Question

A certain population is known to be growing at a rate given by the logistic equation dx/dt=x(b-ax),where a and b are positive constants .
Show that the maximum rate of growth will occur when the population is equal to half the equilibrium size, that is, when the population is (b/2a).

Answer 1

The logistic equation is given by $\frac{dx}{dt}=x(b-ax)$ where $\frac{dx}{dt}$ is the rate of growth .

Let $y=\frac{dx}{dt}$ ..so we have to show that y is maximum when the population $(x) = \frac{b}{2a}$ .

Now

$y= x(b-ax)$

Now differentiation both sides w.r.t $x,$ we get

$\frac{dy}{dx}=x(-a)+(b-ax)*1 \space$ (using product rule of differentiation.)

$or, \frac{dy}{dx}=b-2ax\\$

for maxima /minima we must have $\frac{dy}{dx}=0$ ,( to get the critical points)

$\therefore b-2ax = 0\\or, x= \frac{b}{2a}\\ Now \space \frac{d^2y}{dx^2}= -2a \\i.e \space \frac{d^2y}{dx^2}\space is\space negative \space ( because \space a \space is \space positive \space constant )\\ \therefore y \space is \space maximum \space at \space x=\frac{b}{2a}$

Hence the maximum rate of growth will occur when the population is equal to half the equilibrium size, that is, when the population is (b/2a).

Hence PROVED.