# Semelparous organisms breed only once during their lifetime. Examples of this type of...

## Question:

Semelparous organisms breed only once during their lifetime. Examples of this type of reproduction can be found in Pacific salmon and bamboo. The per capita rate of increase, r, can be thought of as a measure of reproductive fitness. The greater the value of r, the more offspring an individual produces. The intrinsic rate of increase is typically a function of age, x. Models for age-structured populations of semelparous organisms predict that the intrinsic rate of increase as a function of x is given by

{eq}r(x)= \frac{\ln[l(x)m(x)]}{x} {/eq}

where l(x) is the probability of surviving to age x and m(x) is the number of female offspring at age x. The optimal age of reproduction is the age x that maximizes r(x). Find the optimal age of reproduction for

{eq}l(x)= e^{-ax}\\ m(x)= bx^c {/eq}

where a, b, and c are positive constants.

## Critical Points:

To determine the optimal values of a function, we must first determine the candidates:

that is, the points for which the derivative of the function does not exist or is equal to zero.

First, we need to simplify the function:

{eq}r\left( x \right) = \frac{{\ln \left[ {l\left( x \right)m\left( x \right)} \right]}}{x}\\ l\left( x \right) = {e^{ - ax}},\;m\left( x \right) = b{x^c}\\ \\ r\left( x \right) = \frac{{\ln \left[ {{e^{ - ax}} \cdot b{x^c}} \right]}}{x}\\ r\left( x \right) = \frac{{\ln \left[ {{e^{ - ax}}} \right] + \ln \left[ {b{x^c}} \right]}}{x}\\ r\left( x \right) = \frac{{ - ax\ln \left[ e \right] + \ln \left[ b \right] + \ln \left[ {{x^c}} \right]}}{x}\\ r\left( x \right) = \frac{{ - ax + \ln \left[ b \right] + c\ln \left[ x \right]}}{x} {/eq}

Now, using derivatives, we determine the critical points:

{eq}r'\left( x \right) = \frac{{\left( { - a + c\frac{1}{x}} \right)x - \left( { - ax + \ln \left[ b \right] + c\ln \left[ x \right]} \right) \cdot 1}}{{{x^2}}}\\ r'\left( x \right) = \frac{{ - ax + c + ax - \ln \left[ b \right] - c\ln \left[ x \right]}}{{{x^2}}}\\ r'\left( x \right) = \frac{{c - \ln \left[ b \right] - c\ln \left[ x \right]}}{{{x^2}}} = 0\\ \\ c - \ln \left[ b \right] - c\ln \left[ x \right] = 0\\ c - \ln \left[ b \right] = c\ln \left[ x \right]\\ \ln \left[ x \right] = \frac{{c - \ln \left[ b \right]}}{c}\\ x = {e^{\frac{{c - \ln \left[ b \right]}}{c}}} {/eq}

Taking into account the sign of the derivative:

{eq}0 < x < {e^{\frac{{c - \ln \left[ b \right]}}{c}}} \to r'\left( x \right) > 0\\ x > {e^{\frac{{c - \ln \left[ b \right]}}{c}}} \to r'\left( x \right) < 0 {/eq}

We can affirm. the critical point is a maximum. So, the optimal age of reproduction is {eq}x = {e^{\frac{{c - \ln \left[ b \right]}}{c}}} {/eq} 