Question

A flower bed will be in the shape of a sector of a circle (a pie-shaped region) of radius \(r\) and vertex angle \(\theta .\) Find \(r\) and \(\theta\) if its area is a constant \(A\) and the perimeter is a minimum.

Short answer

The radius \(r = \sqrt{A}\) and the angle \(\theta = 2\).

Step-by-step solution

Understanding the Problem

We are given a sector of a circle with radius \(r\), vertex angle \(\theta\), and a constant area \(A\). Our goal is to find \(r\) and \(\theta\) so that the perimeter is minimized.

Formula for Area of Sector

The area of the sector is given by the formula \(A = \frac{1}{2} r^2 \theta\). Since \(A\) is constant, this equation allows us to express \(\theta\) in terms of \(A\) and \(r\): \(\theta = \frac{2A}{r^2}\).

Formula for Perimeter of Sector

The perimeter \(P\) of the sector consists of the arc length and the two radii, so it is given by \(P = r\theta + 2r\). Substituting \(\theta = \frac{2A}{r^2}\) into this formula, we get \(P = \frac{2Ar}{r^2} + 2r = \frac{2A}{r} + 2r\).

Differentiating Perimeter for Minimum

To minimize \(P\), we differentiate it with respect to \(r\). The derivative is \(\frac{dP}{dr} = -\frac{2A}{r^2} + 2\).

Setting Derivative to Zero

Setting \(\frac{dP}{dr} = 0\) gives \(-\frac{2A}{r^2} + 2 = 0\). Solving for \(r\), we get \(r^2 = A\), thus \(r = \sqrt{A}\).

Determining \(\theta\)

Using \(r = \sqrt{A}\), substitute into \(\theta = \frac{2A}{r^2}\), hence \(\theta = 2\).

Verification

With \(r = \sqrt{A}\) and \(\theta = 2\), both conditions of the problem (having constant area \(A\) and minimized perimeter) are satisfied by the derived values. This verifies our solution.

Key concepts

Understanding Calculus in Optimization

Calculus is a branch of mathematics that studies how things change. It is fundamental in solving optimization problems like the one involving the sector of a circle. In this exercise, we use calculus to find the values of

• radius \(r\), and
• vertex angle \(\theta\)

that minimize the perimeter of the sector while maintaining a constant area. Such problems usually involve

• differentiating functions to find critical points
• and testing these points to identify minimum or maximum values.

By setting the derivative of the perimeter function to zero, we can find the optimal values for \(r\) and \(\theta\). This is key in ensuring that we get the smallest possible perimeter while retaining a specific area.

Calculating the Area of a Sector

The area of a sector of a circle is an essential concept in geometry and calculus. The sector is a portion of a circle, similar to a "slice of pizza." The formula for the area \(A\) of a sector with radius \(r\) and angle \(\theta\) (in radians) is given by:\[ A = \frac{1}{2} r^2 \theta \]This formula makes it clear that the area is directly proportional to the square of the radius \(r\) and the angle \(\theta\). In our problem, because the area \(A\) is constant, we can rearrange this formula to express \(\theta\) in terms of \(A\) and \(r\):\[ \theta = \frac{2A}{r^2} \]This relationship is crucial as it allows us to substitute \(\theta\) into the formula for the perimeter, simplifying the optimization process.

Understanding the Perimeter of a Sector

The perimeter of a sector involves both the arc length and the straight sides (radiuses) of the sector. The formula is:\[ P = r\theta + 2r \]Where:

• \(r\theta\) is the arc length
• and \(2r\) accounts for the two radii.

In the problem, we substitute \(\theta = \frac{2A}{r^2}\) into the perimeter formula to get:\[ P = \frac{2A}{r} + 2r \]The goal is to minimize \(P\), meaning we need to find the value of \(r\) that produces the smallest perimeter \(P\) while meeting the given area condition. Finding this minimal value involves calculus, specifically differentiation.

Differentiation: A Tool for Optimization

Differentiation is a core tool in calculus that helps us find rates of change. When dealing with optimization problems, we use differentiation to

• Find the critical points of a function
• Determine where a function reaches its minimum or maximum.

For this exercise, after expressing the perimeter in terms of \(r\), we use differentiation to find the derivative \(\frac{dP}{dr}\). The expression for this derivative is:\[ \frac{dP}{dr} = -\frac{2A}{r^2} + 2 \]Setting \(\frac{dP}{dr} = 0\) allows us to solve for \(r\), giving us \(r^2 = A\), and hence \(r = \sqrt{A}\). By substituting back, we confirm \(\theta = 2\). This process illustrates how differentiation helps us efficiently find solutions to optimization problems, ensuring both conditions of area and minimized perimeter are met.