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 occure when the population is equal to half the equilibrium size, that is when the population is b/2a.

Answer 1

Answer on Question #74057 – Math – Differential Equations

Question

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

$$\frac{dx}{dt} = x(b - ax)$$

. 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

$$\frac{b}{2a}$$

.

Solution

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

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

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

Answer: the minimum rate of growth will occur when population is

$$\frac{b}{2a}$$

.

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