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 at which the area of a sector is maximized given a fixed perimeter \( p \), we can follow these steps:

### Step 1: Understand the Perimeter of the Sector
The perimeter \( P \) of a sector of a circle is given by the formula:
\[ P = 2r + r\theta \]
where \( r \) is the radius and \( \theta \) is the angle in radians.

### Step 2: Express \( \theta \) in terms of \( r \) and \( p \)
From the perimeter formula, we can rearrange it to express \( \theta \):
\[ p = 2r + r\theta \]
This can be rearranged to find \( \theta \):
\[ r\theta = p - 2r \implies \theta = \frac{p - 2r}{r} \]

### Step 3: Write the Area of the Sector
The area \( A \) of the sector is given by:
\[ A = \frac{1}{2} r^2 \theta \]
Substituting \( \theta \) from Step 2:
\[ A = \frac{1}{2} r^2 \left(\frac{p - 2r}{r}\right) = \frac{1}{2} r(p - 2r) \]
This simplifies to:
\[ A = \frac{p}{2} r - r^2 \]

### Step 4: Differentiate the Area with respect to \( r \)
To find the maximum area, we need to differentiate \( A \) with respect to \( r \) and set the derivative equal to zero:
\[ \frac{dA}{dr} = \frac{p}{2} - 2r \]
Setting the derivative to zero for maximization:
\[ \frac{p}{2} - 2r = 0 \]

### Step 5: Solve for \( r \)
From the equation \( \frac{p}{2} - 2r = 0 \), we can solve for \( r \):
\[ 2r = \frac{p}{2} \implies r = \frac{p}{4} \]

### Conclusion
The radius \( r \) at which the area of the sector is maximized is:
\[ \boxed{\frac{p}{4}} \]