A rectangular recreational field needs to be built outside of a gymnasium. Three walls of fencing are needed and the fourth wall is to be a wall of the gymnasium itself. The ideal area for such a field is exactly 250000ft2. In order to minimize costs, it is necessary to construct the fencing using the least amount of material possible. Assuming that the material used in the fencing costs $53/ft, what is the least amount of money needed to build this fence of ideal area? Round your answer to the nearest two decimal places.
Well, I must say, building a recreational field outside a gymnasium is a great idea. I always love a good game of dodgeball. Now, let's calculate the least amount of money needed for this fencing masterpiece.
We know that the area of the field is 250000ft². Since it is a rectangular field, we can assume that the length and width are unknown but they multiply to give us 250000ft².
Let's call the length "L" and the width "W". Therefore, L × W = 250000ft².
Now, keep in mind that we only need to construct three walls of fencing. The fourth wall is already taken care of by the gymnasium wall itself. So, we only need to calculate the perimeter of the rectangular field.
The perimeter of a rectangle is given by the formula: P = 2L + 2W.
Since we already know that L × W = 250000ft², we can rewrite the formula as:
P = 2(L + W).
Now, let's minimize the costs by using the least amount of material possible. The cost of the fencing material is $53/ft.
To find the least amount of money needed, we need to minimize the perimeter of the fencing.
Since the perimeter is P = 2(L + W), we want to minimize L + W.
To minimize the sum L + W while keeping the product L × W constant, we need to make L and W as close to each other as possible. In other words, we need to make the field as close to a square shape as we can.
So, let's find the square with an area of 250000ft². The square root of 250000 is approximately 500. Therefore, we can assume that L = 500ft and W = 500ft.
Plugging these values into the formula for the perimeter, we get:
P = 2(500 + 500) = 2 × 1000 = 2000ft.
Now, we can calculate the cost of the fencing using the least amount of material possible:
Cost = $53/ft × 2000ft.
Calculating that gives us $106,000.
Therefore, the least amount of money needed to build this fence of the ideal area is approximately $106,000.
And while we're at it, can I join in on the games at the recreational field? I promise I won't juggle any dodgeballs.
To find the least amount of money needed to build the fence, we need to determine the dimensions of the rectangular field.
Let's assume the length of the field is "L" and the width is "W".
Since three walls need to be fenced and the fourth wall is an existing wall, the perimeter of the field is given by:
Perimeter = 2W + L
The area of the rectangular field is given by:
Area = L * W
Given that the ideal area for the field is 250,000 ft^2, we can set up the following equation:
250,000 = L * W --(1)
To minimize the cost, we need to minimize the perimeter, as that will minimize the amount of fencing material needed.
To minimize the perimeter, let's solve equation (1) for L in terms of W:
L = 250,000 / W
Substituting this expression for L in the perimeter equation, we get:
Perimeter = 2W + (250,000/W)
Next, we'll differentiate the perimeter equation with respect to W and set the derivative equal to zero to find the value of W that minimizes the perimeter:
d(Perimeter) / dW = 2 - 250,000 / W^2 = 0
Solving this equation for W, we get:
2 = 250,000 / W^2
W^2 = 250,000 / 2
W^2 = 125,000
W = sqrt(125,000) ≈ 353.55
Since W represents the width of the field, it cannot be negative. Therefore, we discard the negative square root.
So, the width of the field (W) is approximately 353.55 ft.
Now, we can substitute this value back into equation (1) to find the length (L):
250,000 = L * 353.55
L = 250,000 / 353.55 ≈ 707.11 ft
The length of the field (L) is approximately 707.11 ft.
Finally, we can calculate the perimeter by substituting the values of L and W into the perimeter equation:
Perimeter = 2W + L = 2 * 353.55 + 707.11 ≈ 1414.22 ft
Now, we can calculate the cost of the fencing:
Cost = Perimeter * $53/ft = 1414.22 * 53 ≈ $74,999.66
Therefore, the least amount of money needed to build the fence of ideal area is approximately $74,999.66.
To determine the least amount of money needed to build the fencing, we need to find the dimensions of the rectangular field that result in the ideal area.
Let's assume the length of the rectangular field is L and the width is W.
We know that the area of a rectangle is given by the formula A = L * W.
Given that the ideal area is 250000ft^2, we have the equation:
250000 = L * W
To minimize costs, we want to minimize the amount of fencing material used. Since we only need to build three walls of fencing, the amount of fencing required can be calculated as follows:
Fencing = 2L + W
We need to find the values of L and W that satisfy the area equation while minimizing the amount of fence required.
To do this, we can use techniques of calculus, specifically optimization. We can rewrite the equation for the amount of fencing as:
Fencing = 2L + 250000/L
To minimize this expression, we take the derivative with respect to L and set it equal to zero:
d(Fencing)/dL = 2 - 250000/L^2 = 0
2 = 250000/L^2
L^2 = 250000/2
L^2 = 125000
L = √125000
L ≈ 353.55 ft
Since the rectangular field does not have a fence on one side (due to the gymnasium wall), the dimensions are L ≈ 353.55 ft and W = 250000 / L ≈ 250000 / 353.55 ≈ 706.91 ft.
Now, we can calculate the amount of fencing required:
Fencing = 2L + W
Fencing ≈ 2(353.55) + 706.91 ≈ 1413.82 ft
Finally, to find the cost of the fencing, we multiply the total length of the fencing by the cost per foot ($53):
Cost = Fencing * $53
Cost ≈ 1413.82 ft * $53 ≈ $74,926.46
Therefore, the least amount of money needed to build the fence of the ideal area is approximately $74,926.46, rounded to the nearest two decimal places.
along building = w
out from building = L
total length = w + 2 L = p
area = A = w L =250,000
so L = 250,000/w
then the fencing perimeter
p = w + 2 L = w + 500,000/w
dp/dw = 0 for minimum
dp/dw = 1 - 500,000/w^2 = 0
w = sqrt (500,000) = 707.1
L = 353.6
p = w + 2 L = 1414.2
cost = 1414.2*53 = 74,952.96
If the width is w, the length is 250000/w, so the perimeter is
p = 2w + 250000/w
dp/dw = 2 - 250000/w^2
dp/dw=0 when w = 500/√2 = 353.6 ft
So, the least fencing needed is 1414.2 ft