Question

A farmer is building THREE side by side rectangular enclosures of equal size by first building a rectangular fence and then adding two additional parallel fences down the middle. If the farmer has 600 feet of fencing material available to him, what dimensions should he choose to maximize the total area of the entire enclosure.

Answer 1

To maximize the total area with 600 feet of fencing, the farmer should build three side-by-side rectangular enclosures. The optimal dimensions are: length 450 feet and width 150 feet for each section.

Let's denote the dimensions of the rectangular enclosure as follows:

- Length of the entire enclosure: L

- Width of one rectangular section: W

Since there are three side-by-side rectangular enclosures, the total length L is divided into three equal parts. Therefore, the length of one section is L/3.

The perimeter of one rectangular section is given by the sum of the lengths of all four sides:

`\[ P = 2\left((L)/(3)\right) + 2W \]`

The total fencing material available is given as 600 feet:

P = 600

1. Express the Perimeter (P) in Terms of L and W:

`\[ P = 2\left((L)/(3)\right) + 2W \]\[ 600 = (2L)/(3) + 2W \]`

2. Solve for L in Terms of W:

`\[ (2L)/(3) = 600 - 2W \]\[ 2L = 1800 - 6W \]\[ L = 900 - 3W \]`

3. Express the Area A in Terms of L and W:

`\[ A = L * W \]\[ A = (900 - 3W) * W \]\[ A = 900W - 3W^2 \]`

4. Maximize A by Finding the Critical Points:

`\[ (dA)/(dW) = 900 - 6W \]`

Setting the derivative equal to zero:

900 - 6W = 0

W = 150

5. Find the Corresponding L Value:

L = 900 - 3W

L = 900 - 3(150)

L = 450

So, the dimensions that maximize the total area are:

- Width W: 150 feet

- Length L: 450 feet

This means the farmer should use 450 feet of fencing for the length of the entire enclosure and 150 feet for the width of each of the three rectangular sections.

Answer 2

150 ft x 75 ft

Explanation:

A farmer is building 3 side-by-side rectangular enclosures of equal size by first building a rectangular fence and then adding two additional parallel fences down the middle.

Let x be the width of each rectangular closure.

Let y be the length of each rectangular closure.

(See the attached diagram).

Since the width of the entire enclosure is equal to 3x, the area (A) of the entire enclosure can be expressed as:

`A = 3xy`

The total amount of fencing available (600 ft) will be used for 6 widths and 4 lengths, so:

`6x + 4y = 600`

Rearrange this equation to isolate y:

`6x + 4y = 600`

`4y=600-6x`

`y=(600-6x)/(4)`

`y=150-(3)/(2)x`

Substitute this into the area equation:

`A = 3x\left(150-(3)/(2)x\right)`

`A = 450x-(9)/(2)x^2`

To find the value of x that maximizes A, take the derivative of A with respect to x:

`\frac{\text{d}A}{\text{d}x} = 450-9x`

Set it to zero and solve for x:

`450-9x=0`

`9x=450`

`x=50`

Therefore, the width of each rectangular enclosure that will maximize the total area of the entire enclosure is 50 ft.

To find the corresponding length (y), we can substitute x = 50 into the equation for the perimeter:

`6(50) + 4y = 600`

`300 + 4y = 600`

`4y = 300`

`y=75`

As the width of the entire enclosure is equal to 3x, then the dimensions that maximize the total area of the entire enclosure are:

• Width: 3x = 3 × 50 ft = 150 ft
• Length: y = 75 ft