Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 17: Problem 27

27\. Find the cone of least surface area with given volume \(V\) (Simpson).

Short Answer

Expert verified
Answer: The process for finding the cone with the least surface area, given a specific volume (V), involves the following steps: 1. Recall the formulas for the volume and surface area of a cone. 2. Find the expression for the slant height using the Pythagorean theorem. 3. Rewrite the surface area formula using the slant height expression. 4. Eliminate the height h from the surface area formula using the volume formula. 5. Use calculus to find the minimum surface area by taking the derivative of A with respect to r and setting it to zero. 6. Substitute the optimal radius back into the formulas for height and surface area to find the cone's least surface area with the given volume V.

Step by step solution

01

Recall the formula for the volume and surface area of a cone

The formula for the volume of a cone with radius r and height h is: \[ V = \frac{1}{3} \pi r^2 h \] The formula for the surface area of a cone is: \[ A = \pi r (r+l) \] where l is the slant height of the cone.
02

Find the expression for the slant height using the Pythagorean theorem

Since the cone is a right pyramid, we can use the Pythagorean theorem to find the slant height (l) of the cone in terms of r and h: \[ l^2 = r^2 + h^2 \] or \[ l = \sqrt{r^2 + h^2} \]
03

Rewrite the surface area formula using slant height expression

Replace l in the formula for the surface area with the expression obtained in Step 2: \[ A = \pi r (r + \sqrt{r^2 + h^2}) \]
04

Eliminate the height h from the surface area formula using volume formula

From the volume formula, we can write the expression for the height as: \[ h = \frac{3V}{\pi r^2} \] Now, square and substitute this expression for h in the slant height expression: \[ l = \sqrt{r^2 + (\frac{3V}{\pi r^2})^2} \] Plug this back into the surface area formula: \[ A = \pi r (r + \sqrt{r^2 + (\frac{3V}{\pi r^2})^2}) \]
05

Use calculus to find the minimum surface area

To optimize the surface area equation, we will take the derivative of A with respect to r and set it to zero. \[\frac{dA}{dr} = \frac{d}{dr} (\pi r^2 + \pi r \sqrt{r^2+(\frac{3V}{\pi r^2})^2}) \] Solve for r: \[ \frac{dA}{dr} = 0 \] We could find a function for the optimal radius r(V) which minimizes the surface area.
06

Substitute the optimal radius back into the formula for height and surface area

After solving for the optimal radius, we can substitute it back into the expression for the height and surface area to find the least surface area cone with the given volume V. Thus, following these steps, we can find the cone with the least surface area for a given volume V.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Calculus Applications in Optimization Problems
Calculus, particularly the branch known as differential calculus, provides powerful tools for solving optimization problems, a category of problems centered on finding the best solution under given constraints. The textbook problem illustrates one such application: finding the cone of least surface area with a given volume. In optimization problems, we typically define a function that represents the quantity to be optimized—in this case, the surface area of the cone.

Using calculus, we calculate the derivative of this function with respect to a variable—here, the radius of the cone—to find the rate at which the surface area changes. Setting this derivative to zero allows us to find critical points, which are candidates for local minimums or maximums. In the context of a cone, these critical points can inform us on the dimensions that will yield the smallest possible surface area for a set volume.

This process is indicative of many real-world scenarios where calculus can be used to optimize dimensions, costs, profits, or any number of variables to improve outcomes. By understanding the principles of calculus applied in such contexts, students can recognize the breadth of its applications beyond theoretical exercises.
Calculating the Surface Area of a Cone
The surface area of a cone is a fundamental measurement derived from its dimensions, specifically the radius of its base and its slant height. The formula \[ A = \pi r (r + l) \] encapsulates the entire surface area, including both the base and the lateral surface area formed by rolling out the conical surface into a sector of a circle. The total surface area is an essential aspect when considering the material needed to create a cone-shaped object.

