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The problem you're working on involves a rectangular region enclosed by 400 feet of fencing. You need to express the area of the region, \( A(w) \), as a function of its width, \( w \).

Here’s how you can set it up:

### 1. Understanding the perimeter constraint:
The perimeter \( P \) of a rectangle is given by the formula:
\[
P = 2l + 2w
\]
where \( l \) is the length and \( w \) is the width. You're given that the total perimeter (fencing) is 400 feet, so:
\[
2l + 2w = 400
\]
Simplifying:
\[
l + w = 200
\]
This implies:
\[
l = 200 - w
\]

### 2. Expressing the area:
The area \( A \) of the rectangle is given by:
\[
A = l \times w
\]
Substitute the expression for \( l \) from earlier:
\[
A(w) = (200 - w) \times w
\]
Simplifying further:
\[
A(w) = 200w - w^2
\]

Thus, the area as a function of width \( w \) is:
\[
A(w) = 200w - w^2
\]

This is the required function.

Would you like further explanation or have any other questions?

### Here are some related questions:
1. How do you find the width that maximizes the area of the rectangle?
2. What is the maximum possible area for this rectangle?
3. How would the function change if you were given more or less than 400 feet of fencing?
4. Can this problem be extended to non-rectangular shapes?
5. What happens to the area when the width \( w \) approaches 0 or 200?

### Tip:
To maximize the area, you could take the derivative of the function \( A(w) = 200w - w^2 \), set it equal to zero, and solve for \( w \).

You want to enclose a rectangular region with 400 feet of fencing. Express A(w), the area of the region as a function of its width.

Learn how to express the area A(w) of a rectangular region enclosed by 400 feet of fencing as a function of its width, w. This problem covers key algebraic and geometric concepts and explores quadratic functions. Suitable for students in Grades 9-11.

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