Calculate Size of Small Rectangles to Fill Max Area in Encompassing Rectangle

  • Context: Undergrad 
  • Thread starter Thread starter kleinma
  • Start date Start date
  • Tags Tags
    Geometry
Click For Summary
SUMMARY

The discussion focuses on calculating the dimensions of smaller rectangles that can fit within a larger rectangle while maintaining their height/width ratio. The formula derived is based on the area of the larger rectangle (L x W) and the number of smaller rectangles (n). The dimensions of each smaller rectangle are determined using the equation (L/sqrt(n)) x (W/sqrt(n)), ensuring that the aspect ratio remains constant. This approach provides a practical solution for real-world applications where maximizing area is essential.

PREREQUISITES
  • Understanding of basic geometry and area calculations
  • Familiarity with algebraic equations and ratios
  • Knowledge of aspect ratios in rectangles
  • Ability to manipulate square roots in mathematical expressions
NEXT STEPS
  • Research geometric area optimization techniques
  • Learn about aspect ratio preservation in design
  • Explore algebraic methods for solving area-related problems
  • Investigate practical applications of rectangle packing in real-world scenarios
USEFUL FOR

Mathematicians, engineers, designers, and anyone involved in optimizing space utilization within defined boundaries.

kleinma
Messages
92
Reaction score
0
The issue at hand, is I will have a rectangle of size x. This rectangle will contain n smaller rectangles. Each of these smaller rectangles is the same size though.

I need to figure out how to calculate the SIZE that the smaller rectangles should be, based on

1) the size of the encompassing rectangle
2) the number of smaller rectangles that go inside it

so that I can fill the maximum area inside the main rectangle, without altering the height/width ratio of the smaller rectangles (ie they can grow in size, but they can't be sized in a way that it distorts their height/width ratio, so some area in the encompassing rectangle will not be filled in, and this is ok)

both these values are known when I need to do the calculation.


Here are a few images to illustrate this (note sizes of red rectangles in each image are always the same). Is there any specific geometric formula to do this?
 

Attachments

  • example1.gif
    example1.gif
    1.5 KB · Views: 520
  • example2.gif
    example2.gif
    1.9 KB · Views: 501
  • example3.gif
    example3.gif
    4.4 KB · Views: 512
Mathematics news on Phys.org
I think you could fit an infinite number of rectangles in the larger, either the smaller shave to be infitly small or the larger infinitly big =]
 
That wasn't what I was asking...

This is for a real world problem, so I need a real world answer.. not theoretical assumptions.
 
Well, the "size" of rectangle is called area and its two dimensional. So let's call the rectangle's area L x W. And suppose we want to fit n smaller equivalent rectangles inside the big one.

The dimension of each of the smaller rectangles can be (L/sqrt(n)) x (W/sqrt(n)).
 
kleinma said:
That wasn't what I was asking...

This is for a real world problem, so I need a real world answer.. not theoretical assumptions.

I was just joking around a bit =]

but I think you could go along the lines of the ratio of the original rectangle and then just solve alegraicly. such as

(L1)(W1)=x(L2)(W2)
where L1/W1=L2/W2

x represents the number of smaller rectangles you want
L1 is the length of big rectangle
W1 is the width of the big rectangle
L2 is the length of smaller rectangles
W2 is the width of the smaller rectangle

the L1/W1=L2/W2 should assure the proper ratios

I think this should be right if I read this right, but you said something about theoretically assumptions, and depending on what you want x to be, it could be infinitely small. But I g2g so I have to stop here.
 
Last edited:
You're asking what are the different ways to arrange a bunch of videos onto a screen without changing their aspect ratios right?
 
Alex48674 said:
(L1)(W1)=x(L2)(W2)
where L1/W1=L2/W2

I think this should be right if I read this right,

Thats correct. I got the same results too...
 

Similar threads

Replies
8
Views
5K
  • · Replies 5 ·
Replies
5
Views
9K
  • · Replies 24 ·
Replies
24
Views
4K
Replies
8
Views
3K
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
31
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K