High School

Lyon is attempting to build a fence for the largest possible rectangular play area at the local dog park. He can use up to 500 linear feet of fencing. He determines that if he makes a rectangular play area with a length of 103 or 147 feet, the play area would cover 15,504 square feet. To maximize the area, he can create a play area that is 25,625 square feet with a length of 125 feet. The relationship between the length, [tex]x[/tex], and the area, [tex]y[/tex], of the rectangular play area can be modeled with a quadratic equation.

9. Identify the vertex and another point from the context.
- Vertex: [tex](125, 25,625)[/tex]
- Point: [tex](147, 15,504)[/tex]

10. Find the value of [tex]a[/tex] for the quadratic equation that models the relationship.

11. Write the quadratic equation that models the relationship in vertex form.

Answer :

Sure! Let's solve the given problem step by step to find the quadratic equation that models the relationship between the length and area of the rectangular play area.

### Step 1: Identify the Vertex and a Point

From the problem, we have:
- Vertex: The largest possible area occurs when the length is 125 feet, and the area is 25,625 square feet. So, the vertex is [tex]\((125, 25625)\)[/tex].
- Another Point: When the length is 147 feet, the area is 15,504 square feet. So, we have another point [tex]\((147, 15504)\)[/tex].

### Step 2: Set Up the Vertex Form of the Quadratic Equation

The vertex form of a quadratic equation is given by:
[tex]\[ y = a(x - h)^2 + k \][/tex]
where [tex]\((h, k)\)[/tex] is the vertex. Here, [tex]\(h = 125\)[/tex] and [tex]\(k = 25625\)[/tex], so our equation becomes:
[tex]\[ y = a(x - 125)^2 + 25625 \][/tex]

### Step 3: Use Another Point to Find the Value of [tex]\(a\)[/tex]

We will use the point [tex]\((147, 15504)\)[/tex] to find the value of [tex]\(a\)[/tex].

Substituting [tex]\(x = 147\)[/tex] and [tex]\(y = 15504\)[/tex] into the equation, we have:
[tex]\[ 15504 = a(147 - 125)^2 + 25625 \][/tex]

Now solve for [tex]\(a\)[/tex]:
1. Calculate the difference: [tex]\(147 - 125 = 22\)[/tex].
2. Square the difference: [tex]\(22^2 = 484\)[/tex].
3. Substitute back: [tex]\(15504 = a \cdot 484 + 25625\)[/tex].
4. Subtract 25625 from both sides: [tex]\(15504 - 25625 = a \cdot 484\)[/tex].
5. Simplify: [tex]\(-10121 = a \cdot 484\)[/tex].
6. Solve for [tex]\(a\)[/tex]: [tex]\(a = \frac{-10121}{484}\)[/tex].

[tex]\[ a \approx -20.911 \][/tex]

### Step 4: Write the Quadratic Equation

Using the found value of [tex]\(a\)[/tex], write the equation in vertex form:
[tex]\[ y = -20.911(x - 125)^2 + 25625 \][/tex]

This quadratic equation models the relationship between the length [tex]\(x\)[/tex] and the area [tex]\(y\)[/tex] of the rectangular play area.

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