Largest product What is the largest possible product of a collection of positive real numbers which sums to exactly 27?
I got (3)^9=19683
.Is this correct?
Thanks in Advance.
 A: Observe first of all that the product of two numbers whose sum is fixed is maximum if the numbers are equal. Hence we can restrict ourselves to the case when we have $n$ equal numbers: each of them is $27/n$ and their product is $(27/n)^n$.
To find the value of $n$ for which this is maximum, consider the function
$$
f(x)=\ln\left({27\over x}\right)^x=x(\ln27-\ln x).
$$
This reaches a maximum when $f'(x)=0$, that is when
$$
\ln27-\ln x-1=0
\quad\hbox{and}\quad
x={27\over e}\approx9.93.
$$
As we need an integer value for $x$, we can check $x=9$ and $x=10$ to find that $f(10)>f(9)$.
A: If a sum of $m$ terms equals $\sum a_i = 27$ then by AM-GM $\sqrt[m]{\prod a_i}\le \frac {\sum a_i}m$ with equality holding only if the $a_i$ terms are equal.
So we may assume to get the maximum product that the terms are equal.
So $a_i = \frac {27}m$.  And we need to find the maximum value of $(\frac {27}{m})^m$
This will occur precisely when $f(m)=\ln({27}{m})^m = m(\ln 27 - \ln m)$ is maximum.
Which occurs when $f'(m)=(\ln 27 - \ln m) - m\frac 1m=\ln 27 - 1 - \ln m$ is closest to zero and $f''(m) = -\frac 1m < 0$.
$\ln 27 - \ln m -1=0 \implies \ln m = \ln 27 - 1 \implies m = 27/e \approx 10$.
So $2.7*10 = 27$ and $2.7^{10} = 20589.1132094649$ is higher than $3^9 = 19683$.  However I would guess that if we require the terms be integers then $3^9$ would be the highest possible value. 
However if we do not require the sum have an integer number of terms and can involve a poduct as well, we could have $m = 27/e$ and the terms are $e*\frac {27}e = 27$ and $e^{\frac {27}e} = 20593.791206982240416202028351902$ would be the absolute highest.  (But that really can't qualify as a "sum", can it?)
