If by optimal age, you mean producing the maximum offspring, then yes, you want to find where dr/dx = 0
r(x) = log(e^-.1x * 4x^1.5)/x
That's kind of messy, and not tractable to algebraic solution. Netter use a graphical or numeric method. Looks like dr/dx=0 at x=1.07875
http://www.wolframalpha.com/input/?i=derivative+log%28e^%28-.1x%29+*+4x^1.5%29%2Fx%2C+x+%3D+0.9+to+10
Semelparous organisms breed only once during their lifetime. Examples of this type of reproduction strategy can be found with Pacific salmon and bamboo. The per capita rate of increase, r, can be thought of as a measure of reproductive fitness. The greater 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
r(x)=ln[L(x)M(x)] / x,
where L(x) is the probability of surviving to age x and M(x) is the number of female births at age x.
Suppose that
L(x)=e^(−0.1x)
and
M(x)=4x^(1.5).
Find the optimal age of reproduction.
<correct me if i'm wrong with my procedure, but do i take the derivative, set it equal to zero and solve for x?, if so, i'm still stuck~> I appreciate your help!
1 answer
1 answer