Question

A certain population is known to be growing at a rate given by the logistic equation

dx/dt =x(b - ax)

Show that the minimum 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

Let's find the minimum rate of growth, so the function $x(b-ax)$ should be minimized. The

derivative of the rate change must be zero:

$\dfrac{d(bx-ax^2)}{dx} = 0$

$b-2ax = 0$

$x = \dfrac{b}{2a}$

$\dfrac{d^2(bx-ax^2)}{dx^2} = -2a<0$ if $a>0$ , hence in this case we get a maximum;

$\dfrac{d^2(bx-ax^2)}{dx^2} = -2a>0$ if $a<0$ , hence in this case we get a minimum.

Such a population x is equal to half the equilibrium size.