These days, people from all over the world learn how to live with the covid-19 as part of our life. In my country we are back to school but need to keep 2 meters between one student to another. This reality makes me think about these questions: What is the minimum area required for a classroom with X students in it? And what is the most efficient shape for this classroom? This is my solution to these questions, but I would be grateful to hear your thoughts about it.
In attempt to find the most efficient shape I started thinking about a circle. If we will draw a circle with radius of 2m around a student, the closest other students have to be on this circle. We can put max 6 students on this circle to keep 2m between them. If we will continue and drew a circle (with radius of 2m) around each student and put 6 students on each circle (symmetrically located), we will get pattern like this- a hexagon. I think this equilateral hexagon is the most efficient shape.
By drawing lines we can divide the hexagon to many equilateral triangles with side length of 2m, where the vertex represents students. Assume that we want to enter X students to an equilateral hexagon shape classroom.
N- the number of rings around the center (for example- at the image N=3)
How many students N rings can contain? the number of students at the 1st,2st,3st...N ring: 1,6,12,18...6N or 1, 6(1,2,3...N). the total number of students at N rings: X=1+6(1+2+3+...N). 1+2+3+...N = (1+N)N/2 so X=1 +3N(N +1) or X=3N^2 +3N +1 .
How many triangles there is in N rings? the number of tringles at the 1st,2st,3st...N ring: 6,18,30... . 6 is the first term and them each term is +12 to the previous one. the total number of triangles at N rings: Tri = N/2 * (12+12N-12) = 6N^2
the area of each triangle: A= √3 . the total area of N rings: area= Tri * A = 6√3 N^2 .
the connection between X (number of students) and N (number of rings): X=3N^2 +3N +1 or 3N^2 +3N +1 -X =0 . we need the positive solution of the quadratic equation- N= (√(12X-3) -3)/6
the connection between X (number of students) and area (the are of the hexagon class): area= 6√3 N^2 = 6√3 * [(√(12X-3) -3)/6]^2
after some algebra we get the final formula: area = 2√3 X -3√(4X-1) +1