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

Let x-----> the length of rectangle y-----> the width of rectangle we know that  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-x² using a graph tool see the attached figure the vertex is the point (20.5,420.25) that means  for x=20.5 m ( length of rectangle) the area is 420.25 m² 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)  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 m² 400 m² < 420.25 m² so the answer Part 2) is  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 m² area using the barriers=400 m² 420.25-400=20.25 m² the answer part 3) is 20.25 m²


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Mathematics, added 2020-11-23 20:23:55

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