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The problem asks to find the length and width of a rectangle with an area of 200 square feet and whose perimeter is minimized.

Let's denote:
- \( l \) as the length of the rectangle,
- \( w \) as the width of the rectangle.

### Step 1: Use the area constraint
The area of the rectangle is given by:
\[
A = l \cdot w = 200 \, \text{square feet}.
\]
This gives the relation:
\[
l \cdot w = 200.
\]
Solve for \( l \):
\[
l = \frac{200}{w}.
\]

### Step 2: Express the perimeter
The perimeter \( P \) of a rectangle is given by:
\[
P = 2l + 2w.
\]
Substitute \( l = \frac{200}{w} \) into the perimeter equation:
\[
P(w) = 2\left(\frac{200}{w}\right) + 2w = \frac{400}{w} + 2w.
\]

### Step 3: Minimize the perimeter
To minimize the perimeter, we take the derivative of \( P(w) \) with respect to \( w \):
\[
\frac{dP}{dw} = -\frac{400}{w^2} + 2.
\]
Set the derivative equal to zero to find the critical points:
\[
-\frac{400}{w^2} + 2 = 0.
\]
Solve for \( w \):
\[
\frac{400}{w^2} = 2 \quad \Rightarrow \quad w^2 = 200 \quad \Rightarrow \quad w = \sqrt{200} \approx 14.14 \, \text{feet}.
\]

### Step 4: Find the corresponding length
Now that we have \( w = \sqrt{200} \), substitute it back into the equation \( l \cdot w = 200 \) to find \( l \):
\[
l = \frac{200}{\sqrt{200}} = \sqrt{200} \approx 14.14 \, \text{feet}.
\]

### Conclusion
The rectangle that minimizes the perimeter while having an area of 200 square feet is a square with both the length and width approximately 14.14 feet.

Would you like further details or clarifications?

Here are some related questions you might find interesting:
1. How can we confirm that this solution gives the minimum perimeter?
2. What other shapes might minimize perimeter for a fixed area?
3. What is the significance of the derivative in optimization problems?
4. How can you generalize this problem for any area?
5. What is the geometrical interpretation of a rectangle with minimum perimeter?

**Tip:** When minimizing or maximizing quantities, always check the second derivative to ensure it corresponds to a minimum or maximum value.

Find the length and width of a rectangle that has an area of 200 square feet and whose perimeter is a minimum.

Learn how to find the length and width of a rectangle with an area of 200 square feet while minimizing its perimeter. This problem involves key geometry and optimization concepts, including the use of derivatives to find the critical points. Suitable for high school students in Grades 10-12.

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