Marie wishes to enclose a rectangular region in her backyard using part of her 50-foot long house as 1 side...

1. Translate the problem requirements: Marie has a 50-foot house wall as one side of a rectangle. She uses 80 feet of fencing for the other 3 sides. We need to find the length of the side along the house that gives maximum area.
2. Set up the constraint relationship: Express how the 80 feet of fencing is distributed among the three sides that need fencing.
3. Express area in terms of one variable: Use the constraint to write area as a function of a single dimension.
4. Find the maximum area configuration: Determine which dimensions give the largest possible area within our constraints.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we have and what we're looking for in everyday terms.

Marie has a house that's 50 feet long, and she wants to use part of this house wall as one side of a rectangular garden area. Think of it like this: she's creating a rectangle where one side is already "free" because it uses the existing house wall.

For the other three sides of this rectangle, she has exactly 80 feet of fencing material. So she needs to fence in three sides: two sides that are perpendicular to the house, and one side that's parallel to (and opposite from) the house wall.

We need to find the length of the side that runs along the house - this is the side that will give her the maximum possible area for her rectangular region.

Process Skill: TRANSLATE - Converting the real-world scenario into mathematical relationships

2. Set up the constraint relationship

Now let's think about how the 80 feet of fencing gets used up.

Let's call the length along the house \(\mathrm{x}\) feet, and the width (perpendicular to the house) \(\mathrm{w}\) feet.

The fencing is used for:

- Two sides of width \(\mathrm{w}\) (the sides perpendicular to the house)
- One side of length \(\mathrm{x}\) (the side parallel to the house, opposite the house wall)

So our constraint is: \(\mathrm{w} + \mathrm{w} + \mathrm{x} = 80\)
Or more simply: \(2\mathrm{w} + \mathrm{x} = 80\)

This means: \(\mathrm{w} = \frac{80 - \mathrm{x}}{2}\)

Also, since Marie can only use part of her house wall, we have: \(\mathrm{x} \leq 50\)

3. Express area in terms of one variable

The area of any rectangle is length × width.

In our case: \(\mathrm{Area} = \mathrm{x} \times \mathrm{w}\)

Substituting our constraint relationship:
\(\mathrm{Area} = \mathrm{x} \times \frac{80 - \mathrm{x}}{2}\)
\(\mathrm{Area} = \frac{\mathrm{x}(80 - \mathrm{x})}{2}\)
\(\mathrm{Area} = \frac{80\mathrm{x} - \mathrm{x}^2}{2}\)

Now we have area expressed in terms of just one variable, \(\mathrm{x}\).

4. Find the maximum area configuration

To find the maximum area, we need to think about what value of \(\mathrm{x}\) gives us the largest possible area.

Looking at our area formula: \(\mathrm{Area} = \frac{80\mathrm{x} - \mathrm{x}^2}{2}\)

This is a downward-opening parabola (because of the \(-\mathrm{x}^2\) term). For such expressions, the maximum occurs at the midpoint of the range where the expression is positive.

The expression \(80\mathrm{x} - \mathrm{x}^2\) is positive when \(\mathrm{x}\) is between 0 and 80.
So the maximum occurs at \(\mathrm{x} = 40\).

But wait! We have a constraint: \(\mathrm{x} \leq 50\) (Marie's house is only 50 feet long).

Since \(40 < 50\), our optimal value \(\mathrm{x} = 40\) is achievable.

Let's verify: When \(\mathrm{x} = 40\):

- Width \(\mathrm{w} = \frac{80 - 40}{2} = 20\) feet
- Area \(= 40 \times 20 = 800\) square feet

Process Skill: APPLY CONSTRAINTS - Checking that our mathematical solution respects the physical limitations of the problem

5. Final Answer

The length of the side of the rectangular region that is bounded by Marie's house is 40 feet.

This corresponds to answer choice D: 40.

We can verify this makes sense: Marie uses 40 feet of her house wall, and the fencing creates a rectangle that's 40 feet along the house by 20 feet deep, giving her the maximum possible area of 800 square feet.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding which side uses the house wall

Students often confuse which dimension represents the side along the house versus the sides perpendicular to it. They might set up the constraint as \(\mathrm{x} + 2\mathrm{w} = 80\) instead of \(2\mathrm{w} + \mathrm{x} = 80\), thinking the house provides two of the shorter sides rather than one of the longer sides.

2. Forgetting the house length constraint

Many students focus only on the 80 feet of fencing constraint and completely overlook that the house is only 50 feet long, meaning \(\mathrm{x} \leq 50\). This constraint becomes crucial when checking if the optimal mathematical solution is physically achievable.

3. Incorrectly counting the fenced sides

Students sometimes think they need to fence all four sides and set up the constraint as \(2\mathrm{x} + 2\mathrm{w} = 80\), forgetting that one side (along the house) requires no fencing at all.

Errors while executing the approach

1. Algebraic manipulation errors when substituting

When substituting \(\mathrm{w} = \frac{80-\mathrm{x}}{2}\) into the area formula, students often make errors like writing \(\mathrm{Area} = \mathrm{x}(80-\mathrm{x})\) instead of \(\mathrm{Area} = \frac{\mathrm{x}(80-\mathrm{x})}{2}\), dropping the division by 2.

2. Incorrect identification of the maximum point

For the quadratic \(\mathrm{Area} = \frac{80\mathrm{x} - \mathrm{x}^2}{2}\), students may incorrectly find the maximum by setting the derivative to zero but making calculation errors, or they might not recognize that this parabola opens downward and has its maximum at \(\mathrm{x} = 40\).

Errors while selecting the answer

1. Reporting the width instead of the length along the house

After correctly finding \(\mathrm{x} = 40\) and \(\mathrm{w} = 20\), students sometimes get confused about what the question is asking for and select 20 (the width) instead of 40 (the length along the house), especially since the question specifically asks for "the length of the side that is bounded by her house."

2. Selecting the total fencing length

Some students, after seeing their calculation involves the number 80, mistakenly think this represents the answer and might look for 80 in the choices, not realizing they need to find the specific dimension along the house wall.