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.
Please help! (I need explanation aswell)
Notice: i said positive not natural so you can use fractions.
 A: 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 $2.5^4 = 39.0625$.
A: 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 $39,06$
A: 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 $2.5$.
A: 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
