http://primepuzzle.com/tunxis/product-and-sum-problem.txt

lrb - 10/8,9,11/2009

ref: http://www.yofx.org/?p=460

Product and sum problem (and its continuous analog)

*** Please note: For an .html version of this, please visit ***

http://primepuzzle.com/tunxis/product-and-sum-problem.html

Pick positive integers that sum to 20, like 2, 3, 3, 6, 6. Form the product
of these integers. In this case, the product would be 2*3*3*6*6, which
is equal to 648. What would you have to pick so that the sum is still 20
but the product is as large as possible?

The answer to this problem ends up being 3, 3, 3, 3, 3, 3, 2.

3*3*3*3*3*3*2=1458

Notice that these numbers are (almost) all the same number, namely 3.
And that there are 20/3 of them, where / stands for "integer division" (i.e.
drop remainders). That is, 20/3 is equal to 6.

So, 6*3+2=20 and 3^6*2 is maximal. 

The product is maximal when you pick the same number (almost) every
time.

Here's the "continuous analog" to the above problem.
 
Don't limit yourself to positive integers.

Can we find a single non-integral number, call it x, such that repeated
addition (we call this multiplication) of it a certain (probably
non-integral) number of times, namely 20/x "times," would give us the
same sum as before, namely

(20/x)*x=20

and the "product" of all these (now all identical) terms, namely

x^(20/x)

is maximal?

This "product" would not have an integral number of terms.

Since we're now dealing with a continuous function of a real variable we
can use calculus.

Let f(x)=x^(20/x)

It can be seen that f(x)=e^((20/x)*ln(x))   (take ln of both sides to see this)

f'(x)=e^((20/x)*ln(x))*((20/x)/x+ln(x)*(-20/(x^2))    (chain rule, product rule, etc.)

Setting this to 0, we get

20/(x^2)-ln(x)*20/(x^2)=0

So, ln(x)=1

So x=e

The maximal "product" is e^(20/e)=1568.05115

Recall that 3*3*3*3*3*3*2=1458

In a sense, we shrank each 3 down a bit (to 2.71828) and bumped the
single 2 up some (to 2.71828) to get a higher product. The integral
number of terms in the sum used to be 7 (six 3's and one 2). It is now
20/e, which equals 7.35758, which of course is not integral.

It's not hard to see that 20 has no real importance in this problem. Any
number will work. You'll just get more (or fewer) terms in the sum and the maximal
product will be bigger (or smaller). The cool thing is that the base of natural
logarithms, e, 2.71828, will always be the best value to pick for the
repeating term.