[Kevin Watson]

Question: Is there an optimum stack height?

Answer: Yes, there is.

Let n represent the total number of containers
in our scoring area. For the sake of our
calculations, this is a constant value.

Let x represent the number of containers that we
use to make our stack.

Then n-x equals the number of containers in the
scoring area not used in building the stack.

Our score, excluding "king-of-the-hill" points,
can then be calculated by multiplying our stack
height by the number of containers not used in
building our stack. Creating a mathematical
expression for this we get:

score = x(n-x)
= nx-x^2
= -x^2 + nx

So the (oft quoted) quadratic coefficents are:

a=-1
b=number of containers in our alliance's scoring area
c=number of "king-of-the-hill" points

** Calculus Alert **

Now to optimize our score we take the derivative
of the above equation...

score' = -2x + n

...and then solve for x when score' is held at zero.

-2x + n = 0
-2x = -n
x = -n/-2
x = n/2

What this expression says is that we will get the
maximum score if we use half of the available
containers in the scoring area to build our stack.

Homework question:

This falls apart if n is odd. Afterall, how do you
stack half of a container?!? So my question is simply:
Given an odd number of containers in your scoring
area, do you put that last, odd-numbered container on
the stack or not?

Hint:

Well, first you might just make a table that relates
the two possible scores for each case (n=3,5,7,etc).
Then if you're still curious, you might want to do a
mathematical proof to convince yourself that this
still holds true for a million and one containers in
the scoring area.

-Kevin