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.