I am trying to find the extreme value of this function $$f(x_1,\ldots,x_n) = x_1x_2^2\cdots x_n^n(1-x_1-2x_2 - \cdots - nx_n),$$ with $x_1,\ldots,x_n > 0$.

I computed the partial derivatives and set them equal to $0$ which gives me $x_i = 1$ so the extreme value is $1- n(n+1)/2$ but when I plot for the case $n=2$ it doesn't look like the extreme value occur at $x_1 = x_2 = 1$

  • $\begingroup$ Could you show your work? $\endgroup$
    – Alex Silva
    Jan 13, 2015 at 16:23

1 Answer 1


You have $f(x,1,1,...,1) = x(1-x -(2+3+\cdots+n))$, so it is easy to see that $\inf_{x >0} f(x,1,1,...,1) = -\infty$.

Since $x_k>0$, we see that $f(x_1,...,x_k) \ge 0$ iff $\langle (1,2,...,n)^T, x \rangle \le 1$. Since the set of $x$ for which $f$ is strictly positive is contained in the compact set $S=\{x | x_k \ge 0, \langle (1,2,...,n)^T, x \rangle \le 1 \}$, we see that $f$ is bounded above.

If we take $x_k = {1 \over n(n+1)}$, it is easy to check that $f(x_1,...,x_n) >0$. Furthermore, if any $x_k = 0$, we see that $f(x_1,...,x_k) = 0$, and so if $\max_{x \in S} f(x_1,...,x_k)$ is attained at some point $x$, we see that $x_k >0$.

Combining these facts shows that the $\max$ of $f$ (with $x_k>0$) is attained at a local unconstrained maximum of $f$, and so ${\partial f(x) \over \partial x} = 0$ at a maximizer.

We have ${\partial f(x) \over \partial x_k} = k x_1 x_2^2\cdots x_k^{k-1}\cdots x_n^n (1-x_1-\cdots -n x_n) - k x_1 x_2^2\cdots x_k^{k-1}\cdots x_n^n$, setting this to zero and using the fact that $x_k>0$ reduces to $x_k = (1-x_1-\cdots -n x_n)$, in particular, each $x_k$ has the same value.

Hence the maximizer is also a local maximizer of $f(x,x,...,x) = x^{{1 \over 2}n (n+1)} (1-{1 \over 2}n (n+1)x)$, setting the derivative to zero gives $x = {2 \over 2+n+n^2}$. Hence we see that $(x,....,x)^T$ is the maximizer.

However, if one were to ask me which value is 'extreme', I think I would go for $-\infty$ :-).

  • 1
    $\begingroup$ @AlexSilva: Thanks! It bothers me a little that answering many of these questions is 99% justification and 1% solving :-). $\endgroup$
    – copper.hat
    Jan 13, 2015 at 17:23

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