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

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** Calculus Alert **
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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