Maximum area of a reactangle which can be inscribed in a circle of a given radius R is

Options

A. `(piR^(2))/(3)`
B. `(3R^(2))/(sqrt(2))`
C. `2R^(2)`
D. `3R^(2)`

Answer

To find the maximum area of a rectangle that can be inscribed in a circle of radius \( R \), we can follow these steps:

### Step 1: Understand the Geometry
Consider a circle with radius \( R \). The diameter of the circle is \( 2R \). Let the rectangle inscribed in the circle have length \( l \) and breadth \( b \).

### Step 2: Use the Pythagorean Theorem
Since the rectangle is inscribed in the circle, the diagonal of the rectangle is equal to the diameter of the circle. According to the Pythagorean theorem, we have:
\[ l^2 + b^2 = (2R)^2 = 4R^2 \]

### Step 3: Express Length in Terms of Breadth
From the equation above, we can express \( l \) in terms of \( b \):
\[ l^2 = 4R^2 - b^2 \implies l = \sqrt{4R^2 - b^2} \]

### Step 4: Write the Area of the Rectangle
The area \( A \) of the rectangle can be expressed as:
\[ A = l \cdot b = b \cdot \sqrt{4R^2 - b^2} \]

### Step 5: Differentiate the Area
To find the maximum area, we need to differentiate \( A \) with respect to \( b \) and set the derivative equal to zero. First, we can rewrite the area:
\[ A = b \sqrt{4R^2 - b^2} \]

Using the product rule for differentiation:
\[ \frac{dA}{db} = \sqrt{4R^2 - b^2} + b \cdot \frac{d}{db}(\sqrt{4R^2 - b^2}) \]
Using the chain rule:
\[ \frac{d}{db}(\sqrt{4R^2 - b^2}) = \frac{1}{2\sqrt{4R^2 - b^2}} \cdot (-2b) = -\frac{b}{\sqrt{4R^2 - b^2}} \]
Thus,
\[ \frac{dA}{db} = \sqrt{4R^2 - b^2} - \frac{b^2}{\sqrt{4R^2 - b^2}} \]
Setting the derivative equal to zero for maximization:
\[ \sqrt{4R^2 - b^2} - \frac{b^2}{\sqrt{4R^2 - b^2}} = 0 \]

### Step 6: Solve for \( b \)
Multiplying through by \( \sqrt{4R^2 - b^2} \) gives:
\[ (4R^2 - b^2) - b^2 = 0 \implies 4R^2 - 2b^2 = 0 \implies 2b^2 = 4R^2 \implies b^2 = 2R^2 \implies b = R\sqrt{2} \]

### Step 7: Find \( l \)
Substituting \( b \) back to find \( l \):
\[ l = \sqrt{4R^2 - (R\sqrt{2})^2} = \sqrt{4R^2 - 2R^2} = \sqrt{2R^2} = R\sqrt{2} \]

### Step 8: Calculate the Maximum Area
Now, the maximum area \( A \) is:
\[ A = l \cdot b = (R\sqrt{2}) \cdot (R\sqrt{2}) = 2R^2 \]

### Conclusion
The maximum area of the rectangle inscribed in a circle of radius \( R \) is:
\[ \boxed{2R^2} \]