Problem 33
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.
Step-by-Step Solution
Verified Answer
For minimum perimeter, \(r = \sqrt{A}\) and \(\theta = 2\) radians.
1Step 1: Area of the Sector
The area of the sector is given as a constant \(A\). The formula for the area \(A\) of a sector with radius \(r\) and angle \(\theta\) (in radians) is \(A = \frac{1}{2} r^2 \theta\).
2Step 2: Rearrange for \(\theta\)
Rearrange the formula from Step 1 to express \(\theta\) in terms of \(A\) and \(r\):\[\theta = \frac{2A}{r^2}.\]
3Step 3: Perimeter of the Sector
The perimeter \(P\) of the sector is the sum of the arc length and the two radii: \(P = r \theta + 2r\).
4Step 4: Substitute \(\theta\) in the Perimeter Formula
Substitute \(\theta = \frac{2A}{r^2}\) from Step 2 into the perimeter formula from Step 3: \[P = r \cdot \frac{2A}{r^2} + 2r = \frac{2A}{r} + 2r.\]
5Step 5: Minimize Perimeter Formula
To find the minimum perimeter, take the derivative of \(P = \frac{2A}{r} + 2r\) with respect to \(r\) and set it equal to zero: \[\frac{dP}{dr} = -\frac{2A}{r^2} + 2 = 0.\]
6Step 6: Solve for \(r\)
Solve the equation from Step 5: \(-\frac{2A}{r^2} + 2 = 0\). This simplifies to \(\frac{2A}{r^2} = 2\) or \(r^2 = A\). Thus, \(r = \sqrt{A}\).
7Step 7: Calculate \(\theta\) for Minimum Perimeter
Use \(r = \sqrt{A}\) in the expression for \(\theta\):\[\theta = \frac{2A}{r^2} = \frac{2A}{A} = 2.\] Therefore, \(\theta = 2\) radians when the perimeter is minimized.
Key Concepts
Calculus ProblemsSector of a CirclePerimeter MinimizationRadian Measure
Calculus Problems
Calculus is a versatile tool that helps solve a variety of problems, including those where optimization is involved. In this exercise, we studied an optimization problem that involves both geometry and calculus principles. The problem focuses on minimizing the perimeter of a flower bed, which is in the shape of a sector of a circle.
Optimization problems in calculus typically require finding maximum or minimum values of a function. To do this, calculus uses concepts like derivatives. Taking the derivative of a function and setting it to zero helps find where the function reaches its extremum points. In the case of this exercise, we derived the formula for the perimeter and used differentiation to find the minimum perimeter.
Understanding calculus problems often requires both analytical and creative thinking. You need to identify relevant formulas and derivatives and apply them in the correct sequence. Once mastered, these concepts are invaluable for solving real-world problems.
Optimization problems in calculus typically require finding maximum or minimum values of a function. To do this, calculus uses concepts like derivatives. Taking the derivative of a function and setting it to zero helps find where the function reaches its extremum points. In the case of this exercise, we derived the formula for the perimeter and used differentiation to find the minimum perimeter.
Understanding calculus problems often requires both analytical and creative thinking. You need to identify relevant formulas and derivatives and apply them in the correct sequence. Once mastered, these concepts are invaluable for solving real-world problems.
Sector of a Circle
A sector of a circle looks like a "slice of pie" and is part of a circle bounded by two radii and an arc. This figure is characterized by its radius and the angle between those radii. In this exercise, the flower bed's shape was described as a sector, and we needed to determine its perfect dimensions.
The area of a sector is calculated using the formula: \[ A = \frac{1}{2} r^2 \theta \] where \(r\) is the radius, and \(\theta\) the angle in radians. This formula is quite intuitive once you break it down:
The area of a sector is calculated using the formula: \[ A = \frac{1}{2} r^2 \theta \] where \(r\) is the radius, and \(\theta\) the angle in radians. This formula is quite intuitive once you break it down:
- \( \frac{1}{2} r^2 \) reflects the base area of the circle taken into half.
- \( \theta \) is the fraction of the circle's full 360 degrees that the sector encompasses.
Perimeter Minimization
An essential goal of the exercise was to minimize the perimeter of the sector-shaped flower bed. This was done by thinking creatively about how to express the problem mathematically and then using calculus techniques to solve it.
The perimeter of a sector is not straightforward like that of a rectangle or triangle. It is the sum of the lengths of its two radii and its arc. The formula is:\[ P = r\theta + 2r \]
Using the property of the constant area, we expressed \(\theta\) in terms of \(r\) and substituted back to form a new equation for perimeter \(P\) in terms of radius \(r\) only.
After this substitution, calculus comes into play. By taking the derivative of the perimeter function and equating it to zero, we can find the optimal value for \(r\) that minimizes \(P\). This is a classic application of differentiation in finding minimum or maximum values of functions in calculus.
The perimeter of a sector is not straightforward like that of a rectangle or triangle. It is the sum of the lengths of its two radii and its arc. The formula is:\[ P = r\theta + 2r \]
Using the property of the constant area, we expressed \(\theta\) in terms of \(r\) and substituted back to form a new equation for perimeter \(P\) in terms of radius \(r\) only.
After this substitution, calculus comes into play. By taking the derivative of the perimeter function and equating it to zero, we can find the optimal value for \(r\) that minimizes \(P\). This is a classic application of differentiation in finding minimum or maximum values of functions in calculus.
Radian Measure
The concept of radian measure is crucial when dealing with sectors of a circle, particularly when angles appear in formulas. Unlike degrees, radians offer a natural way to express angles, directly related to the radius of a circle.
One radian is the angle made by an arc whose length equals the radius of the circle. Important in calculus, as well as trigonometry, calculating angles in radians helps better understand relationships in geometric contexts. For example, when calculating the area of a sector \( A = \frac{1}{2} r^{2} \theta \), \(\theta\) must be in radians to maintain dimensional consistency.
Ultimately, using radian measures simplifies many calculus operations and often leads to formulas that are more elegant and conceptually straightforward than their degree-based counterparts. This is seen in how the minimal perimeter calculations are streamlined and intuitive by adopting radian measure.
One radian is the angle made by an arc whose length equals the radius of the circle. Important in calculus, as well as trigonometry, calculating angles in radians helps better understand relationships in geometric contexts. For example, when calculating the area of a sector \( A = \frac{1}{2} r^{2} \theta \), \(\theta\) must be in radians to maintain dimensional consistency.
Ultimately, using radian measures simplifies many calculus operations and often leads to formulas that are more elegant and conceptually straightforward than their degree-based counterparts. This is seen in how the minimal perimeter calculations are streamlined and intuitive by adopting radian measure.
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