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Magazine Chapter 3: Problem 43
A 24 -in. piece of string is cut in two pieces. One piece is used to form a circle and the other to form a square. How should the string be cut so that the sum of the areas is a minimum? a maximum?
Short Answer
Expert verified
To minimize the area, cut the string at: \[ \frac{24\text{蟺}}{4 + \蟺} \] To maximize the area, either x = 0 or x = 24.
Step by step solution
01
Define the Variables
Let the length of the piece of string used to form the circle be denoted as x inches. Consequently, the length of the piece used to form the square will be (24 - x) inches.
02
Express the Radius of the Circle
The circumference of the circle is given by x inches. Therefore, the radius of the circle, r, can be expressed as: \[ r = \frac{x}{2\pi} \]
03
Express the Side Length of the Square
The perimeter of the square is (24 - x) inches. Hence, each side length of the square, s, can be written as: \[ s = \frac{24 - x}{4} \]
04
Calculate the Area of the Circle
The area of the circle, A_circle, is: \[ A_{\text{circle}} = \pi r^2 = \pi \left( \frac{x}{2\pi} \right)^2 = \frac{x^2}{4\pi} \]
05
Calculate the Area of the Square
The area of the square, A_square, is: \[ A_{\text{square}} = s^2 = \left( \frac{24 - x}{4} \right)^2 = \frac{(24 - x)^2}{16} \]
06
Formulate the Total Area
The total area A as a function of x is: \[ A(x) = A_{\text{circle}} + A_{\text{square}} = \frac{x^2}{4\pi} + \frac{(24 - x)^2}{16} \]
07
Differentiate the Total Area
Differentiate A(x) with respect to x and set the result to zero to find the critical points: \[ \frac{dA}{dx} = \frac{1}{2\pi} x + \frac{1}{8} (x - 24) = 0 \]
08
Solve the Critical Point Equation
Solve the equation: \[ \frac{x}{2\pi} + \frac{x - 24}{8} = 0 \] Multiply through by 8蟺 to clear the fractions: \[ 4x + \pi(x - 24) = 0 \] Combine like terms and solve for x: \[ x(4 + \pi) = 24蟺 \] Therefore, \[ x = \frac{24\pi}{4 + \pi} \]
09
Determine the Second Derivative
Compute the second derivative of A(x): \[ \frac{d^2A}{dx^2} = \frac{1}{2\pi} + \frac{1}{8} \] Evaluate the second derivative at the critical point to confirm it is a minimum: \[ \frac{d^2A}{dx^2} > 0 \] Thus, the critical point corresponds to a minimum area.
10
Determine the Maximum Area
The maximum area occurs when x is either 0 or 24 since the second derivative test confirms the critical point is a minimum; check these endpoints. For x = 0, the entire string forms a square, and for x = 24, the entire string forms a circle.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Minimization problem
Minimization problems in calculus involve finding the minimum value of a function. The given exercise is a classic example where we need to minimize the sum of the areas using a piece of string cut into two parts. One part forms a circle, and the other part forms a square.
When approaching a minimization problem:
When approaching a minimization problem:
- First, identify the quantities to minimize or maximize.
- Express these quantities as functions of a single variable.
- In our case, we needed to minimize the total area formed by the circle and square.
Critical points
Critical points are where the first derivative of a function is zero or undefined. These points are potential locations for local minima, maxima, or points of inflection.
For finding critical points:
For finding critical points:
- Take the first derivative of the function.
- Set the first derivative equal to zero.
- Solve the equation to find the values of the variable.
Second derivative test
The second derivative test helps confirm whether a critical point is a local minimum or maximum.
- If the second derivative is positive at a critical point, the function has a local minimum.
- If the second derivative is negative, the function has a local maximum.
- First, find the second derivative of your function.
- Evaluate the second derivative at the critical points found.
Area calculation
Calculating the area of different shapes involved expressing each area in terms of a single variable so we could form an equation for the total area.
- For the circle, we expressed the radius in terms of x, where x was the length of the string used for the circle. The area is then \[ A_{\text{circle}} = \frac{x^2}{4\pi} \].
- For the square, we expressed the side length in terms of (24 - x), as this was the remaining string length. The area is then \[ A_{\text{square}} = \frac{(24 - x)^2}{16} \].
