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Magazine Chapter 4: Problem 1
Solve each of the maximum-minimum problems in Exercises \(1-20 .\) Some may not have a solution, whereas others may have their solution at the endpoint of the interval of definition. A rectangle is to have perimeter \(100 \mathrm{~m}\). What dimensions will give it the greatest area?
Short Answer
Expert verified
A 25 m x 25 m square gives the greatest area.
Step by step solution
01
Understand the Problem
We need to find the dimensions of a rectangle that will give it the greatest area, with the constraint that its perimeter must be 100 meters.
02
Formulate Equations
Let the length of the rectangle be \( l \) meters and the width be \( w \) meters. The perimeter equation is given by \( 2l + 2w = 100 \). This simplifies to \( l + w = 50 \). The area of the rectangle is \( A = lw \).
03
Express Area in Terms of One Variable
Substitute \( w = 50 - l \) from the perimeter equation into the area equation: \( A = l(50 - l) = 50l - l^2 \). Thus, the area \( A(l) = 50l - l^2 \) is a function of \( l \) only.
04
Calculate the Derivative
Find the derivative of the area with respect to \( l \), using \( A(l) = 50l - l^2 \): \( A'(l) = 50 - 2l \).
05
Find Critical Points
Set the derivative equal to zero to find critical points: \( 50 - 2l = 0 \), which simplifies to \( l = 25 \).
06
Determine Maximum Area
To confirm \( l = 25 \) yields a maximum, check the second derivative \( A''(l) = -2 \), which indicates a concave down parabola, confirming a maximum at \( l = 25 \).
07
Find Corresponding Width
Substitute \( l = 25 \) back into \( l + w = 50 \) to find \( w \). Thus, \( 25 + w = 50 \) leads to \( w = 25 \).
08
Final Dimensions
The rectangle dimensions that give the greatest area are both \( 25 \) meters for length and width, forming a square.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivative Calculus
Understanding derivative calculus is key when solving optimization problems in mathematics. A derivative represents the rate at which a function is changing at any given point, acting as a tool to solve various real-life problems.
In our example, we have the function for the area of a rectangle expressed as a function of one variable: \( A(l) = 50l - l^2 \). Here, \( A(l) \) describes how the area changes with respect to the length \( l \).
In our example, we have the function for the area of a rectangle expressed as a function of one variable: \( A(l) = 50l - l^2 \). Here, \( A(l) \) describes how the area changes with respect to the length \( l \).
- The derivative of \( A \), denoted \( A'(l) \), helps identify where changes occur in the function, providing information on increasing or decreasing areas.
- The derivative formula \( A'(l) = 50 - 2l \) tells us the slope of \( A(l) \) for any \( l \).
Critical Points
Critical points are crucial in calculus and optimization. They are the input values where a function's derivative is zero or undefined. These points often serve as candidates for local maxima, minima, or neither.
In our example, we derive \( A'(l) = 50 - 2l \) and set it to zero to find the critical point. Solving this equation gives us \( l = 25 \).
In our example, we derive \( A'(l) = 50 - 2l \) and set it to zero to find the critical point. Solving this equation gives us \( l = 25 \).
- This value of \( l \) is a critical point as the derivative at this point equals zero, suggesting potential maxima or minima.
- They must be further analyzed, using tests like the second derivative test, to confirm their nature.
Maximum and Minimum Values
Finding the maximum and minimum values of a function is a primary goal in optimization problems. These values tell us the highest or lowest point a function can reach, respectively.
In the context of our rectangle problem, the main aim was to maximize the area. After finding the critical point \( l = 25 \), the next step was determining whether this point offers a maximum or minimum area.
In the context of our rectangle problem, the main aim was to maximize the area. After finding the critical point \( l = 25 \), the next step was determining whether this point offers a maximum or minimum area.
- For the rectangle's dimensions, a maximum value implies the largest possible area under the given constraints.
- Since the final dimensions were equal, forming a 25 by 25 square, this confirmed the maximum area found as 625 square meters.
Second Derivative Test
The second derivative test is a helpful tool in calculus to determine if a critical point is a local maximum or minimum. It involves taking the second derivative of a function and evaluating it at the critical point.
In our rectangle problem, we calculated the second derivative as \( A''(l) = -2 \). The negative value signifies a concave down parabola, indicating that the function reaches a local maximum where \( l = 25 \).
In our rectangle problem, we calculated the second derivative as \( A''(l) = -2 \). The negative value signifies a concave down parabola, indicating that the function reaches a local maximum where \( l = 25 \).
- If \( A''(l) > 0 \), it implies a local minimum, as the graph curves upwards.
- If \( A''(l) < 0 \), as in this case, a local maximum is present since the graph curves downwards.
