Ch. 4 - Applications of Derivatives

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 4 - Applications of Derivatives

Problem 4.5.20a

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 guidance

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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Key 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.

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.

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

Checking Antiderivative Formulas

Right, or wrong? Say which for each formula and give a brief reason for each answer.

∫√(2x + 1) dx = √(x² + x +C)

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Textbook Question

Theory and Examples

In Exercises 51 and 52, give reasons for your answers.

Let f(x) = |x³ − 9x|.

a. Does f'(0) exist?

Textbook Question

52. Two masses hanging side by side from springs have positions s_1 = 2 sin t and s_2 = sin 2t,

respectively.

a. At what times in the interval 0 < t do the masses pass each other? (Hint: sin 2t = 2 sint cost.)

Textbook Question

Theory and Examples

Sketch the graph of a differentiable function y = f(x) that has a local minimum at (1, 1) and a local maximum at (3, 3).

Textbook Question

Checking Antiderivative Formulas

Right, or wrong? Say which for each formula and give a brief reason for each answer.

∫tanθ sec²θ dθ = sec³θ / 3 + C

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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) = √(x² − 2x − 3), 3 ≤ x < ∞