Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Th

Q: Wait, why do we include the two radii in the perimeter? Isn't the perimeter just the arc length?

That is a classic trap! A circular sector is a closed geometric shape, like a slice of pie. The boundary consists of the curved arc *and* the two straight radii that meet at the center. If you only fence the arc, you haven't enclosed the flower-bed. You must include both radii to complete the perimeter.

Q: Why can't we just use the formula for the area of a circle, $\pi r^2$?

Because we aren't fencing a full circle! We are fencing a sector with a fixed perimeter of $20$ meters. The angle $\theta$ is not fixed; it changes as you adjust the radius to maximize the area. We must derive the area based on the specific constraints of the wire length provided.

Q: How do I know if $r=5$ gives a maximum and not a minimum?

Excellent question! To verify, look at the second derivative of our area function $A(r) = 10r - r^2$. The first derivative is $\frac{dA}{dr} = 10 - 2r$. The second derivative is $\frac{d^2A}{dr^2} = -2$. Since the second derivative is negative, the function is concave down, which mathematically guarantees that our critical point is a local maximum.

Question

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:

Options:

(A) 12.5

(B) 10

(C) 25

(D) 30

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of the Flower-Bed

A circular sector is defined by its radius $r$ and the central angle $	heta$ (measured in radians). The perimeter $P$ of this sector consists of two straight radii and the curved arc length $s$.

Mathematically, the perimeter is expressed as:
$$P = 2r + s$$

Given that the total wire available is $20$ meters, we establish the primary constraint:
$$2r + s = 20$$

### The Art of Variable Reduction

We aim to maximize the area $A$ of the sector, which is given by the formula:
$$A = \frac{1}{2}r^2	heta$$

To simplify the optimization, we utilize the relationship between arc length, radius, and angle, defined as $s = r	heta$. Substituting this into our perimeter constraint, we obtain:
$$2r + r	heta = 20$$

By isolating $	heta$, we create a bridge between the perimeter constraint and the area formula:
$$	heta = \frac{20 - 2r}{r} = \frac{20}{r} - 2$$

### Constructing the Area Function

Next, we substitute our expression for $	heta$ into the area formula:
$$A = \frac{1}{2}r^2 \left( \frac{20}{r} - 2 ight)$$

Distributing the $\frac{1}{2}r^2$ term across the parentheses yields:
$$A = \frac{1}{2}r^2 \cdot \frac{20}{r} - \frac{1}{2}r^2 \cdot 2$$

Simplifying this expression, we arrive at the elegant quadratic function:
$$A = 10r - r^2$$

### The Calculus of Maximization

To find the maximum area, we determine the point where the rate of change is zero by differentiating $A$ with respect to $r$:
$$\frac{dA}{dr} = 10 - 2r$$

Setting the derivative to zero to find the critical point:
$$10 - 2r = 0 \Rightarrow r = 5 	ext{ meters}$$

Substituting $r = 5$ back into our area function, we calculate the maximum area:
$$A = 10(5) - (5)^2 = 50 - 25 = 25 	ext{ square meters}$$

### The Pro-Tip for JEE Success

For any circular sector with a fixed perimeter $P$, the area is always maximized when the radius is $r = \frac{P}{4}$ and the arc length is $s = \frac{P}{2}$.

In this specific problem, where $P = 20$, we confirm $r = 5$ and $s = 10$. This shortcut is a powerful tool to keep in your arsenal for competitive examinations.

**The maximum area of the flower-bed is $25 	ext{ m}^2$.**

Animated Solution for Mathematics - Differentiation: 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:

Select Answer:

Visualized Solution

Visualizing the Circular Sector

Defining the Perimeter Equation

Relating Arc Length and Angle

Setting up the Area Function

Substituting $θ$ into Area

Simplifying the Area Expression

Differentiating for Maximum Area

Finding the Critical Point

Calculating the Maximum Area

Conclusion and Pro-Tip

The Sigma Insight: Maxima and Minima

Analyzing the Geometry of the Flower-Bed

The Art of Variable Reduction

Constructing the Area Function

The Calculus of Maximization

The Pro-Tip for JEE Success

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