Question

Write 271 as the sum of positive real numbers so as to maximize their product.

Answer 1

Note that, to maximize the product, all of the numbers should be equal.
This follows from the Arithmetic Mean-Geometric Mean Inequality, which states that, for a set of non-negative real numbers, $x1,\dots ,xn$,
$\frac{(x1+\dots +xn)}{n}>=\frac{(x1\dots xn{)}^1/}{n},(AM>=GM)$
with equality if, and only if, $x1=\dots =xn$.
In this case, the sum of the numbers is fixed at 271. For a given set, S, of positive numbers, therefore, the arithmetic mean equals 27$\frac{1}{n}$. For each such set, the geometric mean, and hence the product, of the numbers is maximized when all of the numbers are equal.
Hence we seek the maximum value of y = ${\left(\frac{271}{x}\right)}^x$, where x is a positive integer.
taking log on both sides,
$lny=x.ln\left(\frac{271}{x}\right)=x(ln271-lnx)\left(\frac{1}{y}\right)y^{\prime }=ln271-lnx-1$
$y^{\prime }=\left(\frac{271}{x}\right)\times (ln271-lnx-1)$
Setting y' = 0, $x=\frac{271}{e}$ 99.7, which is clearly a maximum.
Now we need only try both 100 and 99 to confirm that 100 is the maximum value for y, when x is an integer.
Therefore the maximum product occurs when 271 = 2.71 + 2.71 + ... + 2.71. (100 equal terms.)