Question Find the rectangular field dimensions that maximize the enclosed area given 800 feet of fencing to create 3 identical smaller plots. Express answers as reduced fractions.

Studdy Solution

STEP 1

Assumptions
1. The rancher has 800 feet of fencing.
2. The field is rectangular.
3. The field is subdivided into 3 identical smaller rectangular plots with two additional fences parallel to one of the shorter sides.
4. We need to maximize the enclosed area.
5. We will use calculus to find the maximum area.

STEP 2

Let's denote the length of the shorter side of the entire rectangle as $x$ and the length of the longer side as $y$ . Since there are two additional fences parallel to the shorter side, the total length of fencing used is $3x + 2y$ .

STEP 3

We can write the equation for the perimeter using the total length of fencing:
$3x + 2y = 800$

STEP 4

We want to express $y$ in terms of $x$ to eventually find a function for the area that depends only on $x$ . We can rearrange the perimeter equation to solve for $y$ :
$y = \frac{800 - 3x}{2}$

STEP 5

The area $A$ of the entire rectangular field is given by the product of $x$ and $y$ :
$A = x \cdot y$

STEP 6

Substitute the expression for $y$ from STEP_4 into the area equation:
$A(x) = x \cdot \left(\frac{800 - 3x}{2}\right)$

STEP 7

Simplify the area function:
$A(x) = \frac{800x - 3x^2}{2}$

STEP 8

To maximize the area, we need to find the critical points of $A(x)$ . This is done by taking the derivative of $A(x)$ with respect to $x$ and setting it equal to zero.

STEP 9

Calculate the derivative of $A(x)$ :
$A'(x) = \frac{d}{dx}\left(\frac{800x - 3x^2}{2}\right)$

STEP 10

Apply the derivative:
$A'(x) = \frac{800}{2} - \frac{3 \cdot 2x}{2}$

STEP 11

Simplify the derivative:
$A'(x) = 400 - 3x$

STEP 12

Find the critical points by setting $A'(x)$ equal to zero:
$400 - 3x = 0$

STEP 13

Solve for $x$ :
$x = \frac{400}{3}$

STEP 14

We need to verify that this critical point is a maximum. We can do this by using the second derivative test or by analyzing the sign of the first derivative before and after the critical point.

STEP 15

Calculate the second derivative of $A(x)$ :
$A''(x) = \frac{d}{dx}(400 - 3x)$

STEP 16

Apply the derivative:
$A''(x) = -3$

STEP 17

Since $A''(x)$ is negative, the function $A(x)$ is concave down, which means the critical point at $x = \frac{400}{3}$ is a maximum.

STEP 18

Now that we have the value of $x$ that maximizes the area, we can find the corresponding value of $y$ using the equation from STEP_4:
$y = \frac{800 - 3x}{2}$

STEP 19

Substitute $x = \frac{400}{3}$ into the equation for $y$ :
$y = \frac{800 - 3\left(\frac{400}{3}\right)}{2}$

STEP 20

Simplify the expression for $y$ :
$y = \frac{800 - 400}{2}$

STEP 21

Calculate the value of $y$ :
$y = \frac{400}{2}$
$y = 200$

STEP 22

The dimensions that maximize the enclosed area are $x = \frac{400}{3}$ feet for the shorter side and $y = 200$ feet for the longer side.

STEP 23

To provide the answer in fractions reduced to lowest terms, we express $x$ as a fraction:
$x = \frac{400}{3} = \frac{400}{3} \cdot \frac{1}{1}$

STEP 24

Since $x$ is already in its lowest terms, we can write the final dimensions:
The dimensions that maximize the enclosed area are $x = \frac{400}{3}$ feet and $y = 200$ feet, or in terms of fractions, $x = \frac{400}{3}$ feet and $y = \frac{200}{1}$ feet.

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