A rectangular yard has an area of 24m^2. What are the possible perimeters for this yard?

When given the area of a rectangular yard, we use the formula for the area of a rectangle which is area = length times width. If the rectangular yard is 24 meters squared, then the options for whole number length and width pairs would be the following factor pairs of 24: 1 meter and 24 meters, 2 meters and 12 meters, 3 meters and 8 meters, or 4 meters and 6 meters. There are also rational number factor pairs such as 1.5 meters and 16 meters or 2.4 meters and 10 meters. Factor pairs are numbers that multiply together to get the desired number, in this case, 24.
once you have the factor pairs, you can answer the question about the perimeter of the rectangular yard. The perimeter of a rectangle is twice the length plus twice the width. In the example of the whole number pairs this would result in 2 times 1 plus 2 times 24 = 50 meters, 2 times 2 plus 2 times 12 = 28, 2 times 3 plus 2 times 8 = 22, and 2 times 4 plus 2 times 6 = 20.

We are given a rectangular yard with area of 24 square meters, and we are asked to find the possible perimeters.
Since the yard is a rectangle (a parallelogram with right angles), we can describe the dimensions as length and width. The area is given by A=lw and the perimeter by P=2l+2w.
Solving the equation of the area for the length l, noting that A=24, yields w=24/l .
The domain for l and w is all positive reals.
(1) There are an infinite number of possible perimeters. Start with any positive real number l; then, w is given by 24/l, and the perimeter will be 2l+2w.
For example, if l=6, then w=4, and the perimeter is 20 meters. If l=13, then w=24/13, and the perimeter is 386/13.
Another example would be the following: l=sqrt(2), w=24/sqrt(2), P=26sqrt(2)
(2) Possible perimeters with integer (whole number) sides:
1x24   P=502x12   P=283x8     P=224x6     P=20
These are the factorizations of 24.
(3) There is no largest perimeter, but there is a least perimeter.
We have P=2l+2(24/l) .
Using calculus, we can find the minimum perimeter to occur when the length and width are both equal to 2sqrt(6) .
This gives a perimeter of 8sqrt(6) approx 19.5959 .
(4) We can graph the possible perimeters: P=2l+48/l