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The rate of change (dP)/(dt) of the number of algae in a tank is modeled by the following differential equation: (dP)/(dt)=(2317)/(10614)P(1-(P)/( 662)) 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? Answer:

The rate of change dPdt \frac{d P}{d t} of the number of algae in a tank is modeled by the following differential equation:\newlinedPdt=231710614P(1−P662) \frac{d P}{d t}=\frac{2317}{10614} P\left(1-\frac{P}{662}\right) \newlineAt t=0 t=0 , the number of algae in the tank is 174174 and is increasing at a rate of 2828 algae per minute. At what value of P P is P(t) P(t) growing the fastest?\newlineAnswer:

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Q. The rate of change dPdt \frac{d P}{d t} of the number of algae in a tank is modeled by the following differential equation:\newlinedPdt=231710614P(1−P662) \frac{d P}{d t}=\frac{2317}{10614} P\left(1-\frac{P}{662}\right) \newlineAt t=0 t=0 , the number of algae in the tank is 174174 and is increasing at a rate of 2828 algae per minute. At what value of P P is P(t) P(t) growing the fastest?\newlineAnswer:
  1. Simplify Equation: The growth rate of P(t)P(t) is given by the differential equation dPdt=231710614P(1−P662)\frac{dP}{dt} = \frac{2317}{10614}P\left(1 - \frac{P}{662}\right). To find the value of PP where P(t)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 PP is zero. Let's first simplify the growth rate function.
  2. Find Derivative: Simplify the differential equation by dividing the constant 23172317 by 1061410614 to get a simplified constant factor.\newline231710614≈0.2183\frac{2317}{10614} \approx 0.2183 (rounded to four decimal places for simplicity).\newlineSo, the simplified differential equation is dPdt≈0.2183P(1−P662)\frac{dP}{dt} \approx 0.2183P(1 - \frac{P}{662}).
  3. Set Derivative Equal: Now, let's find the derivative of the growth rate function with respect to PP, which is ddP[dPdt]\frac{d}{dP}\left[\frac{dP}{dt}\right]. This will give us the rate of change of the growth rate with respect to PP. The derivative of 0.2183P(1−P662)0.2183P(1 - \frac{P}{662}) with respect to PP is 0.2183(1−P662)−0.2183P(1662)=0.2183−0.2183×2P6620.2183(1 - \frac{P}{662}) - 0.2183P\left(\frac{1}{662}\right) = 0.2183 - \frac{0.2183\times 2P}{662}.
  4. Eliminate Denominator: Set the derivative equal to zero to find the critical points:\newline0=0.2183−0.2183×2P6620 = 0.2183 - \frac{0.2183\times 2P}{662}.\newlineThis simplifies to 0=2183−2183×2P6620 = 2183 - \frac{2183\times 2P}{662}.
  5. Solve for P: Multiply both sides by 662662 to eliminate the denominator:\newline0=2183×662−2183×2P0 = 2183 \times 662 - 2183 \times 2P.
  6. Confirm Maximum: Divide both sides by 21832183 to solve for PP: 0=662−2P0 = 662 - 2P.
  7. Confirm Maximum: Divide both sides by 21832183 to solve for PP: \newline0=662−2P0 = 662 - 2P. Now, solve for PP: \newline2P=6622P = 662. \newlineP = rac{662}{2}. \newlineP=331P = 331.
  8. Confirm Maximum: Divide both sides by 21832183 to solve for PP: 0=662−2P0 = 662 - 2P. Now, solve for PP: 2P=6622P = 662. P = rac{662}{2}. P=331P = 331. We found that the critical point is at P=331P = 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 P2P^2 is negative), we can conclude that P=331P = 331 is the maximum point without further calculation.

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