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 40000
40000
ft2
2
. 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 $84/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.
cost=84(2W + L)
but LW=40000 to L=40000/W
cost=84(2W+40000/W)
minimum cost for a particular W will be when the derivative of cost with resjpect to W is zero...or
0=84(2 -40000/W^2) or
solve for W. Then solve for L. Then solve for cost at that L,W
You will note that the fence needs to be divided equally between lengths and widths. So, the minimum fence length is for a field that is 200√2 x 100√2
To find the least amount of money needed to build the fence, we need to determine the dimensions of the field that will minimize the amount of fencing used.
Let's assume the length of the rectangular field is L and the width is W.
The area of the rectangular field is given as 40000 ft², so we have the equation:
L * W = 40000
Since we only need to build three walls of fencing, the perimeter of the field (excluding the wall of the gymnasium) will be:
Perimeter = 2 * (L + W)
To minimize the amount of fencing used, we can use calculus. Let's differentiate the perimeter equation with respect to one variable (either L or W), and then find the value of L or W that makes this derivative equal to zero.
Differentiating the perimeter equation with respect to L:
d(Perimeter)/dL = 2 * (dL/dL + dW/dL) = 2 * (1 + dW/dL)
Setting this derivative equal to zero:
2 * (1 + dW/dL) = 0
1 + dW/dL = 0
dW/dL = -1
Similarly, differentiating the perimeter equation with respect to W:
d(Perimeter)/dW = 2 * (dL/dW + dW/dW) = 2 * (dL/dW + 1)
Setting this derivative equal to zero:
2 * (dL/dW + 1) = 0
dL/dW + 1 = 0
dL/dW = -1
From the above two equations, we have:
dW/dL = -1
dL/dW = -1
Simplifying, these equations imply that W/L = -1.
Since we know L * W = 40000, we can substitute W = -L into this equation:
L * (-L) = 40000
-L² = 40000
L² = -40000
L = ±√(-40000)
However, since dimensions cannot be negative, we discard the negative solution. So, L = √(40000) ≈ 200 ft.
Substituting this value of L into the equation W = 40000 / L gives:
W = 40000 / 200 = 200 ft.
Therefore, the length and width of the rectangular field that will minimize the amount of fencing used are both 200 ft.
Now, to find the least amount of money needed to build the fence, we need to calculate the perimeter of the field:
Perimeter = 2 * (L + W) = 2 * (200 + 200) = 2 * 400 = 800 ft.
Finally, we can calculate the cost of the fencing:
Cost = Perimeter * Cost per ft = 800 ft * $84/ft = $67200.
So, the least amount of money needed to build this fence of ideal area is approximately $67200.