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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?

user642796
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user23709
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    no, it's a different question. In the other one the number of elements in the sum is not fixed. – mau Jun 03 '13 at 12:49

4 Answers4

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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.

Milind Hegde
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By symmetry of the expression and because there is no previlege of one variable the maximum is reached at $x_1=\cdots=x_6=10$

  • This argumentation is not true. It also shows, that $x_1 = \ldots = x_6 = 10$ is a local minimum. – gerw Jun 03 '13 at 19:11
  • @gerw To see that $x_1=\cdots=x_6=10$ gives a global maximum compare its value with the value given for example by $x_1=\cdots=x_5=1$ and $x_6=55$ –  Jun 03 '13 at 19:17
  • But you also inferred: "If the problem is symmetric, the solution is symmetric". This is not true. – gerw Jun 04 '13 at 07:07
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(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.

mau
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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.

Douglas B. Staple
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  • Hmm I wrote this thinking the solution was constrainted to be in the integers (which it turns out to be anyway). However for the proof to work with real $x_i$, you have to use the slightly different iteration by mau. – Douglas B. Staple Jun 03 '13 at 13:05