662}$ .
Solve for P: Multiply both sides by $662$ to eliminate the denominator: $\newline$ $0 = 2183 \times 662 - 2183 \times 2P$ .
Confirm Maximum: Divide both sides by $2183$ to solve for $P$ : $0 = 662 - 2P$ .
Confirm Maximum: Divide both sides by $2183$ to solve for $P$ : $\newline$ $0 = 662 - 2P$ . Now, solve for $P$ : $\newline$ $2P = 662$ . $\newline$ P = rac{662}{2}. $\newline$ $P = 331$ .
Confirm Maximum: Divide both sides by $2183$ to solve for $P$ : $0 = 662 - 2P$ . Now, solve for $P$ : $2P = 662$ . P = rac{662}{2}. $P = 331$ . We found that the critical point is at $P = 331$ . To confirm that this is a maximum, we can use the second derivative test or analyze the behavior of the growth rate function. However, since the growth rate function is a parabola that opens downwards (as the coefficient of $P^2$ is negative), we can conclude that $P = 331$ is the maximum point without further calculation.