The area of a rectangle will be maximum for the given perimeter, when rectangle is a

Options

A. parallelogram
B. trapezium
C. square
D. none of these

Explanation

To solve the problem of maximizing the area of a rectangle given a fixed perimeter, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Perimeter**: Let the perimeter of the rectangle be \( P \). For a rectangle with sides \( A \) and \( B \), the perimeter is given by:
\[ P = 2A + 2B \]
Therefore, we can express the relationship as:
\[ A + B = \frac{P}{2} \]

2. **Express One Variable in Terms of the Other**: From the equation \( A + B = \frac{P}{2} \), we can express \( B \) in terms of \( A \):
\[ B = \frac{P}{2} - A \]

3. **Write the Area Function**: The area \( A \) of the rectangle is given by:
\[ \text{Area} = A \times B \]
Substituting for \( B \):
\[ \text{Area} = A \left( \frac{P}{2} - A \right) = \frac{P}{2}A - A^2 \]

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

5. **Set the Derivative to Zero**: To find the critical points, set the derivative equal to zero:
\[ \frac{P}{2} - 2A = 0 \]
Solving for \( A \):
\[ 2A = \frac{P}{2} \implies A = \frac{P}{4} \]

6. **Find the Corresponding Value of B**: Substitute \( A = \frac{P}{4} \) back into the equation for \( B \):
\[ B = \frac{P}{2} - A = \frac{P}{2} - \frac{P}{4} = \frac{P}{4} \]

7. **Conclusion**: Since both sides \( A \) and \( B \) are equal, we conclude that the rectangle is a square. Therefore, the area of the rectangle is maximized when the rectangle is a square.

### Final Answer:
The area of a rectangle will be maximum for the given perimeter when the rectangle is a **square**.

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