what are all the rectangular pens you can build with 16 yds of fencing
What size pens? For turtles, rabbits, cattle, or elephants?
Using only whole numbers for lengths and widths
1 by 7
2 by 6
3 by 5
4 by 4 ---> notice that the sum is always 8
if you allow sides with fractional measurement, then there would be an infinite number of cases
e.g.
2.5 by 5.5
.....
3.412 by 4.588
.....
etc --------> notice that the sum is always 8
To find all the rectangular pens you can build with 16 yards of fencing, you need to determine all the possible combinations of length and width that satisfy the given condition.
To begin with, let's understand the formula to calculate the perimeter of a rectangle. The perimeter of a rectangle is given by the equation: 2 * (length + width).
Since we have 16 yards of fencing, we can set up the equation as follows:
2 * (length + width) = 16
To explore different options, we can set some conditions. For example, we can assume that the length is greater than or equal to the width. Additionally, we can set the range of values for length and width to be positive integers, starting from 1.
Now, let's calculate the different rectangular pens using trial and error.
If length = 1, then 2 * (1 + width) = 16, which results in width = 7. This gives us a rectangular pen with dimensions 1 yard by 7 yards.
If length = 2, then 2 * (2 + width) = 16, which results in width = 6. This gives us a rectangular pen with dimensions 2 yards by 6 yards.
If length = 3, then 2 * (3 + width) = 16, which results in width = 5. This gives us a rectangular pen with dimensions 3 yards by 5 yards.
If length = 4, then 2 * (4 + width) = 16, which results in width = 4. This gives us a rectangular pen with dimensions 4 yards by 4 yards.
If length = 5, then 2 * (5 + width) = 16, which results in width = 3. This gives us a rectangular pen with dimensions 5 yards by 3 yards.
If length = 6, then 2 * (6 + width) = 16, which results in width = 2. This gives us a rectangular pen with dimensions 6 yards by 2 yards.
If length = 7, then 2 * (7 + width) = 16, which does not yield a whole number for width when solving the equation. Therefore, there is no valid solution for this case.
Therefore, considering the given conditions, we can build rectangular pens with the following dimensions using 16 yards of fencing:
- 1 yard by 7 yards
- 2 yards by 6 yards
- 3 yards by 5 yards
- 4 yards by 4 yards
- 5 yards by 3 yards
- 6 yards by 2 yards