Problem 76
Question
A carpenter wishes to make a rain gutter with a rectangular cross-section by bending up a flat piece of metal that is 18 feet long and 20 inches wide. The top of the gutter is open. $$2$$ (a) Write an expression for the cross-sectional area in terms of \(x,\) the length of metal that is bent upward. (b) How much metal has to be bent upward to maximize the cross-sectional area? What is the maximum cross-sectional area?
Step-by-Step Solution
Verified Answer
The maximum cross-sectional area of the gutter is 4320 square inches, and it occurs when no metal is bent upwards, i.e., \(x = 0\) inches.
1Step 1: Defining Variables
Firstly, the bent up sides form the height of the rectangular cross-section (say \(x\) inches) and remaining metal forms the base which is \(20 - 2x\) inches.
2Step 2: Formulate the Area Function
The cross-sectional area \(A\) of a rectangle is given by the product of the length and the width. Here, the length is 18 feet (or 216 inches) and the width (the remaining flat part after bending up the edges) is \(20 - 2x\) inches. So, the area \(A\) can be expressed in terms of \(x\) as \(A = 216(20-2x)\)
3Step 3: Find the Derivative
Next, find the derivative of the area function \(\frac{dA}{dx} = 216(-2) = -432\)
4Step 4: Solving for Optimal Value
Since the derivative doesn't depend on \(x\), the maximum or minimum occurs at the boundary of the definition area of \(x\). \(x\) should be non-negative and less or equal than 10, which gives \(20-2*x \geq 0\). Therefore, there is a maximum at \(x=0\) and \(x=10\).
5Step 5: Finding the Maximum Cross-Sectional Area
Compute \(A(0) = 216*20 = 4320\) square inches and \(A(10) = 216*0 = 0\). Hence, the maximum cross-sectional area is when \(x=0\), and it is \(4320\) square inches
Key Concepts
Rectangular Cross-SectionArea FunctionOptimization in Precalculus
Rectangular Cross-Section
Understanding the rectangular cross-section is pivotal when it comes to tasks involving structures like rain gutters, beams, or any other object where this shape makes an appearance. A rectangular cross-section is a cut through a 3D object that reveals a rectangle. The dimensions of this rectangle are usually defined by two measures: the width (or the base) and the height (or the side).
In the exercise we're discussing, a carpenter aims to create a rain gutter with such a cross-section by bending a flat piece of metal. Think of the metal as clay that can be molded. When you bend the sides upward, each bent section forms the height, leaving the unbent part as the base. It's crucial to visualize this because it helps us understand how changing one dimension (the height, by bending) affects the other (the base), and ultimately, the overall area. This relationship between the dimensions is at the core of optimizing the cross-sectional area.
In the exercise we're discussing, a carpenter aims to create a rain gutter with such a cross-section by bending a flat piece of metal. Think of the metal as clay that can be molded. When you bend the sides upward, each bent section forms the height, leaving the unbent part as the base. It's crucial to visualize this because it helps us understand how changing one dimension (the height, by bending) affects the other (the base), and ultimately, the overall area. This relationship between the dimensions is at the core of optimizing the cross-sectional area.
Area Function
The area function of a shape is a mathematical way to describe how the area changes with respect to different dimensions. It is usually expressed in terms of variables that represent these dimensions. To establish the area function for the gutter in our example, we start with the formula for the area of a rectangle, which is the product of its length and width.
In our case, the length (the part along the 18 feet of metal) remains constant. The width, however, changes based upon how much of the metal is bent upward. The area function, thus, becomes a representation of the area of the cross-section in terms of the variable that defines the height, represented as '\(x\)'. The area function, in this context, is a foundational tool to understand the optimization problem; it quantifies how the area depends on our decision of how much to bend the metal. By manipulating this function, we can explore different scenarios and aim to find the conditions under which the area is maximized.
In our case, the length (the part along the 18 feet of metal) remains constant. The width, however, changes based upon how much of the metal is bent upward. The area function, thus, becomes a representation of the area of the cross-section in terms of the variable that defines the height, represented as '\(x\)'. The area function, in this context, is a foundational tool to understand the optimization problem; it quantifies how the area depends on our decision of how much to bend the metal. By manipulating this function, we can explore different scenarios and aim to find the conditions under which the area is maximized.
Optimization in Precalculus
Optimization is a mathematical technique used extensively in precalculus to find the best solution among many possibilities. It's like discovering the best deal while shopping within a budget or determining the tallest possible height for a building with limited materials. In optimization problems, we often deal with maximization or minimization: We might look for conditions that give us the largest area, the smallest cost, or the shortest travel time.
In our rain gutter problem, optimization comes into play when we want to find the dimensions that maximize the cross-sectional area. The process involves taking the derivative of the area function (to find how fast the area changes with '\(x\)'), and then analyzing it to find the maximum value. As per our exercise, since the derivative is a constant negative, we learn that the area decreases as '\(x\) increases, indicating that the maximum area is obtained when '\(x\) is as small as possible. The boundary conditions of our problem are essential; they dictate the practical limits within which we can 'play.' It's a balance of trade-offs, and precalculus provides us the mathematical language and techniques to find that sweet spot, or in our case, the optimal '\(x\)'.
In our rain gutter problem, optimization comes into play when we want to find the dimensions that maximize the cross-sectional area. The process involves taking the derivative of the area function (to find how fast the area changes with '\(x\)'), and then analyzing it to find the maximum value. As per our exercise, since the derivative is a constant negative, we learn that the area decreases as '\(x\) increases, indicating that the maximum area is obtained when '\(x\) is as small as possible. The boundary conditions of our problem are essential; they dictate the practical limits within which we can 'play.' It's a balance of trade-offs, and precalculus provides us the mathematical language and techniques to find that sweet spot, or in our case, the optimal '\(x\)'.
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