If from a wire of length 36 metre a rectangle of greatest area is made, than its two adjacent sides in metre are

Options

A. 6,12
B. 9,9
C. 10,8
D. 13,5

Explanation

To solve the problem of finding the two adjacent sides of a rectangle with the greatest area that can be formed from a wire of length 36 meters, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We need to form a rectangle using a wire of length 36 meters, and we want to maximize the area of this rectangle.

2. **Define Variables**: Let the lengths of the two adjacent sides of the rectangle be \( A \) and \( B \).

3. **Write the Perimeter Equation**: The perimeter \( P \) of a rectangle is given by the formula:
\[ P = 2A + 2B \]
Since the total length of the wire is 36 meters, we can set up the equation:
\[ 2A + 2B = 36 \]

4. **Simplify the Perimeter Equation**: Dividing the entire equation by 2 gives:
\[ A + B = 18 \]
From this, we can express \( B \) in terms of \( A \):
\[ B = 18 - A \]

5. **Write the Area Function**: The area \( A \) of the rectangle can be expressed as:
\[ \text{Area} = A \times B = A \times (18 - A) = 18A - A^2 \]

6. **Differentiate the Area Function**: To find the maximum area, we need to differentiate the area function with respect to \( A \):
\[ \frac{d(\text{Area})}{dA} = 18 - 2A \]

7. **Set the Derivative to Zero**: To find the critical points, we set the derivative equal to zero:
\[ 18 - 2A = 0 \]
Solving for \( A \):
\[ 2A = 18 \implies A = 9 \]

8. **Find the Corresponding Value of B**: Now, substitute \( A = 9 \) back into the equation for \( B \):
\[ B = 18 - A = 18 - 9 = 9 \]

9. **Conclusion**: The two adjacent sides of the rectangle that maximize the area are:
\[ A = 9 \text{ meters and } B = 9 \text{ meters} \]

### Final Answer:
The two adjacent sides of the rectangle are 9 meters and 9 meters.