Question

Twenty meters of wire is available to fence off a flower bed in the form of a circular sector. What must the radius of the circle be, if the area of the flower bed is greatest?

• A. 10 m
• B. 4m
• C. 5m
• D. 6m

Answer 1

The Correct Option is C

We are given that a total length of 20 meters of wire is used to fence a circular sector, and we need to find the radius of the circle that maximizes the area of the sector.

Step 1: Formula for the perimeter of a sector
The perimeter of a circular sector consists of two radii and the arc length. If \( r \) is the radius of the circle and \( \theta \) is the central angle of the sector (in radians), then the perimeter \( P \) of the sector is: \[ P = 2r + r\theta \] We are given that the perimeter is 20 meters, so: \[ 2r + r\theta = 20 \] This simplifies to: \[ r(2 + \theta) = 20 \] 

Step 2: Formula for the area of the sector
The area \( A \) of a sector is given by: \[ A = \frac{1}{2}r^2 \theta \] We need to maximize this area with respect to \( r \). 

Step 3: Express \( \theta \) in terms of \( r \)
From the perimeter equation, we can solve for \( \theta \): \[ \theta = \frac{20}{r} - 2 \] 

Step 4: Substitute \( \theta \) into the area formula
Substituting the expression for \( \theta \) into the area formula, we get: \[ A = \frac{1}{2} r^2 \left( \frac{20}{r} - 2 \right) \] Simplifying this: \[ A = \frac{1}{2} \left( 20r - 2r^2 \right) \] \[ A = 10r - r^2 \] 

Step 5: Maximize the area
To find the value of \( r \) that maximizes the area, we take the derivative of \( A \) with respect to \( r \): \[ \frac{dA}{dr} = 10 - 2r \] Setting \( \frac{dA}{dr} = 0 \) to find the critical points: \[ 10 - 2r = 0 \quad \Rightarrow \quad r = 5 \] 

Step 6: Verify that this is a maximum
The second derivative of \( A \) is: \[ \frac{d^2A}{dr^2} = -2 \] Since the second derivative is negative, \( r = 5 \) gives a maximum.

Answer:

\[ \boxed{5 \text{ m}} \]