## The Objective Functions to Minimize

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

$F(\lambda )=\int _{0}^{\lambda }f(y)h(y)dy\int _{\lambda }^{1}g(y)h(y)dy$

or

$E(\lambda )=\ln(F(\lambda ))=\ln \left(\int _{0}^{\lambda }f(y)h(y)dy\right)+\ln \left(\int _{\lambda }^{1}g(y)h(y)dy\right)$

*F* is probably the more intuitive function: we can see that there's a trade-off point, $\lambda$, at which we switch from producing males to producing females. If we think of the two function $f(x)$ and $g(x)$ as "value" functions, with $f(x)\geq g(x)$ over the interval $x\in [0,1]$, and $h(x)$ 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 $\lambda =0$ or $\lambda =1$, $F(0)=F(1)=0$; because $F(x)\geq 0$, we know that there is a maximum on the interior of the interval $[0,1]$.

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

${\frac {f(\lambda )}{\int _{0}^{\lambda }f(y)h(y)dy}}={\frac {g(\lambda )}{\int _{\lambda }^{1}g(y)h(y)dy}}$
which we can then solve for $\lambda$, given the functions *f, g, and h*.

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

$F(\lambda )={\frac {\int _{0}^{\lambda }f(y)h(y)dy\int _{\lambda }^{1}g(y)h(y)dy}{\int _{0}^{1}f(y)h(y)dy\int _{0}^{1}g(y)h(y)dy}}=\left({\frac {\int _{0}^{\lambda }f(y)h(y)dy}{f_{T}}}\right)\left({\frac {\int _{\lambda }^{1}g(y)h(y)dy}{g_{T}}}\right)$;

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

## The Minimization with Specific *f, g,* and *h*

Assume that $f(x)=x^{r}$, $g(x)=x$, and $h(x)=\kappa x^{a}(1-x)^{a}$.

We expand $I=\int y^{s}y^{a}(1-y)^{a}dy=\int y^{s}y^{a}\left(\sum _{n=0}^{a}{n \choose k}(-1)^{n}y^{n}\right)dy$

or

$I=\sum _{n=0}^{a}{n \choose k}(-1)^{n}\int y^{n+a+s}dy=\sum _{n=0}^{a}{n \choose k}(-1)^{n}{\frac {y^{n+a+s+1}}{n+a+s+1}}|$

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

$I_{f}=\sum _{n=0}^{a}{n \choose k}(-1)^{n}{\frac {y^{n+a+r+1}}{n+a+r+1}}|_{0}^{\lambda }=\sum _{n=0}^{a}{n \choose k}(-1)^{n}{\frac {\lambda ^{n+a+r+1}}{n+a+r+1}}$

and

$I_{g}=\sum _{n=0}^{a}{n \choose k}(-1)^{n}{\frac {y^{n+a+2}}{n+a+2}}|_{\lambda }^{1}=\sum _{n=0}^{a}{n \choose k}(-1)^{n}{\frac {1-\lambda ^{n+a+2}}{n+a+2}}$

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

$P(\lambda )=\sum _{n=0}^{a}{n \choose k}(-1)^{n}\left({\frac {1-\lambda ^{n+a+2}}{n+a+2}}-\lambda ^{1-r}{\frac {\lambda ^{n+a+r+1}}{n+a+r+1}}\right)=0$

or

$P(\lambda )=\sum _{n=0}^{a}{n \choose k}(-1)^{n}\left({\frac {1-\lambda ^{n+a+2}}{n+a+2}}-{\frac {\lambda ^{n+a+2}}{n+a+r+1}}\right)=0$

or

$P(\lambda )=\sum _{n=0}^{a}{n \choose k}(-1)^{n}\left({\frac {1}{n+a+2}}-\lambda ^{n+a+2}\left({\frac {1}{n+a+2}}+{\frac {1}{n+a+r+1}}\right)\right)=0$

Here's a little script that shows some of those functions *P*.