InterMath
Title
Minimum Perimeters
Problem Statement
If dimensions are whole numbers, find the minimum perimeter for a rectangle with an area of 25
square units. Repeat for rectangles of areas of 30, 35, 40, ... . In general, how can you predict the
dimensions of the rectangle with minimum perimeter as you continue working with larger whole
numbers?
Problem setup
How can I find the minimum perimeter of a rectangle based upon a given area?
Plans to Solve/Investigate the Problem
To solve this investigation, I will need to use my knowledge of the area and perimeter of a
rectangle. I know that the area of a rectangle is equal to the length multiplied by the width.
(A = l * w) The perimeter of a rectangle can be determined by finding the sum of all of the sides.
(P = s1 + s2 + s3 + s4 or P=2l + 2w) I need to look at the relationship between the perimeter and
area to determine which dimensions of a rectangle will yield the smallest perimeter.
This problem relates to an activity that I did with a group of 6th grade gifted students during my
student teaching. The students were told that they were caterers for a party. They were supposed
to determine which arrangement of tables would yield the most seating possibilities. The students
were instructed that each table represents 1 unit. All of the tables have to be arranged in one
figure. They cannot be separated into four or so groups of tables. In other words, they must all
share at least one side with another table. The tables cannot be arranged in a “U” shape either.
They must be arranged as either squares or rectangles. The students were given the number of
tables that they were supposed to arrange. (This number is equal to the area.) They then used
manipulatives to create the constructions and determine the perimeter. The students recorded
their findings in a data chart and were then asked to explain why this occurred.
Investigation/Exploration of the Problem
To solve this investigation I decided to use Microsoft Excel. I constructed a spreadsheet which
would include the length, width, area, and perimeter of a rectangle. I set up the area column by
inserting the formula A = length * width. The perimeter formula is P = (2*length) + (2*width).
As specified in the instructions, I began with a rectangle whose area is 25 units². I listed all
possible dimensions of the rectangle by using my knowledge of the factors. I came up with only
three possible dimensions for a rectangle with an area of 25 units² since the instructions specified
that only whole numbers could be used. Here is the chart that I developed:
Length
1
5
25
Width
Area
25
5
1
25
25
25
Perimeter
52
20
52
For a rectangle with an area of 25 units², I found that its length and width should both be equal to
5 units. This would result in a square with each side equal to 5 units. Therefore, all rectangles
whose area is a square number will yield a square in order to arrive at the smallest perimeter. The
length of each of the sides of the square will be equal to the square root of the area. The
perimeter of these rectangles will equal four times the side, since all sides will be equal. (P = 4s).
For example:
Area (length * width)
Length of Each Side
Perimeter (4 * s)
(s = √Area)
4
2
8
9
3
12
16
4
16
25
5
20
36
6
24
49
7
28
64
8
32
81
9
36
100
10
40
This chart shows how the formula that I developed can be used to find the perimeter of such
figures.
Now, I just have to determine how to find the smallest perimeter from a given area when it is not
a square number. I used Microsoft Excel to explore different areas and to try to find a pattern.
Using the same spreadsheet that I previously used, I began to try new areas. I began by listing all
of the factors of the number. Then, I used the formulas for area and perimeter to determine the
measurements. Here are my findings when given a rectangle with an area of 30 units²:
Length
1
2
3
5
6
10
15
30
Width
Area
30
15
10
6
5
3
2
1
30
30
30
30
30
30
30
30
Perimeter
62
34
26
22
22
26
34
62
I found that there were eight possible dimensions for a rectangle with this area. The smallest
perimeter possible will result from a 5 unit * 6 unit rectangle, or 6 unit by 5 unit rectangle.
Next, I tried a rectangle with an area of 35 units² and arrived with these results:
Length
Width
1
5
7
35
Area
35
7
5
1
35
35
35
35
Perimeter
72
24
24
72
I found four possibilities for the dimensions of the rectangle. However, the smallest perimeter
will result from a 5 unit * 7 unit rectangle, or 7 unit * 5 unit rectangle.
I then found that there are eight possible dimensions for a 40 units² rectangle and these results
can be seen in the chart below:
Length
1
2
4
5
8
10
20
40
Width
Area
40
20
10
8
5
4
2
1
40
40
40
40
40
40
40
40
Perimeter
82
44
28
26
26
28
44
82
After trying all of these different cases, I noticed a pattern in my findings. The smallest perimeter
will be found by using the dimensions which are the closest to result in a square. That is, the two
factors which are closest in number to each other and still multiply together to equal the area.
When listing out all possible dimensions in a chart like I did above, the smallest perimeter will
be those factors which are in the middle of the list. This holds true for those square numbers,
also. If you think of each unit as a block, the smallest perimeter will be acquired when these
blocks are arranged in a square format or in as close to a square format that can be made. The
picture below illustrates this idea:
These factors can also be found by simply listing all of the factors of a number. For example:
168
1 * 168
2 * 84
3 * 56
4 * 42
6 * 28
7 * 24
8 * 21
12 * 14
When using this method, the final factors in the list will yield the smallest perimeter of a
rectangle with that area. For a rectangle whose area is equal to 168 units², the smallest perimeter
will result from a rectangle with dimensions equal to 12 units * 14 units, or 14 units * 12 units.
In order to find the smallest possible perimeter, all that we have to do is factor.
Extensions of the Problem
This problem can be extended by looking at what dimensions would result in the largest
perimeter. I have already found through my investigations that this would be a rectangle that is 1
unit wide and x units long, where x is the number given by the area of the rectangle. If you think
of each unit as a block, the longest perimeter will result when all of the blocks are arranged in
one straight line. The diagram below illustrates this idea:
Another extension of this problem could be to look at the same concept using different figures.
For example, carry out the investigation with other quadrilaterals instead of rectangles.
Author & Contact
Andrea Heath
Andrea_heath@ecats.gcsu.edu