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A rectangular fence is given.
A line of fencing divides the rectangle in half beginning at the top of the outer perimeter and stretching down vertically to the bottom perimeter.
The length of the bottom perimeter of the fence is divided into two equal parts where the dividing fence meets.
Each of the two sections along the bottom perimeter of fencing is labeled x.
The width of the fence is labeled y.
please provide the correct answers

To maximize the area enclosed by the fencing, let's define the problem mathematically.

Given:
- Total fencing available: 160 feet.
- The fence forms two adjacent rectangular corrals with total dimensions of \(2x\) along the length and \(y\) for the width.

### Step-by-Step Solution

1. **Formulate the Perimeter Equation**:
   Since the fence uses two lengths of \(y\) and three segments of \(x\) (two along the bottom and one in the middle), the total fencing length equation is:
   \[
   2y + 3x = 160
   \]

2. **Express \(y\) in Terms of \(x\)**:
   Solving for \(y\):
   \[
   y = \frac{160 - 3x}{2}
   \]

3. **Set Up the Area Function**:
   The area \(A\) of the rectangle is given by:
   \[
   A = 2x \cdot y
   \]
   Substitute \(y\) from the previous step:
   \[
   A = 2x \cdot \frac{160 - 3x}{2}
   \]
   Simplifying:
   \[
   A = x(160 - 3x)
   \]
   \[
   A = 160x - 3x^2
   \]

4. **Maximize the Area**:
   To find the maximum area, we take the derivative of \(A\) with respect to \(x\) and set it to zero:
   \[
   \frac{dA}{dx} = 160 - 6x
   \]
   Setting \(\frac{dA}{dx} = 0\):
   \[
   160 - 6x = 0
   \]
   \[
   x = \frac{160}{6} = \frac{80}{3} \approx 26.67 \text{ feet}
   \]

5. **Calculate \(y\)**:
   Substitute \(x = \frac{80}{3}\) into the equation for \(y\):
   \[
   y = \frac{160 - 3 \cdot \frac{80}{3}}{2}
   \]
   Simplifying:
   \[
   y = \frac{160 - 80}{2} = \frac{80}{2} = 40 \text{ feet}
   \]

### Answer
The correct dimensions to maximize the enclosed area are:
- \( x = \frac{80}{3} \approx 26.67 \) feet
- \( y = 40 \) feet

Let me know if you need more details or have any questions!

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Here are some related questions to deepen your understanding:

1. How would the dimensions change if only one rectangular corral was needed?
2. What if the total available fencing was reduced to 100 feet?
3. How does the width-to-length ratio affect the maximized area?
4. How can we apply this approach to maximizing areas of other shapes, like circular pens?
5. How would the answer change if the fence was divided into three sections instead of two?

**Tip:** Always look for symmetry when optimizing area problems, as symmetry often leads to maximum values in such configurations.

A rancher has 160 feet of fencing with which to enclose two adjacent rectangular corrals. What dimensions (in ft) should be used so that the enclosed area will be a maximum?

Explore how to maximize the enclosed area of two adjacent rectangular corrals using 160 feet of fencing. This problem involves algebraic optimization and quadratic functions, perfect for high school students in Grades 10-12.

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