Arectangular 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 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

The volume is 56.8 triangular the