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

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.

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

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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
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
@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