The problem is as shown. I tried using gradient and Hessian but can not make any conclusions from them. Any ideas?

$$\max x_1^{a_1}x_2^{a_2}\cdots x_n^{a_n}$$

subject to

$$\sum_{i=1}^nx_i=1,\quad x_i\geq 0,\quad i=1,2,\ldots,n,$$

where $a_i$ are given positive scalars. Find a global maximum and show that it is unique.

  • $\begingroup$ That looks like an exercise on Lagrange multipliers. And please, post your questions here instead of giving a link. $\endgroup$ – Mark Sep 25 '18 at 16:13
  • $\begingroup$ @Mark Thanks Mark I am new here and please forgive me for this kind of mistakes. Also, I changed the tag as well. $\endgroup$ – Ricky Shao Sep 25 '18 at 16:22
  • $\begingroup$ Take this as advice: it's a good idea to write in your question what you tried (with technical details) and why it doesn't work for you! $\endgroup$ – johnny09 Mar 24 '19 at 22:16

Calling $f = \prod_{k=1}^n x_k^{\alpha_k}$ we have the Lagrangian

$$ L(x,\mu) = f - \mu\left(\sum_{k=1}^n x_k-1\right) $$

then the stationary points are determined by

$$ \alpha_k \frac{f}{x_k}-\mu = 0,\ \ \ k = 1,\cdots, n\\ \sum_{k=1}^n x_k-1=0 $$

or making $\lambda = \frac{\mu}{f}$

$$ \frac{\alpha_k}{x_k}-\lambda = 0\\ \sum_{k=1}^n x_k-1=0 $$

$\lambda$'s value is obtained substituting $x_k = \frac{\alpha_k}{\lambda}$ so

$$ \sum_{k=1}^n \alpha_k-\lambda=0 $$

and also $x_k = \frac{\alpha_k}{\sum_{k=1}^n \alpha_k}$


  • $\begingroup$ Yeah I came up with the same answer as yours, took me so long. Thank you so much. For the uniqueness, do you compute the Hessian of the function? $\endgroup$ – Ricky Shao Sep 25 '18 at 17:04
  • $\begingroup$ @RickyShao Please have a look at en.wikipedia.org/wiki/Hessian_matrix concerning bordered hessians. $\endgroup$ – Cesareo Sep 25 '18 at 19:03

It looks like a homework, so I'll give you a hint instead of a complete solution. To maximize your function is the same as to maximize $$ F(x)=\ln(x_1^{a_1}x_2^{a_2}\ldots x_n^{a_n})=\sum_{i=1}^na_i\ln x_i. $$ The condition $x_i\ge 0$ cannot be active at the maximum, hence, we do not need it in the Lagrange equation for the gradients. Use Lagrange multipliers to conclude the necessary condition as $$ \frac{a_i}{x_i}+v=0. $$ Express each $x_i$, use the equality condition to find $v$, and finally find $x_i$.

The new problem is concave, then the Lagrange condition is also sufficient. The objective function $F$ is strictly concave, hence, the maximum is unique.

  • $\begingroup$ Yeah, thank you for your answer, I used the same method as @Cesareo to solve the first part of the problem. Your solution provides an easy way to prove the uniqueness of the problem. Thank you so much. $\endgroup$ – Ricky Shao Sep 25 '18 at 17:09
  • $\begingroup$ @RickyShao You are welcome. $\endgroup$ – A.Γ. Sep 25 '18 at 17:20

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