A builder has 600 feet of fencing to enclose three adjacent rectangular partioned areas. Find the largest possible enclosed area of the partioned areas.
I will assume the maximum area is achieved with three adjacent rectangular areas with total length (end to end) = y and width = x.
Total area = x * y
2y + 3x = 600
A = x * (300 - 1.5 x) = 300 x -1.5 x^2
dA/dx = 0
300 = 3x
x = 100
y = (600 - 300)/2 = 150
Total area = x*y = 15,000
I just realized that the equation that relates x and y to the total length of material should be
2y + 4x = 600.
You need 4 fence segments of length x to create 3 areas.
Therefore, y = 300 - 2x
A = x*(300 - 2x) = 300x - 2x^2
Since dA/dx = 0,
4x = 300 and x = 75.
y = 150
Maximum total area = x*y = 11,250
You say you are takng a precalculus course, but I have used calculus in this derivation.
The same result can be obtained by "completing the square" with the A(x) function.
To maximize the enclosed area, we need to find the dimensions of the partitioned areas that give us the largest possible combined area.
Let's assume the width of each partitioned area is 'x' feet.
The total length of fencing used for the partitions will be 2x (one on each side) for each partitioned area. Since we have three partitions, the total length used for partitions is 6x feet.
The remaining fencing is used for the outer boundary. So, the length of the outer boundary is 600 - 6x feet.
To find the dimensions that maximize the enclosed area, we need to express the area 'A' in terms of 'x' and find the maximum value of 'A'.
The length of each partitioned area is equal to the length of the outer boundary, which is 600 - 6x feet.
Therefore, the area of each partitioned area 'A_partition' is given by:
A_partition = (600 - 6x) * x
The total enclosed area 'A' is the sum of the areas of the three partitioned areas:
A = 3 * A_partition = 3 * (600 - 6x) * x
Now, let's find the value of 'x' that maximizes 'A'.
To do this, we need to find the critical points. We can take the derivative of 'A' with respect to 'x' and set it to zero:
dA/dx = 0
Differentiating the equation for 'A' with respect to 'x' gives:
dA/dx = 3 * (600 - 6x) + 3 * x(-6) = 0
Simplifying the equation:
1800 - 18x - 18x = 0
Collecting like terms:
-36x + 1800 = 0
-36x = -1800
x = -1800 / -36
x = 50
Now, we need to check whether this critical point corresponds to a maximum or minimum. To do this, we can analyze the second derivative of 'A' with respect to 'x':
d^2A/dx^2 = -36
Since the second derivative is negative, it means the critical point at x = 50 corresponds to a maximum.
Therefore, to maximize the enclosed area, each partitioned area should have a width of 50 feet.
The length of the outer boundary is 600 - 6x = 600 - 6(50) = 600 - 300 = 300 feet.
The area of each partitioned area is (600 - 6x) * x = (600 - 6(50)) * 50 = 300 * 50 = 15000 square feet.
The total enclosed area is 3 * 15000 = 45000 square feet.
To find the largest possible enclosed area, we need to determine the dimensions of the rectangular partitions that would maximize the area.
Let's assume the three rectangular partitioned areas have lengths L1, L2, and L3 and widths W1, W2, and W3, respectively.
To determine the dimensions of the partitions, we need to set up an equation based on the given total length of the fencing.
The total length of the fencing is the sum of the perimeters of the three partitioned areas:
2L1 + 2W1 + 2L2 + 2W2 + 2L3 + 2W3 = 600
Simplifying the above equation, we get:
2(L1 + W1 + L2 + W2 + L3 + W3) = 600
(L1 + W1 + L2 + W2 + L3 + W3) = 300
Now, we need an equation for the area of the partitioned areas:
Area = L1 * W1 + L2 * W2 + L3 * W3
We want to maximize the area, so we need to express the area in terms of one variable without losing generality. Let's express W1, W2, and W3 in terms of L1 and L2:
W1 = (300 - L1 - L2 - W2 - L3 - W3) / 2
W2 = (300 - L1 - L2 - W1 - L3 - W3) / 2
W3 = (300 - L1 - L2 - W1 - L3 - W2) / 2
Now, substitute these expressions for W1, W2, and W3 into the equation for the area:
Area = L1 * ((300 - L1 - L2 - W2 - L3 - W3) / 2) + L2 * ((300 - L1 - L2 - W1 - L3 - W3) / 2) + L3 * ((300 - L1 - L2 - W1 - L3 - W2) / 2)
Simplifying further, we have:
Area = (300L1 - L1^2 - L1L2 - L1W2 - L1L3 - L1W3) / 2 + (300L2 - L1L2 - L2^2 - L2W1 - L2L3 - L2W3) / 2 + (300L3 - L1L3 - L2L3 - L3^2 - L3W1 - L3W2) / 2
Area = 150L1 + 150L2 + 150L3 - (L1^2 + L2^2 + L3^2 + L1L2 + L1L3 + L2L3) / 2 - (L1W2 + L1W3 + L2W1 + L2W3 + L3W1 + L3W2) / 2
To maximize the area, we need to maximize this equation. This can be done by taking partial derivatives with respect to L1, L2, and L3, and setting them equal to zero. Solving these equations will give us the values of L1, L2, and L3 that maximize the area.
However, solving these equations analytically can be quite complex. Therefore, we can use numerical methods or optimization software to find the values of L1, L2, and L3 that maximize the area.
Once we have the values of L1, L2, and L3, we can calculate the corresponding widths W1, W2, and W3 using the equations derived earlier.
Finally, the largest possible enclosed area of the partitioned areas can be calculated as the sum of the individual areas:
Largest Area = L1 * W1 + L2 * W2 + L3 * W3