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

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...

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=50
2x12   P=28
3x8     P=22
4x6     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 `