Problem 21 - Calculus for Biology and Medicine

Question

A circular sector with radius \(r\) and angle \(\theta\) has area \(A\). Find \(r\) and \(\theta\) so that the perimeter is smallest when (a) \(A=2\) and (b) \(A=10\). (Note: \(A=\frac{1}{2} r^{2} \theta\), and the length of the arc \(s=r \theta\), when \(\theta\) is measured in radians; see Figure \(5.57\).)

Step-by-Step Solution

Answer

For both \(A = 2\) and \(A = 10\), the minimal perimeter occurs at \(\theta = 1\) radian; \(r = 2\) for \(A = 2\), and \(r = 2\sqrt{5}\) for \(A = 10\).

Step 1: Understand the Problem

We need to find the radius \(r\) and angle \(\theta\) for a sector of a circle such that the perimeter is minimized while keeping the area \(A\) constant for two given values: 2 and 10. The perimeter \(P\) of a sector is given by \(P = r + r\theta\), and the area \(A\) is given by \(A = \frac{1}{2} r^2 \theta\).

Step 2: Express \(\theta\) in terms of \(r\) and \(A\)

From the area formula \(A = \frac{1}{2} r^2 \theta\), we can solve for \(\theta\) as follows: \[ \theta = \frac{2A}{r^2}. \] Substituting \(\theta\) in terms of \(r\) into the perimeter expression will help us simplify the problem.

Step 3: Substituting \(\theta\) into the Perimeter Equation

Substitute \(\theta = \frac{2A}{r^2}\) into the perimeter formula \(P = r + r\theta\). We get: \[ P = r + r \left(\frac{2A}{r^2}\right) = r + \frac{2A}{r}. \]

Step 4: Differentiate the Perimeter Function

To find \(r\) that minimizes \(P\), differentiate \(P = r + \frac{2A}{r}\) with respect to \(r\): \[ \frac{dP}{dr} = 1 - \frac{2A}{r^2}. \] Set \(\frac{dP}{dr} = 0\) to find critical points: \[ 1 - \frac{2A}{r^2} = 0. \] This gives \(r^2 = 2A\).

Step 5: Solve for \(r\)

Solving \(r^2 = 2A\) gives \(r = \sqrt{2A}\). Thus, our solution for \(r\) in terms of \(A\) is \(r = \sqrt{2A}\).

Step 6: Find \(\theta\) using \(r\)

With \(r = \sqrt{2A}\), substitute back to find \(\theta\) using \(\theta = \frac{2A}{r^2}\): \[ \theta = \frac{2A}{2A} = 1. \] Thus, \(\theta = 1\) radian.

Step 7: Evaluate for \(A = 2\) and \(A = 10\)

For \(A = 2\), \(r = \sqrt{2\times 2} = 2\) and \(\theta = 1\).For \(A = 10\), \(r = \sqrt{2\times 10} = \sqrt{20} = 2\sqrt{5}\) and \(\theta = 1\).

Key Concepts

Understanding Circular Sectors

A circular sector is a portion of a circle defined by two radial lines and the connecting arc. Imagine slicing a pizza into different pieces, each slice resembles a circular sector. Each sector has:

• A radius (), which is the distance from the center of the circle to the perimeter.
• An angle (), measured in radians, that represents the opening of the sector.
• An arc, which is the curved portion of the sector's boundary.

The area () of a circular sector can be calculated using the formula:\[A = \frac{1}{2} r^2 \theta\]where \(\theta\) is the angle in radians. Understanding the relationship between radius, angle, and area is crucial for solving optimization problems involving circular sectors.

Perimeter Minimization

Perimeter minimization involves finding the values of variables that result in the smallest possible perimeter while maintaining a certain condition, such as a fixed area. For a circular sector, the perimeter () consists of the arc length and the straight line segments (the radial lines), calculated as:\[P = r + r\theta\]This optimization problem requires us to minimize this perimeter formula while keeping the area constant. By manipulating the known equation for the sector's area, we can express the angle () in terms of the radius () and the area (A). Substituting this into the perimeter expression enables a simplified form. This simplification sets the stage for differentiation, helping us find the radius where the perimeter is minimal.

Differentiation for Optimization

Differentiation is a powerful tool in calculus used to find rates of change and to identify optimal solutions to problems by locating extrema (minimum or maximum points). For minimizing the perimeter of a circular sector, we differentiate the perimeter function with respect to the radius ():\[\frac{dP}{dr} = 1 - \frac{2A}{r^2}\]By setting the derivative equal to zero, we target the critical points where the function does not increase or decrease, indicating potential minima or maxima. Solving \(\frac{dP}{dr} = 0\) offers the radius that minimizes perimeter, crucial in determining the sector's minimal state. Differentiation reveals insights into function behavior, leading us to solutions that satisfy given conditions.

Understanding Critical Points

Critical points occur in calculus where the derivative of a function equals zero or does not exist. These points are the cornerstone of finding extrema in optimization problems. For the sector with minimized perimeter, the critical points emerge from solving:\[1 - \frac{2A}{r^2} = 0\]Solving this equation reveals \(r^2 = 2A\), leading to the critical radius \(r = \sqrt{2A}\). These critical points determine where the perimeter of the sector shifts towards its smallest possible value.In practical terms, these points help verify we have reached an optimal solution—ensuring the radius and angle provide the resulting smallest perimeter under specified conditions, such as a fixed area.