Solution to puzzle 8: 271
Write 271 as the sum of positive real numbers so as to maximize their product.
Solution 1
Firstly, we 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, ... , xn},
(x1 + ... + xn)/n 
(x1 · ... · xn)1/n,
with equality if, and only if, x1 = ... = 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 271/n, where n is the cardinality of S. 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 = (271/x)x, where x is a positive integer.
Treating x as real for the moment, we can use logarithmic differentiation
ln y = x · ln(271/x) = x (ln 271 − ln x)
(1/y) y' = ln 271 − ln x − 1
y' = (271/x)x (ln 271 − ln x − 1)
Setting y' = 0, x = 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.)
Solution 2
Another approach is to note that we seek the least integer, x, such that
(271/x)x > (271/(x+1))x+1.
Expanding both sides, 271x/xx > 271x+1/(x+1)x+1.
Dividing by 271x and rearranging, (x+1)x+1/xx > 271.
Simplifying, we seek the least x such that x(1 + 1/x)x+1 > 271.
We know that x is reasonably large, in which case (1 + 1/x)x+1 e. So we seek the least x such that x > 271/e 99.7.
This approach gives an intuitive feel as to why the terms (2.71, in this case) are close to e, and can be made more rigorous.
Further reading
Source: Problem of the Week 911 on The Math Forum @ Drexel