# A rectangular enclosure is to be created using 82m rope.

a) What are the dimensions of the rectangular with Maximum area

b) Suppose 41 barriers each 2m long, are used instead. Can the same area be enclosed

C) How much more area can be enclosed if the rope is used instead of the barriers

## Answers

If you are not satisfied with the answer or you canâ€™t find one, then try to use the search above or find similar answers below.

Find similar answersMathematics, added 2021-06-22 23:40:58

If a car can travel 435 miles on 25 gallons of gas, how many miles per gallon does the car get? ...

Mathematics, added 2021-06-22 21:52:12

Mathematics, added 2021-06-22 20:37:22

Mathematics, added 2021-06-22 12:00:15

Which equation describes the function shown on the graph? ...

Mathematics, added 2021-06-22 10:21:41

Which equation describes the function shown on the graph? ...

Mathematics, added 2021-06-21 01:23:49

This is one page worth 10 points there will be more send answers asap ...

Mathematics, added 2021-06-20 22:24:34

This is one page worth 10 points there will be more send answers asap ...

Mathematics, added 2021-06-20 10:18:47

Mathematics, added 2021-06-20 09:04:00

Answered by

calculistaLet x-----> the length of rectangle y-----> the width of rectangle we know thatÂ perimeter of rectangle=2*[x+y] perimeter of rectangle=82 m 82=2*[x+y]---> divide by 2 both sides---> 41=x+y--> y=41-x---> equation 1 Area of rectangle=x*y substitute equation 1 in the area formula Area=x*[41-x]----> 41x-xÂ² using a graph tool see the attached figure the vertex is the point (20.5,420.25) that means Â for x=20.5 m (Â length of rectangle) the area is 420.25 mÂ² y=420.25/20.5----> 20.5 m the dimensions are 20.5 m x 20.5 m------> is a square the answer part 1)Â the dimensions of the rectangular with Maximum area is a square with length side 20.5 meters Part 2)b) Suppose 41 barriers each 2m long, are used instead. Can the same area be enclosed? divide the length side of the square by 2 so 20.5/2=10.25--------> 10 barriers the dimensions are 10 barriers x 10 barriers 10 barriers=10*2---> 20 m the area enclosed with barriers is =20*20----> 400 mÂ² 400 mÂ² < 420.25 mÂ² so the answer Part 2) isÂ the area enclosed by the barriers is less than the area enclosed by the rope Part 3)How much more area can be enclosed if the rope is used instead of the barriers area using the rope=420.25 mÂ² area using the barriers=400 mÂ² 420.25-400=20.25 mÂ² the answer part 3) is 20.25 mÂ²