Math Problem Statement

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.

Solution

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)A(w), as a function of its width, ww.

Here’s how you can set it up:

1. Understanding the perimeter constraint:

The perimeter PP of a rectangle is given by the formula: P=2l+2wP = 2l + 2w where ll is the length and ww is the width. You're given that the total perimeter (fencing) is 400 feet, so: 2l+2w=4002l + 2w = 400 Simplifying: l+w=200l + w = 200 This implies: l=200wl = 200 - w

2. Expressing the area:

The area AA of the rectangle is given by: A=l×wA = l \times w Substitute the expression for ll from earlier: A(w)=(200w)×wA(w) = (200 - w) \times w Simplifying further: A(w)=200ww2A(w) = 200w - w^2

Thus, the area as a function of width ww is: A(w)=200ww2A(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 ww approaches 0 or 200?

Tip:

To maximize the area, you could take the derivative of the function A(w)=200ww2A(w) = 200w - w^2, set it equal to zero, and solve for ww.

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Math Problem Analysis

Mathematical Concepts

Algebra
Geometry
Quadratic Functions

Formulas

Perimeter of a rectangle: P = 2l + 2w
Area of a rectangle: A = l × w
Quadratic function: A(w) = 200w - w^2

Theorems

Optimization of quadratic functions

Suitable Grade Level

Grades 9-11