Find the dimensions of a rectangle (in m) with area 216 m2 whose perimeter is as small as possible. (Enter the dimensions as a comma separated list.)

To minimize the perimeter of a rectangle with a given area, the rectangle should be a square. Here's how we can solve this:

### Given:
- Area of the rectangle \( A = 216 \, \text{m}^2 \)
- We want to minimize the perimeter.

Let the dimensions of the rectangle be \( x \) and \( y \) where:
\[
A = x \cdot y = 216
\]

### Step 1: Express \( y \) in terms of \( x \).
\[
y = \frac{216}{x}
\]

### Step 2: Write the perimeter function \( P \).
The perimeter \( P \) of a rectangle is given by:
\[
P = 2(x + y)
\]
Substituting \( y = \frac{216}{x} \) into the equation:
\[
P = 2\left( x + \frac{216}{x} \right)
\]
\[
P = 2x + \frac{432}{x}
\]

### Step 3: Minimize the perimeter function.
To find the value of \( x \) that minimizes \( P \), we take the derivative of \( P \) with respect to \( x \) and set it equal to 0:
\[
\frac{dP}{dx} = 2 - \frac{432}{x^2}
\]
Set this equal to 0:
\[
2 - \frac{432}{x^2} = 0
\]
\[
\frac{432}{x^2} = 2
\]
\[
x^2 = 216
\]
\[
x = \sqrt{216} \approx 14.7 \, \text{m}
\]

### Step 4: Find the corresponding value of \( y \).
Using the relation \( y = \frac{216}{x} \):
\[
y = \frac{216}{14.7} \approx 14.7 \, \text{m}
\]

Thus, the dimensions of the rectangle are approximately \( 14.7 \, \text{m} \times 14.7 \, \text{m} \).

### Final Answer:
The dimensions of the rectangle are \( 14.7 \, \text{m}, 14.7 \, \text{m} \).

Would you like more details or have any further questions?

Here are 5 related questions to expand your understanding:
1. How does the shape of a rectangle affect its perimeter for a fixed area?
2. Why is the square the most efficient shape in terms of minimizing the perimeter for a given area?
3. What is the general formula for the perimeter of a rectangle?
4. How do you find the second derivative to confirm that the minimum value is correct?
5. What is the relationship between the derivative and optimization problems?

**Tip:** When minimizing or maximizing a function, always check the first derivative to find critical points, and the second derivative to confirm if it's a minimum or maximum.

Find the dimensions of a rectangle (in m) with area 216 m² whose perimeter is as small as possible. (Enter the dimensions as a comma-separated list.)

Discover how to find the dimensions of a rectangle with an area of 216 m² that minimizes the perimeter. This problem involves geometry and optimization using derivatives, perfect for students in Grades 10-12.

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