# 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

Let 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

Only authorized users can leave an answer!

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.

More questions