A rectangular enclosure is to be created using 82m rope.
A) What are the dimensions of the max area
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
A)20.5m by 20.5m
B)No. 20.5 m cannot be created using 2-m barriers.
c) 20.25 m 2
Ouch. Ms Sue's answer revealed my typo: 20.5 x 20.5
A) To find the dimensions of the rectangular enclosure with the maximum area using an 82m rope, we can start by considering that the perimeter of a rectangle is given by the formula P = 2(l + w), where "l" represents the length and "w" represents the width.
In this case, we know that the total length of the rope is 82m, which corresponds to the perimeter of the rectangle. So we can write the equation as follows:
82m = 2(l + w)
To find the dimensions that will maximize the area, we can express the area of the rectangle as A = lw and then rewrite it in terms of a single variable. In this case, we can use the equation for the perimeter to isolate one of the variables, and substitute it into the area equation.
Let's solve for l in the perimeter equation:
82m = 2(l + w)
Divide both sides by 2:
41m = l + w
Subtract w from both sides:
l = 41m - w
Now substitute this expression for l into the area equation:
A = l × w
A = (41m - w) × w
To find the maximum area, we can differentiate the area equation with respect to w and set it equal to zero. This will give us the critical points where the area has a maximum value.
dA/dw = 41m - 2w
Setting this expression equal to zero:
41m - 2w = 0
Rearranging the equation:
2w = 41m
Divide both sides by 2:
w = 41m/2
w = 20.5m
Now substitute this value of w back into the equation for l:
l = 41m - w
l = 41m - 20.5m
l = 20.5m
Therefore, the dimensions of the rectangular enclosure with the maximum area using an 82m rope are length (l) = 20.5m and width (w) = 20.5m.
B) If 41 barriers, each 2m long are used instead, we can calculate the total length of the barriers needed for the same enclosure.
Each barrier is 2m long, so the total length of the barriers will be:
Total length = 41 barriers × 2m/barrier
Total length = 82m
Since the total length of the barriers is equal to the total length of the rope, it is possible to enclose the same area with the 41 barriers as with the 82m rope.
C) To calculate the additional area that can be enclosed with the rope instead of the barriers, we need to find the difference in the areas of the two enclosures.
The area of the enclosure using the 82m rope is given by:
Area = length × width
Area = 20.5m × 20.5m
Area = 420.25m²
The area of the enclosure using the barriers can be calculated using the same formula but with the dimensions adjusted for the barriers. Since each barrier is 2m long, the length and width of the enclosure will be reduced by twice the number of barriers used.
Adjusted length = length - 2 barriers
Adjusted width = width - 2 barriers
Therefore, the area of the enclosure using the barriers is:
Area = (20.5m - 2 barriers) × (20.5m - 2 barriers)
Area = (20.5m - 2(41)) × (20.5m - 2(41))
Area = (20.5m - 82m) × (20.5m - 82m)
Area = (-61.5m) × (-61.5m)
Area = 3,772.25m²
The additional area that can be enclosed by using the rope instead of the barriers is the difference between the two areas:
Additional area = Area with rope - Area with barriers
Additional area = 420.25m² - 3,772.25m²
Additional area = -3,352m²
Therefore, the use of the rope instead of the barriers results in 3,352 square meters less area being enclosed.
A) a square: 21.5 x 21.5
B) clearly not, since the square is not a multiple of 2 on a side. Also, 41 cannot be divided evenly by 4.