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== The Objective Functions to Minimize ==

To get the equation that Frank cites on page 317, we can minimize either of the following functions:

<math>F(\lambda)=\int_0^\lambda f(y)h(y)dy\int_\lambda^1 g(y)h(y)dy</math>

or

<math>E(\lambda)=\ln(F(\lambda))=\ln\left(\int_0^\lambda f(y)h(y)dy\right)+\ln\left(\int_\lambda^1 g(y)h(y)dy\right)</math>

''F'' is probably the more intuitive function: we can see that there's a trade-off point, <math>\lambda</math>, at which we switch from producing males to producing females. If we think of the two function <math>f(x)</math> and <math>g(x)</math> as "value" functions, with <math>f(x) \ge g(x)</math> over the interval <math>x \in [0,1]</math>, and <math>h(x)</math> as the probability density of the resources distribution (taken as a beta density by Frank), then we're saying that we want to maximize the product of the value of the males '''along with'''  the value of the females. If we take <math>\lambda=0</math> or <math>\lambda=1</math>, <math>F(0)=F(1)=0</math>; because <math>F(x)\ge 0</math>, we know that there is a maximum on the interior of the interval <math>[0,1]</math>.

If we differentiate either expression, we will ultimately obtain the equation that Frank cites in his text:

<center><math>\frac{f(\lambda)}{\int_0^\lambda f(y)h(y)dy}=\frac{g(\lambda)}{\int_\lambda^1 g(y)h(y)dy}</math></center>

which we can then solve for <math>\lambda</math>, given the functions ''f, g, and h''.

<math>F(\lambda)</math> can be normalized without changing the maximizing value of <math>\lambda</math>: hence we could use instead

<math>F(\lambda)=\frac{\int_0^\lambda f(y)h(y)dy\int_\lambda^1 g(y)h(y)dy}{\int_0^1 f(y)h(y)dy\int_0^1 g(y)h(y)dy}=\left(\frac{\int_0^\lambda f(y)h(y)dy}{f_T}\right)\left(\frac{\int_\lambda^1 g(y)h(y)dy}{g_T}\right)</math>; 

hence we can think of <math>F(\lambda)</math> as the product of the fraction of male value achieved by <math>\lambda</math> and the fraction of female value achieved from <math>\lambda</math> on out to the total resource allocation.


== The Minimization with Specific ''f, g,'' and ''h'' ==

Assume that <math>f(x)=x^r</math>, <math>g(x)=x</math>, and <math>h(x) = \kappa x^a(1-x)^a</math>.

We expand <math>I=\int y^s y^a (1-y)^a dy = \int y^s y^a \left(\sum_{n=0}^a {n \choose k} (-1)^n y^{n}\right) dy </math>

or

<math>I=\sum_{n=0}^a {n \choose k} (-1)^n \int y^{n+a+s} dy =\sum_{n=0}^a {n \choose k} (-1)^n \frac{y^{n+a+s+1}}{n+a+s+1}| </math>

Given our choices of ''f'' and ''g'', we want to evaluate

<math>I_f=\sum_{n=0}^a {n \choose k} (-1)^n \frac{y^{n+a+r+1}}{n+a+r+1}|_0^\lambda = \sum_{n=0}^a {n \choose k} (-1)^n \frac{\lambda^{n+a+r+1}}{n+a+r+1}</math>

and

<math>I_g=\sum_{n=0}^a {n \choose k} (-1)^n \frac{y^{n+a+2}}{n+a+2}|_\lambda^1=\sum_{n=0}^a {n \choose k} (-1)^n \frac{1-\lambda^{n+a+2}}{n+a+2} </math>

Now the equation that we're to solve for <math>\lambda</math> can be written <math>\lambda^r I_g = \lambda I_f</math> which we can reduce by dividing by <math>\lambda^r</math> (since <math>\lambda=0</math> is not a maximum); thus our equation becomes <math>I_g - \lambda^{1-r} I_f =0</math>, or

<math>P(\lambda) = \sum_{n=0}^a {n \choose k} (-1)^n \left(\frac{1-\lambda^{n+a+2}}{n+a+2} - \lambda^{1-r}\frac{\lambda^{n+a+r+1}}{n+a+r+1} \right) = 0</math>

or

<math>P(\lambda) = \sum_{n=0}^a {n \choose k} (-1)^n \left(\frac{1-\lambda^{n+a+2}}{n+a+2} - \frac{\lambda^{n+a+2}}{n+a+r+1} \right) = 0</math>

or

<math>
P(\lambda) = 
\sum_{n=0}^a {n \choose k} (-1)^n 
\left(
\frac{1}{n+a+2} - \lambda^{n+a+2}
\left(
\frac{1}{n+a+2}+
\frac{1}{n+a+r+1} 
\right)\right) = 0
</math>

[http://www.nku.edu/~longa/research/CURM/sexallocation.html Here's a little script] that shows some of those functions ''P''.

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