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 Mar 12 comment for what arguments the function reaches maximum? @Chris, do you need a proof or are you satisfied with a guess. By symmetry, I would also guess that $w_1=\ldots=w_n$ and $x_1 = \ldots = x_n$. An exact analysis anwering your question can be done with Lagrange multipliers, see en.wikipedia.org/wiki/Lagrange_multiplier Mar 6 awarded Editor Mar 6 comment for what arguments the function reaches maximum? @Chris: see my edit. Mar 6 revised for what arguments the function reaches maximum? added 140 characters in body; added 1 characters in body Mar 6 comment for what arguments the function reaches maximum? @Chris: your edit has not changed the message of your post at all. This I have answered. Your last comment says you have only two parameters, namely $w$ (= $w_i$ for all $i$) and $w/x$ (= $w_i/x_i$ for all $i$), which you could easily discuss by yourself. Mar 6 comment for what arguments the function reaches maximum? @Chris: but your problem seems ill-posed: already for $n=2$ you get $f=\infty$ for the values I state above. This can be already guessed as your domain is not compact, and so f need not attain a maximum. - Please let the original post, and write your reformulation as "Ed: .." as an addition to your present post. Mar 5 comment On the notion of partial semigroup @Salvo: mhm, does this also hold if we take condition 2) in your list and (x y) z is defined, but y z is not defined? - ED: ok, you mean that we know that we must interpret xyz as (xy)z because the other possibility makes no sense. Ok, I see. Good argument. It will be hard to work with such expressions practically, but it's interesting. Maybe they should indeed also be considered. Mar 5 answered for what arguments the function reaches maximum? Mar 5 comment What are the most attractive Turing undecidable problems in mathematics? nice, but does it answer the question Mar 5 awarded Commentator Mar 5 comment On the notion of partial semigroup @Salvo: Yes, it depends on what you like. There is probably no final choice, it depends. That is why we have also groups, groupoids etc. - A stronger reason for the above partial semigroup definition by @Andreas is that you do not need brackets any more in word expressions. I think that's the point for that definition. - On the other hand it is also somewhat limiting. For example, most (arbitrary) subsets of groups are not partial semigroups in this sense. Think about it in Z. Mar 4 comment Taking direct sums in $K$-theory in Kirchberg-Phillips classification @Ulrich: yes this fact I had in mind. One can even choose to be A a graph Cā-algebra, if one likes. ā Mar 4 comment Measure Preserving Maps from the Square to the Cube I am not familiar with that stuff but maybe you could approximate 2/3 binary and combine this with that what you know. For instance: $id \times f: R^2 \times R \rightarrow R^2 \times R^2$ brings you $R^3 \rightarrow R^4$. Doubling it with $f$ brings with bring $R^6 \rightarrow R^8$. Concatenating one $f$ brings $R^6 \rightarrow R^9$. Taking third root brings a first approximation $R^2 \rightarrow R^3$. Etc. - Maybe its a phantasy. Mar 4 asked Taking direct sums in $K$-theory in Kirchberg-Phillips classification Mar 3 comment What morphisms / Morita equivalences induce the 2-periodicity isomorphisms of KK-theory? How are $KK_{*+2k}$ for $k \neq 0$ are defined at all? Mar 2 awarded Teacher Mar 2 comment Mathematicians whose works were criticized by contemporaries but became widely accepted later @Joel: I still think the Black--Scholes formula has this mathematical flavour with fine and difficult probability theory. Your last sentence: ??? Mar 2 answered Mathematicians whose works were criticized by contemporaries but became widely accepted later Mar 2 answered On the descent homomorphsim of Kasparov equivariant KK theory Mar 2 awarded Nice Question