Respuesta :
Answer:
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
Step-by-step explanation:
