The perimeter of a sector is p. The area of the sector is maximum when its radius is

Options

A. `sqrtp`
B. `(1)/(sqrtp)`
C. `(p)/(2)`
D. `(p)/(4)`

Explanation

To solve the problem of finding the radius \( r \) of a sector when its area is maximized given that the perimeter is \( p \), we can follow these steps:

### Step 1: Understand the parameters of the sector
A sector has:
- Radius \( r \)
- Central angle \( \theta \)
- Length of the arc \( l \)

The perimeter \( p \) of the sector can be expressed as:
\[ p = 2r + l \]

### Step 2: Express the area of the sector
The area \( A \) of the sector can be expressed in terms of the radius and the angle:
\[ A = \frac{1}{2} r^2 \theta \]

### Step 3: Relate the angle \( \theta \) to the arc length \( l \)
The length of the arc \( l \) can be expressed as:
\[ l = r \theta \]

### Step 4: Substitute \( l \) in the perimeter equation
From the perimeter equation, we can express \( l \) in terms of \( r \):
\[ l = p - 2r \]

### Step 5: Substitute \( l \) into the area formula
Substituting \( l = r \theta \) into the perimeter equation gives:
\[ r \theta = p - 2r \implies \theta = \frac{p - 2r}{r} \]

Now substituting this expression for \( \theta \) into the area formula:
\[ A = \frac{1}{2} r^2 \left(\frac{p - 2r}{r}\right) = \frac{1}{2} r (p - 2r) \]

### Step 6: Simplify the area expression
The area can now be simplified as:
\[ A = \frac{1}{2} (pr - 2r^2) \]

### Step 7: Differentiate the area with respect to \( r \)
To find the maximum area, we differentiate \( A \) with respect to \( r \):
\[ \frac{dA}{dr} = \frac{1}{2} (p - 4r) \]

### Step 8: Set the derivative equal to zero
To find the critical points:
\[ \frac{1}{2} (p - 4r) = 0 \implies p - 4r = 0 \implies r = \frac{p}{4} \]

### Step 9: Verify that this is a maximum
To confirm that this critical point is a maximum, we can check the second derivative:
\[ \frac{d^2A}{dr^2} = -2 \]
Since this is negative, it confirms that \( r = \frac{p}{4} \) is indeed a maximum.

### Conclusion
The radius \( r \) of the sector when its area is maximum is:
\[ \boxed{\frac{p}{4}} \]