Find when the population is growing the fastest, under the logistic model

Question

The population $P$ of an island $y$ years after colonization is given by the function: $\displaystyle P = \frac{250}{1 + 4e^{-0.01y}}$. After how many years was the population growing the fastest?

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I tried taking the second derivative and setting it equal to zero. From there I solved for $y$ but kept getting $y = 0$. Which doesn't make too much sense so I was wondering how anyone would go about answering this problem.

Answer 1

This is a Logistic Growth problem. So... $$P'(y) = kP(L - P)$$ and $$P(y) = {L \over {1 + C{e^{ - ky}}}}$$

If we add in the specifics of your problem...

$$\eqalign{ & P = {{250} \over {1 + 4{e^{ - 0.01y}}}} \cr & P'(y) = kP(L - P) = 0.01P(250 - P) = 2.5P(y) - .01P{(y)^2} \cr & P'' = 2.5 - .02P = 0 \cr & P = 125 \cr & 125 = {{250} \over {1 + 4{e^{ - 0.01y}}}} \cr & 125 + 500{e^{ - 0.01y}} = 250 \cr & {e^{ - 0.01y}} = .25 \cr & y = {{\ln (.25)} \over { - 0.01}} = 138.62 \cr} $$

For future reference, in these types of problems the population is always growing the fastest when $$P=L/2$$

Answer 2

Your idea was good but, may be, there was some mistake in the calculation of the derivatives. Using, as Jake Lebovic, $$P = {L \over {1 + C{e^{ - ky}}}}$$ you have by standard differentiation $$\frac {dP}{dy}=\frac{C k L e^{-k y}}{\left(C e^{-k y}+1\right)^2}$$ $$\frac {d^2P}{dy^2}=\frac{2 C^2 k^2 L e^{-2 k y}}{\left(1+C e^{-k y}\right)^3}-\frac{C k^2 L e^{-k y}}{\left(1+C e^{-k y}\right)^2}=\frac{C k^2 L e^{k y} \left(C-e^{k y}\right)}{\left(C+e^{k y}\right)^3}$$ So, the second derivative cancels if $e^{ky}=C$; putting this in the expression of $P$ shows that this will happen when $P=\frac L2$ whatever could be the values of parameters $L,C,k$.