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The problem you're working on involves DeShawn's design for two adjacent pens for his puppies. Let's break it down step by step:

### Step 1: Relationship Between the Combined Area and Width of the Pens

1. **Key Information**:
   - The total fencing available is **120 feet**.
   - The pens have a length of \(\frac{y}{2}\) each, and the width is \(x\).
   - The pens are adjacent, so the total length of both pens combined is \(y = \frac{y}{2} + \frac{y}{2}\).

2. **Perimeter Equation**: 
   The fencing includes two widths \(x\) for each pen (so \(2x\)) and three lengths \(\frac{y}{2}\) (since the middle length is shared). Thus, the total perimeter is:
   \[
   2x + y = 120
   \]
   Solve for \(y\) in terms of \(x\):
   \[
   y = 120 - 2x
   \]

3. **Area Equation**:
   The combined area of the two pens is:
   \[
   A = 2 \times \left(\frac{y}{2} \times x\right) = yx
   \]
   Substituting \(y = 120 - 2x\) from the perimeter equation:
   \[
   A = x(120 - 2x) = 120x - 2x^2
   \]
   This quadratic equation represents the relationship between the area \(A\) and the width \(x\).

### Step 2: Finding Maximum Area

To find the dimensions that maximize the area, we'll take the derivative of the area function and set it equal to 0.

1. **Derivative of \(A(x) = 120x - 2x^2\)**:
   \[
   \frac{dA}{dx} = 120 - 4x
   \]
   Set the derivative equal to 0 to find the critical points:
   \[
   120 - 4x = 0
   \]
   \[
   x = 30
   \]

2. **Find \(y\) when \(x = 30\)**:
   Substituting \(x = 30\) into the perimeter equation:
   \[
   y = 120 - 2(30) = 60
   \]

3. **Maximum Area**:
   The maximum combined area is:
   \[
   A = 30(60) = 1800 \text{ square feet}
   \]

### Step 3: Type of Function

The function \(A(x) = 120x - 2x^2\) is a **quadratic function**. Since the coefficient of \(x^2\) is negative, the parabola opens downwards, meaning the function has a maximum point. This is why it makes sense that the area is maximized at a certain width.

### Step 4: Impossible Lengths and Widths

There are certain constraints on the values of \(x\) and \(y\):

1. Since the total fencing is 120 feet, neither \(x\) nor \(y\) can be negative.
2. If \(x = 0\), the pens would have no width, which is impossible.
3. Similarly, if \(x = 60\), the length \(y = 120 - 2(60)\) would be zero, resulting in no length, which is also impossible.

Thus, \(x\) must lie between 0 and 60 for the design to be feasible.

---

Would you like me to generate a table or graph for the function \(A(x)\)? If yes, let me know the preferred range for \(x\).

### Related Questions:
1. What is the range of valid values for \(x\) based on the fencing limit?
2. How would the area change if the available fencing were increased to 150 feet?
3. Can you verify that the critical point \(x = 30\) gives the maximum area by checking the second derivative?
4. How would the design change if there were three adjacent pens instead of two?
5. What other shapes could DeShawn use for the pens to maximize the area with the same amount of fencing?

**Tip**: When maximizing area, quadratic functions often represent such problems, and the maximum occurs at the vertex of the parabola.

DeShawn's dog, Almondine, has a growing litter of puppies. DeShawn wants to separate them into two pens that are adjacent, with the same length and width. He has 120 feet of fencing available. The pens have a length of y/2 each and a width of x. Model the relationship between the combined area (A) of the two pens and the width of the pens (x). Include a table, graph, and equation. Additionally, find the dimensions that maximize the area, and determine the type of function A is.

Explore how to model the relationship between the area and dimensions of two adjacent pens using quadratic functions. This problem involves deriving equations for perimeter and area, optimizing the area using derivatives, and graphing the results. Suitable for high school students (Grades 9-12).

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