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Magazine Chapter 3: Problem 19
Show that the rectangle of fixed area whose perimeter is a minimum is a square.
Short Answer
Expert verified
A rectangle of fixed area has minimum perimeter when it is a square.
Step by step solution
01
Define the rectangle's properties
Let the length of the rectangle be \( l \) and the width be \( w \). The area of the rectangle is \( A = lw \) and is given as a constant. The perimeter of the rectangle is given by \( P = 2(l + w) \).
02
Express the perimeter in terms of one variable
Since the area is fixed, express one variable in terms of the other and a constant. Solve for \( l \) in terms of \( w \): \[ l = \frac{A}{w} \]Substitute this expression into the perimeter formula to get:\[ P = 2 \left( \frac{A}{w} + w \right) \]
03
Find the critical points
To minimize the perimeter, take the derivative of \( P \) with respect to \( w \):\[ \frac{dP}{dw} = 2 \left( -\frac{A}{w^2} + 1 \right) \]Set \( \frac{dP}{dw} = 0 \) to find critical points:\[ -\frac{A}{w^2} + 1 = 0 \]\[ \frac{A}{w^2} = 1 \]\[ A = w^2 \]\[ w = \sqrt{A} \].
04
Determine the rectangle is a square
Since \( A = w \times l \) and we found \( w = \sqrt{A} \), substitute \( w \) back to find \( l \): \[ l = \frac{A}{w} = \frac{A}{\sqrt{A}} = \sqrt{A} \]Thus, \( l = w \). This shows the rectangle is a square.
05
Verify that this is a minimum
To ensure it is a minimum, consider the second derivative \( \frac{d^2P}{dw^2} \):\[ \frac{d^2P}{dw^2} = 2 \left( \frac{2A}{w^3} \right) = \frac{4A}{w^3} \]Since \( \frac{4A}{w^3} > 0 \) for all positive \( w \), the second derivative is positive indicating a minimum at \( w = \sqrt{A} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Critical Points
Critical points are pivotal in identifying the points where a function may attain a maximum or minimum value. In optimization problems, critical points are the values of a variable where the derivative of a function is zero or does not exist.
To find critical points, follow these steps:
To find critical points, follow these steps:
- Take the derivative of the function you are interested in optimizing. This gives you the rate of change of the function with respect to its variable.
- Set the derivative equal to zero and solve for the variable. This zeroing out process helps locate where the function's slope is flat, indicative of a maximum, minimum, or saddle point.
- Evaluate the critical points if necessary by using the second derivative test to determine their nature (minimum, maximum, or neither).
Perimeter Minimization
Perimeter minimization is a type of optimization problem aimed at finding the shape with the least boundary length for a given constraint. In this case, the challenge is to minimize the perimeter of a rectangle for a given area.
Here's how you can approach perimeter minimization:
Here's how you can approach perimeter minimization:
- Define the functions involved. For a rectangle, identify the formulas for area and perimeter, knowing that these will direct the optimization process.
- Express the perimeter solely as a function of one variable by utilizing the fixed area constraint. Using the relationship between length and width derived from the area, replace one variable in the perimeter equation.
- Apply calculus techniques to find the critical points of the perimeter function, which reveal possible points of minimization.
- Verify whether these points indeed minimize the perimeter using additional tests, such as the second derivative test.
Derivative
Derivatives are fundamental instruments in calculus used to understand how a function changes. Understanding derivatives is essential for solving optimization problems like minimizing a rectangle's perimeter.
Here are the key aspects of derivatives in this context:
Here are the key aspects of derivatives in this context:
- First Derivative: Represents the rate of change of a function. For our perimeter function, \(&\frac{dP}{dw}\) reveals where the slope is zero, indicating critical points.
- Setting First Derivative to Zero: Finds critical points by solving \(&\frac{dP}{dw} = 0\). Shows potential spots for minimum perimeter by finding where it stops changing.
- Second Derivative: Further insights the nature of critical points. If \(&\frac{d^2P}{dw^2}\) is positive at a critical point, it ensures the perimeter is minimized at that point, indicating a concave up slope.
