A flower in the form of a sector has been fenced by a wire of 40 m length. If the flower-bed has the greatest possible area, then what is the radius of the sector?

Options

A. 25 m
B. 20 m
C. 10 m
D. 5 m

Answer

To solve the problem, we need to maximize the area of a sector of a circle that is fenced by a wire of length 40 meters. Let's break it down step by step.

### Step 1: Understand the problem
We have a sector of a circle, which is fenced by a wire of length 40 meters. The length of the wire corresponds to the arc length of the sector plus the lengths of the two radii.

### Step 2: Set up the equation
Let:
- \( r \) = radius of the sector
- \( l \) = length of the arc of the sector

The total length of the wire can be expressed as:
\[ l + 2r = 40 \]
From this, we can express \( l \) in terms of \( r \):
\[ l = 40 - 2r \]

### Step 3: Area of the sector
The area \( A \) of a sector is given by the formula:
\[ A = \frac{1}{2} \times l \times r \]
Substituting \( l \) from the previous step:
\[ A = \frac{1}{2} \times (40 - 2r) \times r \]
This simplifies to:
\[ A = \frac{1}{2} \times (40r - 2r^2) = 20r - r^2 \]

### Step 4: Differentiate the area function
To find the maximum area, we need to differentiate \( A \) with respect to \( r \):
\[ \frac{dA}{dr} = 20 - 2r \]

### Step 5: Set the derivative to zero
To find the critical points, set the derivative equal to zero:
\[ 20 - 2r = 0 \]
Solving for \( r \):
\[ 2r = 20 \implies r = 10 \]

### Step 6: Verify it's a maximum
To ensure this value gives a maximum area, we can check the second derivative:
\[ \frac{d^2A}{dr^2} = -2 \]
Since the second derivative is negative, this confirms that \( r = 10 \) gives a maximum area.

### Conclusion
The radius of the sector that gives the greatest possible area is:
\[ \boxed{10 \text{ meters}} \]