My Math Forum Calculating the Maximum Number of Dominoes

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February 18th, 2019, 04:02 PM   #1
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Calculating the Maximum Number of Dominoes

Figure out the relationships between the columns to solve for the Layers, x, y, base, and dominoes values for the maximum amount of dominoes. I will say in advance that some or all of the numbers are not whole. I was able to do this by setting a derivative equal to 0. Somebody else did it with the vertex form of a parabola. I want to see how people here do it.
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 February 18th, 2019, 05:25 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,373 Thanks: 1276 If you know ahead of time you've got a polynomial fit then if your data points are evenly spaced successive differences can get you the order of that fit. $\begin{array}{cccc} 1428\\ 1500 &-72\\ 1508 &-8 &-64\\ 1452 &56 &-64\\ 1332 &120 &-64 \end{array}$ This shows there is a quadratic relationship between the dominoes and the rings. There are a variety of ways to solve for the coefficients. The relationship turns out to be $dominoes = -32 ring^2 + 168 ring + 1292$ Thanks from EvanJ
 February 18th, 2019, 06:38 PM #3 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 637 Thanks: 85 You did some of it. I want to see how people solve for the maximum amount of dominoes.
February 18th, 2019, 07:06 PM   #4
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Quote:
 Originally Posted by EvanJ You did some of it. I want to see how people solve for the maximum amount of dominoes.
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 February 19th, 2019, 09:02 AM #5 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 637 Thanks: 85 I wasn't looking for the origin of your equation. Here's what I did.: 1. Base = 4y + 4 2. Layers = (54 - y)/2 3. Dominoes = Base * Layers = -2y^2 + 106y + 108 4. Derivative = -4y + 106 5. Set the derivative equal to 0. It makes y = 26.5. 6. Substitute this into Step 3 to get 1,512.5 dominoes. Somebody else did: "With each new ring added, you decreased the amount of layers by 2, and added 4 dominoes to each side of the ring's base. Since the rings had 4 sides, you added a total of 16 dominoes to the ring's base with each 2 layer decrease. This can be simplified down to an addition of 8 dominoes for each 1 layer decrease. By these deductions, I created the formula *y = (17 - x)(84 + 8x)*, with the 17 and 84 coming from the amount of layers and dominoes in the base layer of the 1st ring, as shown in the table. By distributing, this formula becomes y = -8x^2 + 52x + 1428, and by converting this quadratic equation into its vertex form, it becomes *y = -8(x - 3.25)^2 + 1512.5*. This equation's vertex lies at (3.25, 1512.5), showing that a layer decrease of 3.25 (or 13.75 layers in total), creates the ring with the most dominoes, which would have 1512.5 dominoes in total. Of course, you can't have a quarter of a layer, or half of a domino, so by rounding up, the ring with the most dominoes would have a layer decrease of 3 (or 14 layers in total) and 1512 dominoes in total.﻿" What method do you like better?

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