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Stack Height Scoring Analysis
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 |
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