Questions: A rancher has 800 feet of fencing to put around a rectangular field and then subdivide the field into 3 identical smaller rectangular plots by placing two fences parallel to one of the field's shorter sides. Find the dimensions that maximize the enclosed area. Write your answers as fractions reduced to lowest terms.

Solution Steps

To maximize the enclosed area of the rectangular field with a given perimeter, we can use calculus and the concept of optimization. The total length of the fencing is 800 feet. Let the length of the field be \( L \) and the width be \( W \). The perimeter constraint is given by \( 2L + 4W = 800 \) because the field is subdivided into 3 smaller plots with two additional fences parallel to the width. We need to express the area \( A = L \times W \) in terms of one variable using the perimeter constraint, and then find the critical points by taking the derivative and setting it to zero to find the maximum area.

We have a rectangular field with a total fencing of 800 feet, which is subdivided into 3 identical smaller rectangular plots. Let \( L \) be the length and \( W \) be the width of the field. The perimeter constraint can be expressed as: \[ 2L + 4W = 800 \]

From the perimeter equation, we can solve for \( L \): \[ L = 400 - 2W \]

The area \( A \) of the rectangular field can be expressed as: \[ A = L \times W = (400 - 2W) \times W = 400W - 2W^2 \]

To find the maximum area, we differentiate \( A \) with respect to \( W \): \[ \frac{dA}{dW} = 400 - 4W \]

Setting the derivative equal to zero to find critical points: \[ 400 - 4W = 0 \implies W = 100 \]

Substituting \( W = 100 \) back into the expression for \( L \): \[ L = 400 - 2(100) = 200 \]

Final Answer