Optimizing $\prod x_i^{ax_i+b}$ subject to $\sum x_i = C$

Question

I would like to obtain a bound for the maximum value of the function $$ F(x_1,\dots,x_n) = \prod_{j=1}^n x_j^{ax_j+b} $$ in the simplex $S_n$ $$ S_n = \{(x_1,\dots,x_n)\in \mathbb{R}^n \mid x_j \geq 0 \text{ and }\sum_{j=1}^n x_j = C\}. $$ Here $a,b,C>0$ are some positive constants.

To optimize $F$ it suffices to optimize $\log F$, $$ \log F = \sum_{j=1}^n (ax_j+b) \log(x_j), $$ and then the Lagrange multiplier condition for $\log F$ to have a critical point is that there exists a constant $\lambda$ such that $$ \partial_{x_i} (\log F) = \lambda $$ for all $1\leq i\leq n$. This leads to the equation $$ \partial_{x_i} (\log F) = a\log(x_i) + \frac{ax_i+b}{x_i} = \lambda. $$ If the function $a\log(x_i) + \frac{ax_i+b}{x_i}$ were injective, one could conclude that the only critical point is at the value $(C/n,C/n,\dots,C/n)$, and since the function vanishes at the boundary of $S_n$ (and is positive in the interior) it must be the case that this is the maximum.

However this function is not injective, and this is not the only critical point. In fact, in the simple case $n=2$ and $a=b=1$ I found (with numerical help) the example $(x_1,x_2) = (0.318, 6.305)$, which satisfies $$ x_1^{x_1+1} x_2^{x_2+1} > m^{2m+2} $$ where $m = (x_1+x_2)/2$.

Given this example it seems difficult to characterize the location of the maxima of $F(x)$, and I am curious whether there still might be a relatively simple bound for the value of $$ \sup_{(x_1,\dots,x_n)\in S_n} F(x_1,x_2,\dots,x_n). $$ Any help would be appreciated!

Answer 1

$$\partial_{x_i} (\log F) = a\log(x_i) + \frac{ax_i+b}{x_i} - \lambda=0$$ gives all $x_i=\frac C n$ but this corresponds to a minimum value of $F$.

Using your example $a=b=1$, $n=2$, $C=\frac{6623}{1000}$, I also find $x_1=0.317849$, $x_2=6.30515$, $\log(F_{max})=11.9410$ while $\log(F_{min})=10.3252$.

Edit

I have a serious problem. From a lot of numerical studies, varying $(n,a,b,C)$, it appears (no proof at all) that, at the maximum, all $x_i$'s but one are identical. The other one is close to $C$.

For example, using $n=4$, $a=2$, $b=3$, $C=123$, I obtained $$x_1=122.279 \qquad x_2=x_3=x_4=0.240 \qquad\qquad F_{max}=1174.95$$

Admit that this is true. Let $$x_1=x_2=\cdots=x_{n-1}=t \qquad \text{and} \qquad x_n=C-\sum_{i=1}^{n-1}x_i=C-(n-1)t$$ This makes $$\log(F)=(n-1) (a t+b)\log (t)+ (a u+b)\log (u)\qquad \text{with} \qquad u=C-(n-1)t$$ So, we need to solve for $t$ the derivative which can be easily solved by Newton method starting with $t_0=\epsilon$. Since starting close to $t=0$, cross multiply to remove the denominator. This makes the equation to solve $$G(t)=b (c-n t)+a t u \log \left(\frac{t}{u}\right)=0\qquad \text{with} \qquad u=C-(n-1)t$$

For the above example, this would give $$4 t-2 t(t-41) (\log (123-3t)-\log (t))-123=0$$ and Newton iterates will be $$\left( \begin{array}{cc} n & t_n \\ 0 & 0.001000 \\ 1 & 0.139207 \\ 2 & 0.234742 \\ 3 & 0.240184 \\ 4 & 0.240197 \end{array} \right)$$

Answer 2

I prefer to add another answer to throw in a few ideas, remarks and preliminary results.

First remark

At least to me, the key problem is to first find the value of $x$ which makes the function maximum. So, we can already get rid of parameter $a$ writing

$$F= \prod_{j=1}^n x_j^{ax_j+b}=\Bigg[\prod_{j=1}^n x_j^{x_j+\beta} \Bigg]^a\quad \text{with} \quad \beta=\frac b a$$ $$G=\prod_{j=1}^n x_j^{x_j+\beta}\implies \log(G)=\sum_{j=1}^n (x_j+\beta)\log(x_j)$$ $$\frac d{dx}\log(G)=\sum_{j=1}^n \Big[1+\frac \beta {x_j} +\log(x_j) \Big]$$ As said in the previous answer and justified by @achillehui in comments, at the maximum, all $x_j$'s but one are identical. So, as before, let $x_1=x_2=\cdots=x_{n-1}=t$ to make $$H=\frac d{dx}\log(G)=(n-1) \left(1+\frac{\beta }{t}+\log (t)\right)+\left(1+\frac{\beta }{x_n}+\log (x_n)\right)\tag 1$$ with $x_n=C-(n-1)t$

Solving $(1)$ for $t$ using Newton method should not present any problem. To save iterations, what would be required is a "good" estimate of the starting point (which would be a function of $n$, $\beta$ and $C$).

Second remark

Numerically, for given values of $\beta$ and $C$, the solution of $(1)$ does not seem to be very sensitive to the value of $n$.

For the two examples worked in the question $\left(t_n^{(1)}\right)$ and the first answer $\left(t_n^{(2)}\right)$ some results

$$\left( \begin{array}{ccc} n & t_n^{(1)} & t_n^{(2)}\\ 2 & 0.317849 & 0.240020 \\ 3 & 0.324726 & 0.240108 \\ 4 & 0.332738 & 0.240197 \\ 5 & 0.342311 & 0.240286 \\ 6 & 0.354172 & 0.240376 \end{array} \right)$$

Many other numerical tests seem to show that $n=2$ provide a lower bound of the solution.

Third remark

The plot of $H$ is not pleasant because of the infinite branch when $t\to 0^+$ but the product $(t\,H)$ is much more interesting. Using a Taylor expansion of it around $t=0$ gives

$$t \,H=\beta C (n-1)-(n-1) t (-C \log (t)+C \log (C)+\beta n)+O\left(t^2\right)$$ Using this expansion gives, as an estimate, $$t_0=-\frac{\beta }{W_{-1}\left(-\frac{n\beta }{C}e^{-\frac{\beta }{C}}\right)}\tag 2$$ where $W_{-1}(.)$ is the second branch of Lambert function.

Applied to the worked cases, this would give $t_0^{(1)}=0.292136$ and $t_0^{(2)}=0.238273$; this does not seem too bad.