Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the maximum area (in sq. m) of the flower-bed, is:

We begin by introducing the two quantities that completely describe a circular sector:

Radius  $$r$$  and central angle  $$\theta$$ (in radians).

The flower-bed is fenced by a piece of wire that runs along the two straight sides (the radii) and also along the curved side (the arc). Therefore its total length is

$$\text{Perimeter}= \underbrace{r\theta}_{\text{arc length}} + \underbrace{2r}_{\text{two radii}}.$$

Because the available wire is 20 m long, we have the constraint

$$r\theta + 2r = 20.$$

Next, we recall the standard formula for the area of a sector:

$$\text{Area}= \dfrac12 r^{2}\theta.$$

Our aim is to maximise this area subject to the perimeter condition. From the perimeter equation we first solve for the radius:

$$r(\theta+2)=20 \quad\Longrightarrow\quad r=\dfrac{20}{\theta+2}.$$

Now we substitute this expression for $$r$$ into the area formula:

$$A = \dfrac12\left(\dfrac{20}{\theta+2}\right)^2\theta = \dfrac12 \cdot \dfrac{400\,\theta}{(\theta+2)^2} = \dfrac{200\,\theta}{(\theta+2)^2}.$$

So the area is now a function of the single variable $$\theta$$:

$$A(\theta)=\dfrac{200\theta}{(\theta+2)^2}.$$

To find the maximum we differentiate $$A(\theta)$$ with respect to $$\theta$$ and set the derivative equal to zero. Using the quotient rule (or the product rule with $$(\theta+2)^{-2}$$) we write

$$\dfrac{dA}{d\theta} =200\;\dfrac{(\theta+2)^2\cdot 1-\theta\cdot 2(\theta+2)}{(\theta+2)^4}.$$

Now we simplify the numerator step by step:

$$\begin{aligned} (\theta+2)^2\cdot 1-\theta\cdot 2(\theta+2) &=(\theta+2)^2-2\theta(\theta+2)\\ &=(\theta+2)\left[(\theta+2)-2\theta\right]\\ &=(\theta+2)(2-\theta). \end{aligned}$$

Consequently

$$\dfrac{dA}{d\theta}=200\,\dfrac{(\theta+2)(2-\theta)}{(\theta+2)^4} =200\,\dfrac{2-\theta}{(\theta+2)^3}.$$

Setting the derivative equal to zero gives

$$2-\theta=0 \quad\Longrightarrow\quad \theta=2\;\text{radians}.$$

(The factor $$(\theta+2)^3$$ in the denominator can never be zero for positive $$\theta$$, so it does not influence the solution.)

Because $$A(\theta)$$ increases up to $$\theta=2$$ and then decreases (one can check the second derivative or simply note the sign change of $$2-\theta$$), this value indeed yields the maximum area.

We now compute the corresponding radius:

$$r=\dfrac{20}{\theta+2}=\dfrac{20}{2+2}=5\;\text{m}.$$

Finally, we calculate the maximal area:

$$\begin{aligned} A_{\max}&=\dfrac12 r^{2}\theta =\dfrac12\,(5)^2\,(2)\\ &=\dfrac12\,(25)\,(2)\\ &=25\;\text{square metres}. \end{aligned}$$

Hence, the correct answer is Option C.