The dimensions of the enclosure that maximize the area are 100 feet by 50 feet. The area of the maximized enclosure is 5,000 ft².
How the dimensions are area are computed:
The requirement is to maximize the area of the rectangular enclosure. To achieve this, we need to divide the available 400 feet of fence into five equal parts.
Remember that one side of the enclosure will be formed by a building, requiring 0 feet of fence. Now we have 4 parts of 400/4 = 100 feet each for the exterior fences.
Let the length of the enclosure = L
Let the width = W
The two interior fences partitioning the enclosure into equal size rectangles will be parallel to the width of the enclosure. Therefore, the length of each rectangle will be L/3 and the width will be W/2.
The total length of the fence = L + 2W/3 + 2L/3 + 2W/3 = 100 * 4
Equation Simplification:
L + W = 200
The area of the enclosure, A = LW
Substituting L = 200 - W in the above equation:
A = W(200 - W)
To maximize the area, we need to differentiate the above equation with respect to W and equate it to zero:
dA/dW = 200 - 2W = 0
The solution to the above equation gives:
W = 100
The length of the enclosure becomes:
L = 200 - W = 100
Thus, we can conclude that the dimensions of the enclosure that maximize the area are 100 ft x 50 ft with the area of the maximized enclosure as A = LW = 100 * 50 = 5,000 ft².
Complete Question:
A total of 400 feet of fence is available for constructing a rectangular enclosure. One side of the enclosure will be formed by a building and will require 0 feet of fence. There will be three exterior fences and two interior fences partitioning the enclosure into equal size rectangles. What dimensions, length and width, of the enclosure maximizes the area of the enclosure? Calculate the area of the maximized enclosure.