Maximum value of a product How to write the number $60$ as $\displaystyle\sum^{6}_{i=1} x_i$ such that $\displaystyle\prod^{6}_{i=1} x_i$ has maximum value?

Thanks to everyone :)
Is there a way to solve this using Lagrange multipliers?
 A: Hint: Use the AM-GM Inequality, which says:
$$\frac{\sum_{i=1}^{n}x_i}{n} \geq\sqrt[n]{\prod_{i=1}^{n}x_i}$$
Here the left hand side is a constant, 60/6=10, and the right hand side is what you want to maximise. Equality holds when all the $x_i$'s are equal.
A: By symmetry of the expression and because there is no  previlege of one variable the maximum is reached at $x_1=\cdots=x_6=10$
A: (of course the $x_i$ must be positive, otherwise the product may be as great as you want)
Hint: if you have $x_i \ne x_j$, substitute both with their arithmetic mean.
A: You can prove it without explicitly invoking the AM-GM inequality by first supposing that the product is not maximal when $x_i = 10\ \forall i$. Then, when the product is maximal, $x_i \neq x_j$ for some $i,j$. Then one of $x_i$ and $x_j$ is larger; the other smaller; suppose $x_i < x_j$. The proof is completed by showing that letting $x_i \rightarrow x_i+1$ and $x_j \rightarrow x_j-1$ increases the product, which contradicts the assumption that the product was maximal.
