Problem 45 - A Complete Resource Book in Mathematics for JEE Main

Question

Twenty metre of wire is available to fence off a flower bed in the form of a sector. If the flower bed has the maximum surface then radius is (A) 10 (B) \(5 / 2\) (C) 5 (D) \(15 / 2\)

Step-by-Step Solution

Answer

The radius that maximizes the surface area is 5 meters.

Step 1: Understanding the Problem

We have 20 meters of wire to fence a flower bed that is a sector of a circle. We need to find the radius that maximizes the area of the sector.

Step 2: Constrain the Perimeter

The perimeter of the sector includes the arc length and twice the radius. So if the arc length is \(l\), then the total perimeter is given by \(l + 2r = 20\).

Step 3: Arc Length Formula

The arc length \(l\) is related to the radius \(r\) and the angle \(\theta\) by the formula \(l = r\theta\). Substituting this into the perimeter equation gives \(r\theta + 2r = 20\).

Step 4: Solve for \(\theta\)

Rearrange the equation \(r\theta + 2r = 20\) to find \(\theta\): \(\theta = \frac{20 - 2r}{r}\).

Step 5: Area of the Sector

The area \(A\) of the sector is given by the formula \(A = \frac{1}{2}r^2\theta\). Substituting for \(\theta\), the area becomes \(A = \frac{1}{2}r^2\left(\frac{20 - 2r}{r}\right)\).

Step 6: Simplify the Area Expression

Simplify the area expression to get \(A = 10r - r^2\). This quadratic expression can help us find the maximum area.

Step 7: Find the Maximum Area by Calculus

Take the derivative of \(A = 10r - r^2\) with respect to \(r\): \(\frac{dA}{dr} = 10 - 2r\). Set the derivative equal to zero to find critical points: \(10 - 2r = 0\) which gives \(r = 5\).

Step 8: Determine Maximum Value

To confirm that \(r = 5\) is a maximum, assess the second derivative. The second derivative \(\frac{d^2A}{dr^2} = -2\) is negative, confirming a maximum.

Key Concepts

Geometry

Geometry is about understanding shapes, sizes, and the properties of space. In this problem, we are dealing with a sector of a circle—an essential geometric shape. A sector is like a "slice" of a circle, bounded by two radii and an arc. Imagine cutting a slice out of a round cake; that's similar to a sector.

The problem specifies that the flower bed is a sector with a total perimeter of 20 meters. This perimeter includes two radii and the arc length. Understanding these components is crucial because they determine the shape and size of the sector. The radii are simply two straight lines from the circle's center to its boundary, whereas the arc is the curve joining the endpoints of these radii. By setting the perimeter equation as \(l + 2r = 20\), we are establishing a relationship necessary to solve our problem.

Calculus

Calculus is a powerful tool for solving maximization problems. It allows us to find the points at which a function reaches its highest or lowest values. In this problem, the surface area of the sector is represented by the function \(A = 10r - r^2\).

To determine the maximum area, we need to use the concept of derivatives from calculus. The first derivative \(\frac{dA}{dr} = 10 - 2r\) helps us find the critical points where changes in area stagnate—potential maximum or minimum points. By setting the first derivative equal to zero, we can solve for these points.

• The equation \(10 - 2r = 0\) gives us \(r = 5\).

To confirm this is a maximum point, the second derivative \(\frac{d^2A}{dr^2} = -2\) is considered. A negative second derivative indicates a maximum, ensuring the radius that provides the largest possible area is indeed 5 meters.

Quadratic Functions

Quadratic functions are expressions of the form \(ax^2 + bx + c\). The function \(A = 10r - r^2\) is a classic quadratic function, which is part of why calculus can be easily applied to solve this problem. These functions form parabolas when graphed and have distinct features like a vertex, which in optimization problems such as this, represents the maximum or minimum value.

In this specific example, we see the quadratic function \(A = 10r - r^2\) open downwards due to the negative coefficient of the \(r^2\) term. This shape confirms that the vertex of the parabola is indeed the maximum area, and the symmetry of parabolas aids in quickly identifying the vertex once the equation is solved for \(r\). The calculations confirm that the radius at which this maximum occurs is \(r = 5\), with the parabola peaking at this point, further supported by calculus derivation.