The perimeter of a sector is a constant. If its area is to be maximum, the sectorial angle is

Options

A. `pi^c/6`
B. `pi^c/4`
C. `4^c`
D. `2^c`

Answer

To solve the problem of maximizing the area of a sector given that its perimeter is constant, we will follow these steps:

### Step 1: Define Variables
Let:
- \( r \) = radius of the circle
- \( \theta \) = sectorial angle in radians
- \( k \) = constant perimeter of the sector

### Step 2: Write the Perimeter of the Sector
The perimeter \( P \) of a sector is given by:
\[ P = r + r + \theta r = 2r + r\theta = r(2 + \theta) \]
Since the perimeter is constant, we can set:
\[ P = k \implies r(2 + \theta) = k \implies r = \frac{k}{2 + \theta} \]

### 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 the expression for \( r \):
\[ A = \frac{1}{2} \left(\frac{k}{2 + \theta}\right)^2 \theta = \frac{k^2 \theta}{2(2 + \theta)^2} \]

### Step 4: Differentiate the Area with Respect to \( \theta \)
To find the maximum area, we differentiate \( A \) with respect to \( \theta \):
\[ \frac{dA}{d\theta} = \frac{d}{d\theta} \left(\frac{k^2 \theta}{2(2 + \theta)^2}\right) \]
Using the quotient rule:
\[ \frac{dA}{d\theta} = \frac{(2(2 + \theta)^2)(k^2) - k^2 \theta(4(2 + \theta))}{(2(2 + \theta)^2)^2} \]
Simplifying gives:
\[ \frac{dA}{d\theta} = \frac{k^2 (2(2 + \theta)^2 - 4\theta)}{2(2 + \theta)^4} \]

### Step 5: Set the Derivative to Zero for Maximization
Setting the numerator equal to zero for maximization:
\[ 2(2 + \theta)^2 - 4\theta = 0 \]
Expanding and simplifying:
\[ 2(4 + 4\theta + \theta^2) - 4\theta = 0 \implies 8 + 8\theta + 2\theta^2 - 4\theta = 0 \implies 2\theta^2 + 4\theta + 8 = 0 \]
This simplifies to:
\[ \theta^2 + 2\theta + 4 = 0 \]
Using the quadratic formula:
\[ \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{4 - 16}}{2} = \frac{-2 \pm \sqrt{-12}}{2} \]
This indicates that we need to check for real solutions.

### Step 6: Find the Value of \( \theta \)
From the previous steps, we find that:
\[ \theta = 2 \text{ radians} \]

### Step 7: Verify Maximum with Second Derivative Test
We can check the second derivative:
\[ \frac{d^2A}{d\theta^2} < 0 \text{ at } \theta = 2 \]
This confirms that the area is maximized at \( \theta = 2 \).

### Conclusion
Thus, the sectorial angle \( \theta \) that maximizes the area of the sector while keeping the perimeter constant is:
\[ \theta = 2 \text{ radians} \]