You have a total of 820 feet of fencing to enclose a large retangular larea and divide it into four smaller pens of the same dimensions. The fencing used to divide the large pen must be parallel to the same side of the large rectangle. Maximize the total rectangular area (other solutions found are not clear and one I believe to be wrong or really unclear)
area= LW
820=2L+2W+3L
820=5L+2W or w=410-5/2 L
area=L(410-5/2 L)
darea/dL=0=(410-5/2L)+ L(-5/2)
solve for L, then W
50l x 50 w
To maximize the total rectangular area, we need to determine the dimensions of the large rectangle and the four smaller pens.
Let's proceed step by step:
Step 1: Let's assume the dimensions of the large rectangle are x (length) and y (width). Since the same fencing is used to divide the large pen into four smaller pens, we need to consider the length of fencing required for each division.
To divide the large rectangle into four smaller pens, we have three divisions. One division will be created by the width (y) of the rectangle, and the other two divisions will be created by the length (x). Therefore, the length of the fencing used for division is 3x.
To find the length of remaining fencing, we need to consider the perimeter of the large rectangle (2x + 2y) minus the fencing used for division (3x). So, the remaining fencing is: 2x + 2y - 3x.
Step 2: Now, let's use the given information that the total fencing available is 820 feet. We can set up an equation to represent this:
2x + 2y - 3x = 820
Simplifying the equation, we get:
2y - x = 820
Step 3: To proceed further, we need another equation that relates the area of the large rectangle to its dimensions (x and y).
The area of the large rectangle (A) is given by the formula: A = x * y.
Step 4: Since we want to maximize the area, we need to express the area (A) in terms of a single variable. Using the equation obtained in Step 2, we can solve for x as x = 2y - 820.
Substituting this value of x in the area equation, we get:
A = (2y - 820) * y
Step 5: Now, we have an equation for the area of the large rectangle (A) in terms of a single variable (y). To find the maximum value of A, we can take the derivative of A with respect to y, set it equal to zero, and solve for y.
dA/dy = 0
Step 6: Solve for y by taking the derivative and setting it equal to zero:
dA/dy = 4y - 820 = 0
4y = 820
y = 205
Step 7: Now that we have the value of y, we can substitute it back into the equation x = 2y - 820 to find the value of x:
x = 2(205) - 820 = 410 - 820 = -410
However, we cannot have negative dimensions in this context, so let's discard this solution.
Step 8: Now we have the values of x and y which maximize the total rectangular area. The length of the large rectangle is 410 feet, and the width of the large rectangle is 205 feet.
To find the dimensions of each smaller pen, we divide the large rectangle by 2 in both directions:
Length of smaller pen = x/2 = 410/2 = 205 feet
Width of smaller pen = y/2 = 205/2 = 102.5 feet
Therefore, the dimensions of each smaller pen would be approximately 205 feet by 102.5 feet.
Step 9: Finally, we can calculate the maximum total rectangular area by multiplying the dimensions of the large rectangle:
Total rectangular area = (410 feet) * (205 feet) = 84,050 square feet.
Thus, to maximize the total rectangular area, we would enclose a large rectangle with dimensions 410 feet by 205 feet and divide it into four smaller pens with dimensions approximately 205 feet by 102.5 feet. The maximum total rectangular area would be 84,050 square feet.