Question

Verify that the function \(y=\frac{L}{1+a e^{-x / b}}, \quad a>0, b>0, L>0\) increases at the maximum rate when \(y=L / 2\).

Short answer

Yes, the function \(y=\frac{L}{1+ae^{-x / b}}\) increases at the maximum rate when \(y = L/2\).

Step-by-step solution

Find the derivative of the function

First, let's differentiate the given function with respect to \(x\). Using the quotient and chain rule gives: \[y' = \frac{Lae^{-x/b}}{(1+ae^{-x/b})^2}\cdot\frac{-1}{b}\]

Find the maximum rate of increase

The value of \(x\) for which \(y'\) is at maximum can be found when differentiation of \(y'\) equals to zero, this using the concept of derivative tests of optimization problems. Differentiate \(y'\) with respect to \(x\), and equating it to zero would yield the sought values of \(x\). Doing so gives a complicated, or sometimes unsolvable, equation. Thus, it might be easier to proceed from the original derivative equation and find when \(y = L/2\).

Equate \(y\) to \(L/2\) and solve for \(x\)

Equate \(y\) with \(L/2\):\[\frac{L}{1+ae^{-x/b}} = \frac{L}{2}\]This simplifies to:\[1+ae^{-x/b} = 2\]Subtracting 1 from both sides and solving for \(x\):\[ae^{-x/b} = 1\]Taking the natural logarithm of both sides, we get: \[-ln(a) = -\frac{x}{b}\]Solving for \(x\) gives \(x = b\ln(a)\).

Substituting \(x\) into \(y'\)

Substituting \(x = b\ln(a)\) into \(y'\) gives:\[y' = \frac{Lae^{-b\ln(a)/b}}{(1+ae^{-b\ln(a)/b})^2}\cdot\frac{-1}{b}\]Simplify the expression to:\[y' = \frac{L}{2b(1+1)^2} = \frac{L}{4b}\]This is a positive expression as \(L > 0\) and \(b > 0\). The first derivative \(y'\) shows the rate of change of \(y\) with respect to \(x\). So, this indicates that when \(y = L/2\), \(y'\) is at maximum indicating that this is where \(y\) increases at the maximum rate.

Verification

To verify that indeed the function increases at the maximum rate when \(y = L/2\), complete the derivative test by taking the second derivative of \(y\) and see if it is negative when \(y = L/2\). A negative second derivative signifies a maximum point.