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To find the dimensions that yield the maximum area for a rectangular room with a fixed perimeter, we can use some principles from optimization.

1. **Define the Problem**:
   The perimeter \( P \) of a rectangle with length \( L \) and width \( W \) is given by:
   \[
   P = 2L + 2W
   \]
   Here, \( P = 324 \) feet.

2. **Set Up the Equation**:
   Using the given perimeter, we can rewrite this as:
   \[
   2L + 2W = 324
   \]
   Dividing both sides by 2, we get:
   \[
   L + W = 162
   \]
   Now, solve for \( W \):
   \[
   W = 162 - L
   \]

3. **Express the Area**:
   The area \( A \) of the rectangle is:
   \[
   A = L \times W
   \]
   Substitute \( W = 162 - L \):
   \[
   A = L \times (162 - L)
   \]
   Expanding this gives:
   \[
   A = 162L - L^2
   \]

4. **Find the Maximum Area**:
   To maximize the area, we can take the derivative of \( A \) with respect to \( L \) and set it to zero:
   \[
   \frac{dA}{dL} = 162 - 2L = 0
   \]
   Solving for \( L \):
   \[
   2L = 162
   \]
   \[
   L = 81
   \]
   Substitute \( L = 81 \) back into \( W = 162 - L \):
   \[
   W = 162 - 81 = 81
   \]

5. **Conclusion**:
   The dimensions that yield the maximum area are \( L = 81 \) feet and \( W = 81 \) feet, making the shape a square.

**Answer**:
- **Length (ft) = 81**
- **Width (ft) = 81**

Would you like more details or have any questions?

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### Additional Related Questions
1. What would be the area of this room with these dimensions?
2. Why does a square maximize the area for a given perimeter?
3. How would the dimensions change if the perimeter were 400 feet instead?
4. What would the length and width be if the room had to be twice as long as it is wide?
5. How can this method be applied to maximize other shapes, like circles or triangles?

**Tip**: For rectangles with a fixed perimeter, the square configuration always provides the maximum area.

A carpenter is building a rectangular room with a fixed perimeter of 324 ft. What dimensions would yield the maximum area?

Explore how to find the dimensions that yield the maximum area for a rectangular room with a fixed perimeter of 324 feet. Learn the optimization techniques, key algebraic and geometric concepts, and understand why a square configuration maximizes area.

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