Skip to main content
Ch. 4 - Applications of the Derivative
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 4 - Applications of the DerivativeProblem 4.R.39
Chapter 4, Problem 4.R.39

Optimization A right triangle has legs of length h and r and a hypotenuse of length 4 (see figure). It is revolved about the leg of length h to sweep out a right circular cone. What values of h and r maximize the volume of the cone? (Volume of a cone = (1/3) πr²h.) <IMAGE>

Verified step by step guidance
1
First, express the relationship between the sides of the right triangle using the Pythagorean theorem: h² + r² = 4².
Solve the Pythagorean equation for one of the variables, say r, in terms of h: r = sqrt(16 - h²).
Substitute the expression for r into the volume formula of the cone: V = (1/3)π(sqrt(16 - h²))²h.
Simplify the volume expression: V = (1/3)π(16 - h²)h.
To find the maximum volume, take the derivative of V with respect to h, set it equal to zero, and solve for h. Then, use this value to find r using the expression from step 2.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
5m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Volume of a Cone

The volume of a cone is calculated using the formula V = (1/3)πr²h, where r is the radius of the base and h is the height. This formula derives from the geometric properties of cones and is essential for determining the maximum volume in optimization problems. Understanding how to manipulate this formula is crucial for solving the given problem.
Recommended video:
08:29
Example 5: Packaging Design

Optimization

Optimization in calculus involves finding the maximum or minimum values of a function. In this context, we need to maximize the volume of the cone by adjusting the dimensions h and r. Techniques such as taking derivatives and applying critical point analysis are fundamental to solving optimization problems effectively.
Recommended video:
10:13
Intro to Applied Optimization: Maximizing Area

Right Triangle Relationships

In this problem, the right triangle formed by the legs h and r and the hypotenuse (length 4) is governed by the Pythagorean theorem, which states that a² + b² = c². This relationship allows us to express one variable in terms of the other, facilitating the optimization process by reducing the number of variables involved.
Recommended video:
05:59
Estimating the Area Under a Curve with Right Endpoints & Midpoint
Related Practice
Textbook Question

Find the critical points of the following functions on the given intervals. Identify the absolute maximum and absolute minimum values (if they exist).

ƒ(x) = sin 2x + 3 on [-π , π]

211
views
Textbook Question

Find the critical points of the following functions on the given intervals. Identify the absolute maximum and absolute minimum values (if they exist).

ƒ(x) = 4x¹⸍² - x⁵⸍² on [0, 4]

192
views
Textbook Question

{Use of Tech} Newton’s method Use Newton’s method to approximate the roots of ƒ(x) = e⁻²ˣ + 2eˣ - 6 to six digits.

183
views
Textbook Question

Locating extrema Consider the graph of a function ƒ on the interval [-3, 3]. <IMAGE>

d. Give the approximate coordinates of the zero(s) of f.

207
views
Textbook Question

82–89. Comparing growth rates Determine which of the two functions grows faster, or state that they have comparable growth rates.


2ˣ and 4ˣ⸍²

275
views
Textbook Question

24–34. Curve sketching Use the guidelines given in Section 4.4 to make a complete graph of the following functions on their domains or on the given interval. Use a graphing utility to check your work.


{Use of Tech} ƒ(x) = x (x -1)e⁻ˣ

267
views