Maximizing Efficiency in Two-Dimensional Packing: A Comprehensive Guide

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The discussion focuses on maximizing efficiency in two-dimensional packing for a school project, specifically relating to area rather than volume. Key considerations include the shapes being packed, such as circles, rectangles, or triangles, which influence packing efficiency. Participants suggest exploring packing efficiencies of lattice networks, which have been extensively studied in 3D contexts. Understanding the basic outline of the packing shapes is crucial for gathering relevant data. This foundational knowledge will aid in effectively addressing the project requirements.
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There is this school project on efficiency in packing but i can't actually find any good data .

The main idea is the amount of goods that can be fixed in a place but i want it in just 2 dimentions. It has to be related to area and not to volume as that becomes too complex.

Any help in understanding this topic is welcome.
 
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It depends on what basic outline you want to use. As in, do you want to pack circles in a square or rectangles or triangles etc. The 3D problem has been studied to a large extent in chemistry. Look up packing efficiencies of lattice networks.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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