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Of all the possible combinations of positive numbers that sum to 10, which has the largest multiplication?

I had also got a clue: it's related to e.

Please help! (I need explanation aswell)

Notice: i said positive not natural so you can use fractions.

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4 Answers 4

Suppose we have $a_1$, $a_2$, $\dotsc$, $a_n$ all positive and summing to $10$. Then by the AM–GM inequality their product is maximized when $a_1 = a_2 = \dotsb = a_n$. Since for $n \geq 10$ you'd have a product of numbers less than or equal to $1$, you only have to compute $(10/n)^n$ for $1 \leq n \leq 9$ and you find that the maximum occurs for $n=4$ with $2.5^4 = 39.0625$.

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The answers above give the usual methods. Here's a method I've come up with taking up your clue of a link with $e$.

You know that $$x+y+z+t = 10 \:\:\;\:\;\;\;\;\; x,y,z,t > 0$$ and wish to maximise $P=xyzt$. Take the natural logarithm to get $$\log P = \log x + \log y + \log z + \log t.$$ Now $P$ is maximised exactly when $\log P$ is maximised. So we may rephrase the problem as follows:

Find when the maximum of $f(x,y,z,t) = x + y + z + t$ occurs given the constraint $e^x + e^y + e^z + e^t = 10$.

Following the technique of Lagrange multipliers, we define the function $$g(x,y,z,t,\lambda) = x + y + z + t + \lambda(e^x + e^y + e^z + e^t - 10).$$ Then $$\partial_x g = 1 + \lambda e^x$$ $$\partial_y g = 1 + \lambda e^y$$ $$\partial_z g = 1 + \lambda e^z$$ $$\partial_t g = 1 + \lambda e^t$$ and $$\partial_{\lambda} g = e^x + e^y + e^z + e^t - 10.$$

To find when the maximum occurs we set the partial derivatives to be zero. From the first four partial derivatives we have $e^x=e^y=e^z=e^t = \frac{-1}{\lambda}$. Substituting this into the fifth partial derivative gives $-\frac{4}{\lambda}-10=0$, which gives $\lambda = -\frac{2}{5}$. This then gives $e^x=e^y=e^z=e^t = \frac{5}{2} = 2.5$, returning the result that the product is maximised when all four numbers are $2.5$.

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It is a simple sort of constrained optimization problem. Assume we have two positive numbers adding upto $10$, $x+y=10$, find $x$ and $y$ subject to $\max_{\forall x,y}\, xy$. If you rewrite this $\max_{\forall x,y}x(10-x)$ we have $\frac{d(10x-x^2)}{dx}=0$ then we have $x=5$. If we have 3 numbers or four numbers etc.. one can show that the product is maximized when $x=y=z=t=....$. Therefore one needs to check the number of numbers which will maximize the product. For $2$, namely $x+y=10$ we have $25$ and for $3$, namely $x+y+z=10$ we have $3x=10$ and $x=10/3$ so we have $10^3/3^3=37.04$. when we have $4$ numbers we have $10^4/4^4=39,06$ and $10^5/5^5=32$ as the function $10^k/k^k$ goes to $0$ when $k\rightarrow\infty$ we have $k=4$ which is the optimum solution which gives $39,06$

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Thanks alot! can you please explain to me what it has to do with e? –  Dan Barzilay Sep 13 '12 at 11:45
    
I think for this problem it doesnt have something to do with $e$. However for some sort of limits or maximization problems $e$ can give the optimum solution. –  Seyhmus Güngören Sep 13 '12 at 11:47
    
forget about the e, i dont understand how you knew that when the numbers are equal their multimplication is maximized? –  Dan Barzilay Sep 13 '12 at 11:51
    
$e$ has two definitions. $\lim_{n\rightarrow\infty}(1+\frac{1}{n})^n$ and $\sum_{n=0}^\infty \frac{x^n}{n!}$- May be the second equation looks like similar but I dont have an explicit clue. Sorry. –  Seyhmus Güngören Sep 13 '12 at 11:52
    
you can check en.wikipedia.org/wiki/AM-GM_inequality for that. –  Seyhmus Güngören Sep 13 '12 at 11:56

You were all wrong actually... i checked with my teacher and the asnwer is e 3.67879441 (10/e) times so: e^(10/e) = 39.5986256

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Ok so what are the numbers that you have. First how many numbers will you multiply? and second what are they? –  Seyhmus Güngören Sep 13 '12 at 12:42
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The question does not at all hint that having some of the terms/factors have non-integral multiplicities would be a valid possibility. That is a so nonstandard interpretation of "combinations of positive numbers" that it needs to be said explicitly. –  Henning Makholm Sep 13 '12 at 13:05
    
Also, those numbers multiply to $10$, not sum to $10$, so either your teacher is wrong, or the question/problem was phrased incorrectly. –  Cameron Buie Sep 13 '12 at 13:29
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@Cameron he means he has $e$ but $3.68$ times. Non integer number of elements. They add up to $10$. $4$ times $2.5$ is what we said and $3.68$ times $e$ is what he is saying. The problem is that OP doesn't imply any non integer numbers. $(\frac{10}{k})^k$ is maximized when $k=3.68$ and the corresponding number is $e$ –  Seyhmus Güngören Sep 13 '12 at 14:58

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