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The rate of change (dP)/(dt) of the number of people on a beach is modeled by the following differential equation: (dP)/(dt)=(595)/(2664)P(1-(P)/( 510)) At t=0, the number of people on the beach is 222 and is increasing at a rate of 28 people per hour. 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 people on a beach is modeled by the following differential equation:\newlinedPdt=5952664P(1−P510) \frac{d P}{d t}=\frac{595}{2664} P\left(1-\frac{P}{510}\right) \newlineAt t=0 t=0 , the number of people on the beach is 222222 and is increasing at a rate of 2828 people per hour. At what value of P P is P(t) P(t) growing the fastest?\newlineAnswer:

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Full solution

Q. The rate of change dPdt \frac{d P}{d t} of the number of people on a beach is modeled by the following differential equation:\newlinedPdt=5952664P(1−P510) \frac{d P}{d t}=\frac{595}{2664} P\left(1-\frac{P}{510}\right) \newlineAt t=0 t=0 , the number of people on the beach is 222222 and is increasing at a rate of 2828 people per hour. At what value of P P is P(t) P(t) growing the fastest?\newlineAnswer:
  1. Given Differential Equation: The given differential equation is:\newlinedPdt=5952664⋅P⋅(1−P510)\frac{dP}{dt} = \frac{595}{2664} \cdot P \cdot \left(1 - \frac{P}{510}\right)\newlineTo find the value of PP where P(t)P(t) is growing the fastest, we need to find the maximum point of the rate of change function dPdt\frac{dP}{dt}. This occurs when the derivative of dPdt\frac{dP}{dt} with respect to PP is equal to zero.
  2. Derivative Setup: Let's set up the derivative of dPdt\frac{dP}{dt} with respect to PP:ddP[dPdt]=ddP[5952664⋅P⋅(1−P510)]\frac{d}{dP} \left[\frac{dP}{dt}\right] = \frac{d}{dP} \left[\frac{595}{2664} \cdot P \cdot \left(1 - \frac{P}{510}\right)\right]We apply the product rule and the chain rule to take the derivative.
  3. Derivative Calculation: The derivative is:\newlineddP(dPdt)=5952664×(1−P510)+5952664×P×(−1510)\frac{d}{dP} \left(\frac{dP}{dt}\right) = \frac{595}{2664} \times \left(1 - \frac{P}{510}\right) + \frac{595}{2664} \times P \times \left(-\frac{1}{510}\right)\newlineSimplify the expression.
  4. Simplify Expression: After simplifying, we get:\newlineddP(dPdt)=5952664−2×595×P2664×510\frac{d}{dP} \left(\frac{dP}{dt}\right) = \frac{595}{2664} - \frac{2 \times 595 \times P}{2664 \times 510}\newlineNow, we set this derivative equal to zero to find the critical points.
  5. Find Critical Points: Setting the derivative equal to zero gives us:\newline rac{595}{2664} - rac{2 \times 595 \times P}{2664 \times 510} = 0\newlineSolve for PP.
  6. Solve for P: Multiplying through by (2664×510)(2664 \times 510) to clear the denominators, we get:\newline595×510−2×595×P=0595 \times 510 - 2 \times 595 \times P = 0
  7. Isolate P: Now, we solve for P:\newline595×510=2×595×P595 \times 510 = 2 \times 595 \times P\newlineDivide both sides by (2×595)(2 \times 595) to isolate P.
  8. Calculate Value of P: After dividing, we find:\newlineP = (595×510)/(2×595)(595 \times 510) / (2 \times 595)\newlineThe 595595 terms cancel out.
  9. Calculate Value of P: After dividing, we find:\newlineP = (595×510)/(2×595)(595 \times 510) / (2 \times 595)\newlineThe 595595 terms cancel out.We are left with:\newlineP = 510/2510 / 2
  10. Calculate Value of P: After dividing, we find:\newlineP = (595×510)/(2×595)(595 \times 510) / (2 \times 595)\newlineThe 595595 terms cancel out.We are left with:\newlineP = 510/2510 / 2Finally, we calculate the value of P:\newlineP = 255255\newlineThis is the value of P where P(t)P(t) is growing the fastest.

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