What is the maximum area of a rectangle that can be inscribed in a circle of radius 2 units?

Options

A. 4 square units
B. 6 square units
C. 8 square units
D. 16 square units

Answer

To find the maximum area of a rectangle that can be inscribed in a circle of radius 2 units, we can follow these steps:

### Step 1: Understand the Geometry
We start by visualizing a circle with a radius of 2 units. The rectangle will be inscribed in this circle, meaning all four corners of the rectangle will touch the circle.

### Step 2: Define Variables
Let the length of the rectangle be \( L \) and the breadth be \( B \). The diagonal of the rectangle will be equal to the diameter of the circle.

### Step 3: Use the Pythagorean Theorem
Since the rectangle is inscribed in the circle, we can apply the Pythagorean theorem:
\[ L^2 + B^2 = (2 \cdot \text{radius})^2 = (2 \cdot 2)^2 = 4^2 = 16 \]
Thus, we have:
\[ L^2 + B^2 = 16 \]

### Step 4: Express Area
The area \( A \) of the rectangle can be expressed as:
\[ A = L \cdot B \]
From the equation \( L^2 + B^2 = 16 \), we can express \( B \) in terms of \( L \):
\[ B^2 = 16 - L^2 \quad \Rightarrow \quad B = \sqrt{16 - L^2} \]
Substituting this into the area formula gives:
\[ A = L \cdot \sqrt{16 - L^2} \]

### Step 5: Differentiate to Maximize Area
To find the maximum area, we differentiate \( A \) with respect to \( L \) and set the derivative equal to zero:
\[ A = L \cdot \sqrt{16 - L^2} \]
Using the product rule:
\[ \frac{dA}{dL} = \sqrt{16 - L^2} + L \cdot \frac{d}{dL}(\sqrt{16 - L^2}) \]
The derivative of \( \sqrt{16 - L^2} \) is:
\[ \frac{d}{dL}(\sqrt{16 - L^2}) = \frac{-L}{\sqrt{16 - L^2}} \]
Thus:
\[ \frac{dA}{dL} = \sqrt{16 - L^2} - \frac{L^2}{\sqrt{16 - L^2}} \]
Setting this equal to zero:
\[ \sqrt{16 - L^2} = \frac{L^2}{\sqrt{16 - L^2}} \]
Squaring both sides:
\[ 16 - L^2 = L^4/(16 - L^2) \]
Cross-multiplying and rearranging leads to a polynomial equation.

### Step 6: Solve for L
After simplifying and solving the polynomial, we find:
\[ L^2 = 8 \quad \Rightarrow \quad L = 2\sqrt{2} \]
Substituting \( L \) back into the equation for \( B \):
\[ B = \sqrt{16 - (2\sqrt{2})^2} = \sqrt{16 - 8} = \sqrt{8} = 2\sqrt{2} \]

### Step 7: Calculate Maximum Area
Thus, the maximum area \( A \) is:
\[ A = L \cdot B = (2\sqrt{2}) \cdot (2\sqrt{2}) = 4 \cdot 2 = 8 \text{ square units} \]

### Conclusion
The maximum area of the rectangle that can be inscribed in a circle of radius 2 units is **8 square units**.

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