A wire of length 28 cm is bent to form a circular sector , then the radius (in cm) of the circular sector such that the area of the circular sector is maximum is equal to

Options

A. 5
B. 6
C. 7
D. 8

Answer

To find the radius of the circular sector formed by bending a wire of length 28 cm such that the area of the sector is maximized, we can follow these steps:

### Step 1: Understand the problem
We have a wire of length 28 cm that is bent to form a circular sector. We need to determine the radius \( r \) of this sector such that its area is maximized.

### Step 2: Define the variables
Let:
- \( r \) = radius of the circular sector
- \( s \) = arc length of the sector

### Step 3: Write the perimeter equation
The total length of the wire, which forms the perimeter of the circular sector, can be expressed as:
\[ P = 2r + s = 28 \quad \text{(1)} \]

### Step 4: Express the arc length in terms of radius
From equation (1), we can express the arc length \( s \) in terms of \( r \):
\[ s = 28 - 2r \quad \text{(2)} \]

### Step 5: Write the area of the circular sector
The area \( A \) of the circular sector can be expressed as:
\[ A = \frac{1}{2} \times r \times s \]
Substituting equation (2) into this formula gives:
\[ A = \frac{1}{2} \times r \times (28 - 2r) \]
\[ A = 14r - r^2 \quad \text{(3)} \]

### Step 6: Differentiate the area function
To find the maximum area, we need to differentiate \( A \) with respect to \( r \):
\[ \frac{dA}{dr} = 14 - 2r \]

### Step 7: Set the derivative to zero
To find the critical points, set the derivative equal to zero:
\[ 14 - 2r = 0 \]
Solving for \( r \):
\[ 2r = 14 \implies r = 7 \]

### Step 8: Verify that it is a maximum
To confirm that this value of \( r \) gives a maximum area, we can check the second derivative:
\[ \frac{d^2A}{dr^2} = -2 \]
Since the second derivative is negative, this indicates that the area function has a maximum at \( r = 7 \).

### Conclusion
The radius of the circular sector that maximizes the area is:
\[ \boxed{7 \text{ cm}} \]