Textbook Question
Finding Antiderivatives
In Exercises 1–16, find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation.
(2/3)x⁻¹ᐟ³
1
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Ch. 4 - Applications of Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 4, Problem 4.5.20a
20.The U.S. Postal Service will accept a box for domestic shipment only if the sum of its length and girth (distance around) does not exceed 108 in. a.What dimensions will give a box with a square end the largest possible volume?
Verified step by step guidance1
Step 1: Define the variables. Let the side length of the square end be 'x' (in inches), and the length of the box be 'L' (in inches). The girth is the perimeter of the square end, which is 4x.
Step 2: Write the constraint equation. The sum of the length and girth must not exceed 108 inches. Therefore, the constraint is: L + 4x ≤ 108.
Step 3: Express the volume of the box. The volume of the box is given by the area of the square end multiplied by the length: V = x² * L.
Step 4: Solve for 'L' using the constraint equation. From the constraint, L = 108 - 4x. Substitute this into the volume equation to express volume in terms of 'x': V = x² * (108 - 4x).
Step 5: Maximize the volume. To find the dimensions that give the largest possible volume, take the derivative of V with respect to 'x', set it equal to zero, and solve for 'x'. Then, use the constraint equation to find the corresponding 'L'.
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Video duration:
4mKey Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Volume of a Box
The volume of a box is calculated by multiplying its length by its width and height. In this case, since the box has a square end, if we denote the side length of the square as 'x' and the length as 'L', the volume can be expressed as V = x^2 * L. Understanding how to express volume in terms of the box's dimensions is crucial for maximizing it under given constraints.
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Finding Volume Using Disks
Girth Calculation
Girth is defined as the distance around the box, specifically calculated as 2 times the width plus 2 times the height. For a box with a square end, this simplifies to G = 4x + 2L. The problem states that the sum of the length and girth must not exceed 108 inches, which provides a constraint that must be satisfied when determining the optimal dimensions for maximum volume.
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04:46Determining Error and Relative Error Example 1
Optimization in Calculus
Optimization involves finding the maximum or minimum values of a function subject to certain constraints. In this scenario, we need to maximize the volume function V = x^2 * L while adhering to the constraint given by the girth equation. Techniques such as substitution and taking derivatives to find critical points will be essential in solving this problem effectively.
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Related Practice
Textbook Question
Applications
Suppose that f(x) = d/dx (1 − √x) and g(x) = d/dx (x + 2).
Find:
∫f(x) dx
Textbook Question
Identifying Extrema
In Exercises 41–52:
a. Identify the function’s local extreme values in the given domain, and say where they occur.
k(x) = x³ + 3x² + 3x + 1, −∞ < x ≤ 0
Textbook Question
Roots (Zeros)
a. Plot the zeros of each polynomial on a line together with the zeros of its first derivative.
i. y = x² − 4
Textbook Question
Identifying Extrema
In Exercises 41–52:
a. Identify the function’s local extreme values in the given domain, and say where they occur.
f(x) = (x + 1)², −∞ < x ≤ 0
Textbook Question
Analyzing Functions from Derivatives
Answer the following questions about the functions whose derivatives are given in Exercises 1–14:
a. What are the critical points of f?
f′(x) = (x − 1)(x + 2)(x − 3)