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!
Yes, your procedure is correct. To find the optimal age of reproduction, you need to take the derivative of r(x) with respect to x, set it equal to zero, and then solve for x.
Let's start by finding the derivative of r(x):
r(x) = ln[L(x)M(x)] / x
Taking the derivative of both sides with respect to x:
r'(x) = [L'(x)M(x) + L(x)M'(x)] / [x*L(x)M(x)] - ln[L(x)M(x)] / x^2
Now, let's substitute the given expressions for L(x) and M(x):
L(x) = e^(-0.1x)
M(x) = 4x^(1.5)
L'(x) = (-0.1)e^(-0.1x)
M'(x) = 6x^(0.5)
Substituting these expressions into r'(x):
r'(x) = [(-0.1)e^(-0.1x)*4x^(1.5) + e^(-0.1x)*6x^(0.5)] / [x*e^(-0.1x)*4x^(1.5)] - ln[e^(-0.1x)*4x^(1.5)] / x^2
Simplifying further:
r'(x) = [-0.4x^(0.5)e^(-0.1x) + 6x^(0.5)e^(-0.1x)] / [4x^(2.5)e^(-0.1x)] - ln[4x^(1.5)e^(-0.1x)] / x^2
Now, set r'(x) equal to zero and solve for x:
0 = [-0.4x^(0.5)e^(-0.1x) + 6x^(0.5)e^(-0.1x)] / [4x^(2.5)e^(-0.1x)] - ln[4x^(1.5)e^(-0.1x)] / x^2
Multiplying through by x^2 to eliminate the denominators:
0 = -0.4x^(0.5)e^(-0.1x) + 6x^(0.5)e^(-0.1x) - x^2*ln[4x^(1.5)e^(-0.1x)]
At this point, it is not possible to solve this equation analytically. You will need to use numerical methods or a graphing calculator to approximate the solution. The optimal age of reproduction, x, is the value that satisfies this equation.
Yes, you are on the right track! To find the optimal age of reproduction, you can take the derivative of the function r(x) and set it equal to zero. The critical points where the derivative is zero will give you potential optimal ages of reproduction.
Let's start by taking the derivative of r(x) with respect to x:
r'(x) = [(1 / (L(x)M(x))) * (M'(x)L(x) - L'(x)M(x))] / x^2,
where M'(x) is the derivative of M(x) with respect to x and L'(x) is the derivative of L(x) with respect to x.
First, let's find M'(x) and L'(x):
M'(x) = d/dx (4x^(1.5))
= 6x^0.5
= 6√x,
L'(x) = d/dx (e^(-0.1x))
= -0.1e^(-0.1x).
Now, substitute the expressions for M(x), M'(x), L(x), and L'(x) into the derivative of r(x):
r'(x) = [(1 / (e^(-0.1x) * 4x^(1.5))) * (6√x * e^(-0.1x) - (-0.1e^(-0.1x) * 4x^(1.5)))] / x^2.
Simplifying this expression:
r'(x) = [(6√x * e^(-0.1x) + 0.4e^(-0.1x) * 4x^(1.5))] / (x^2 * e^(-0.1x) * 4x^(1.5)).
To find the critical points, set r'(x) = 0:
[(6√x * e^(-0.1x) + 0.4e^(-0.1x) * 4x^(1.5))] / (x^2 * e^(-0.1x) * 4x^(1.5)) = 0.
Since the denominator cannot be zero, we can ignore it. So, the numerator must be zero:
6√x * e^(-0.1x) + 0.4e^(-0.1x) * 4x^(1.5) = 0.
Divide both sides by e^(-0.1x):
6√x + 0.4 * 4x^(1.5) = 0.
Now we have an equation that we can solve for x. Simplifying further:
6√x + 1.6x^(1.5) = 0.
Unfortunately, this equation doesn't have a simple algebraic solution. We can use numerical methods, such as Newton's method or graphing the equation, to find the optimal age of reproduction.
Once you have the solutions for x, you can evaluate r(x) at each potential optimal age to determine the maximum value of r(x) and therefore find the optimal age of reproduction.
I hope this helps!
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