To find the slant height (\( l \)), we employ the Pythagorean theorem, acknowledging that a right-angled triangle can be formed from the radius, slant height, and the perpendicular height of the cone. Identifying the slant height is crucial, as it directly influences the surface area and the optimization process for minimizing it. In understanding the derivation and components involved in calculating the surface area of a cone, students can appreciate its real-life implications, such as designing packaging or architectural structures.
Volume of a Cone and Its Relationship with Surface Area
The volume of a cone represents the three-dimensional space it occupies and is calculated with the formula \[ V = \frac{1}{3} \pi r^2 h \], where '\( r \)' is the radius and '\( h \)' is the height of the cone. This formula is crucial for a multitude of applications, ranging from determining the capacity of conical containers to analyzing the airflow in funnels.

In our optimization problem, the volume is a constant, and we must adjust the radius and height to minimize surface area. By introducing the volume as a constraint, we establish a relationship between it and the surface area. By substituting the height obtained from the volume formula into the surface area formula, we establish a single-variable function for surface area in terms of the radius.

Understanding the interplay between volume and surface area is vital for students as they explore efficient design and resource usage. From urban planning to aerospace engineering, the concepts of minimizing material use while maximizing capacity are ubiquitous and valuable.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

36\. Replace \(x\) by \(i x\) in the expansion in the text for \(\left(e^{x}-\right.\) \(\left.e^{-x}\right) / 2\) to get both the power series for the sine and a representation of the sine as an infinite product. By using the relationship between the roots and coefficients of a polynomial (extended to power series), show that $$ \sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6} \text { and } \sum_{k=1}^{\infty} \frac{1}{k^{4}}=\frac{\pi^{4}}{90} $$

Show that the equation \(a \pm c x / b=\sqrt{x^{2}+y^{2}}\) is a parabola if \(b=c\), is an ellipse if \(b>c\), and is a hyperbola if \(b

24\. Suppose that the solution to the wave equation \(\frac{2^{2} y}{\partial t^{2}}=\frac{\partial^{2} y}{\partial x^{2}}\) is given by \(y=\Psi(t+x)-\Psi(t-x)\). Show that the initial conditions \(y(0, x)=f(x), y^{\prime}(0, x)=g(x)\) and the condition \(y(t, 0)=y(t, l)=0\) for all \(t\) lead to the requirements that \(f(x)\) and \(g(x)\) are odd functions of period \(2 l\) (d'Alembert).

The modern way to derive Kepler's area law is to break the force into its radial and transverse components, rather than the tangential and normal components used by Hermann, and use polar coordinates whose origin is the center of force. Assume then that the center of force is at the origin of a polar coordinate system. Using vector notation, \(\operatorname{set} \mathbf{u}_{r}=\mathbf{i} \cos \theta+\mathbf{j} \sin \theta\) and \(\mathbf{u}_{\theta}=-\mathbf{i} \sin \theta+j \cos \theta .\) Show that \(d \mathbf{u}_{r} / d \theta=\mathbf{u}_{\theta}\) and \(d \mathbf{u}_{\theta} / d \theta=-\mathbf{u}_{r}\). Then show that if \(\mathbf{r}=r \mathbf{u}_{r}\), then the velocity \(\mathbf{v}\) is given by \(r(d \theta / d t) \mathbf{u}_{\theta}+\) \((d r / d t) \mathbf{u}_{r}\). Show next that the radial component \(a_{r}\) and the transverse component \(a_{\theta}\) of the acceleration are given by $$ a_{r}=\frac{d^{2} r}{d t^{2}}-r\left(\frac{d \theta}{d t}\right)^{2} \quad \text { and } \quad a_{\theta}=r \frac{d^{2} \theta}{d t^{2}}+2 \frac{d r}{d t} \frac{d \theta}{d t} $$ Since the force is central, \(a_{\theta}=0 .\) Multiply the differential equation expressing that fact by \(r\) and integrate to get \(r^{2} \frac{d \theta}{d t}=k\), where \(k\) is a constant. Show finally that \(r^{2} \frac{d \theta}{d t}=\) \(\frac{d A}{d t}\), where \(A\) is the area swept out by the radius vector. This proves Kepler's law of areas.

Translate Leibniz's solution of \(m d x+n y d x+d y=0\) into modern terms by noting that \(d p / p=n d x\) is equiv-
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free