Inflection Point and Maximum Logistic Growth

Growth Rate and the Shape of a Logistic Curve

A logistic population grows slowly when it is small, more rapidly as it increases, and then more slowly as it approaches its carrying capacity. This pattern has a precise turning point: for a nonconstant solution that passes through half the carrying capacity, its growth rate is greatest at that level.

The same population level is where the logistic curve changes concavity, so it is an inflection point. To find when the population reaches that level, we use the explicit logistic solution together with the initial population. The conditions matter: an equilibrium solution does not pass through half capacity, and a time interval that excludes the inflection point may have a different maximum.

Logistic Growth Rate

Let \(P(t)\) be a population at time \(t\). In the logistic model, the population satisfies

where \(r>0\) is the growth-rate parameter and \(K>0\) is the carrying capacity. The instantaneous growth rate is \(P'(t)\), measured in population units per unit time. The factor \(1-P/K\) represents the reduction in growth as the population approaches capacity.

When \(0<P<K\), both \(P\) and \(1-P/K\) are positive, so \(P'(t)>0\). Such a solution increases while remaining below carrying capacity. We will examine the rate along this increasing part of the logistic curve.

Why the Maximum Occurs at Half Capacity

There are two useful ways to see the maximum. First, regard the instantaneous rate as a function of population level:

This is a downward-opening quadratic. Its derivative with respect to \(P\) is \(g'(P)=r-(2r/K)P\), which is zero at \(P=K/2\). The quadratic increases before that value and decreases after it. Evaluating the rate there gives \(rK/4\).

For the curve’s concavity, differentiate the logistic equation with respect to time. The chain rule gives \(P''(t)=g'(P(t))P'(t)\). Substituting the expressions for \(g'\) and \(P'\) yields

On a nonconstant solution with \(0<P(t)<K\), the factor \(P'(t)\) is positive. Consequently, \(P''(t)>0\) when \(P(t)<K/2\), and \(P''(t)<0\) when \(P(t)>K/2\). The population curve is concave up below half capacity and concave down above it. If the solution passes through \(K/2\), its concavity changes there.

Deriving the Time of the Inflection Point

Suppose the initial condition is \(P(t_0)=P_0\), with \(0<P_0<K\). The explicit logistic solution is

At the inflection point, \(P(t)=K/2\). Substituting this value into the solution shows that the denominator must equal \(2\), so \(A e^{-r(t-t_0)}=1\). Taking natural logarithms and solving for \(t\) gives

This is the time relative to the initial condition. It can be earlier than \(t_0\) if \(P_0>K/2\). For a time domain beginning at \(t_0\), the inflection point is part of the observed curve only if \(t_{\mathrm{inf}}\ge t_0\). If it is not, the maximum growth rate on the restricted future interval occurs at its starting time.

Worked Examples

A Common Mistake: Confusing the Rate with Its Derivative

When differentiating the logistic rate with respect to population level, we obtain \(g'(P)=r(1-2P/K)\). A common error is to substitute \(P=K/2\) into this expression and conclude that the growth rate is zero there.

This calculation uses \(g'(P)\), the derivative of the rate as a function of population, in place of \(g(P)\), the rate itself. The correct substitution into the logistic rate is

The rate is positive and maximal at half capacity. What equals zero there is \(g'(K/2)\), indicating that the rate has a maximum as a function of population level.

Conditions and Interpretation

The result relies on the population actually traversing half capacity. For \(r>0\), the constant solutions \(P(t)=0\) and \(P(t)=K\) are equilibria: they have zero growth at all times and do not pass through \(K/2\). Thus it would be incorrect to claim that every logistic solution has an inflection point at half capacity.

For a nonconstant solution with \(0<P_0<K\), the full logistic curve passes through \(K/2\) exactly once. On the full curve, the rate increases before that time and decreases after it. If only a limited time interval is considered, its greatest rate may instead occur at an endpoint, as the third example illustrates.

The result is useful because it gives a clear interpretation of the model parameters. The carrying capacity sets the population level at fastest growth, \(K/2\), while \(rK/4\) sets the size of that maximum rate. Units provide a check: \(r\) has units of inverse time and \(K\) has units of population, so \(rK/4\) has units of population per time.