/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 Find all critical points of the ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 13

Find all critical points of the following functions. \(f(x, y)=x^{4}+y^{4}-16 x y\)

Short Answer

Expert verified
Question: Determine the critical points of the given function \(f(x, y) = x^4 + y^4 - 16xy\). Answer: The critical points of the given function are (0, 0), (8, 4), and (-8, -4).

Step by step solution

01

Calculate the partial derivatives

Calculate the partial derivative of \(f(x, y)\) with respect to both \(x\) and \(y\): \(\frac{\partial f}{\partial x} = 4x^3 - 16y \) \(\frac{\partial f}{\partial y} = 4y^3 - 16x \)
02

Find critical points

Set the partial derivatives equal to zero and solve the system of equations: \(4x^3 - 16y = 0\) \(4y^3 - 16x = 0\) Divide both equations by 4 to simplify: \(x^3 - 4y = 0\) \(y^3 - 4x = 0\) Now, we can rearrange the first equation to get \(x^3\) in terms of \(y\) and rearrange the second equation to get \(y^3\) in terms of \(x\): \(x^3 = 4y\) \(y^3 = 4x\) Substitute the first equation into the second equation to eliminate \(x\): \(y^3 = 4(4y)\) \(y^3 = 16y\) Then, to find the critical points, factoring and solving for \(y\): \(y^3 - 16y = y(y^2 - 16) = 0\) We get 3 possible values for \(y\): \(y = 0, y = 4, y = -4\) For each value of \(y\), we plug in the obtained values into the first equation to find the corresponding \(x\) values: For \(y = 0\): x^3 = 4 * 0 鈬� x = 0 For \(y = 4\): x^3 = 4 * 4 鈬� x = 8 For \(y = -4\): x^3 = 4 * (-4) 鈬� x = -8 So, the critical points for the given function are: \((0, 0), (8, 4), (-8, -4)\)

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Partial Derivatives
Partial derivatives are a foundational concept in multivariable calculus. When working with functions of multiple variables鈥攍ike f(x, y) in your exercise鈥攖he partial derivative with respect to one variable, say x, is found by differentiating with x while treating all other variables as constants.

Simply put, the partial derivative represents the rate at which the function changes as one particular variable changes, while the others are held fixed. In the context of your function f(x, y) = x^4 + y^4 - 16xy, we have the partial derivatives:
  • \frac{\(\partial f \)}{\(\partial x \)} = 4x^3 - 16y
  • \frac{\(\partial f \)}{\(\partial y \)} = 4y^3 - 16x
These equations tell us how the function's value changes in response to changes in x and y independently.
Critical Point Calculation
A critical point occurs where the partial derivatives of a function are zero or undefined. They are points where the function's rate of change switches direction or becomes flat, which often correspond to maxima, minima, or saddle points. The process of finding critical points involves two crucial steps:

Finding the Partial Derivatives


In your problem, the first step was computing the partial derivatives of the function f(x, y) with respect to x and then y.

Setting Equations to Zero


The next step is to set these partial derivatives equal to zero since we're looking for where the rate of change is flat. Solving the resulting system of equations typically gives us the x and y values for the critical points. In your example, after setting the partial derivatives equal to zero and simplifying, you arrive at a system that can be solved by substitution or other methods to find the actual critical points, which are (0, 0), (8, 4), and (-8, -4).
Multivariable Calculus
Multivariable calculus extends the tools of single-variable calculus to functions of several variables. It involves partial derivatives, multiple integrals, and more complex applications, like finding the maxima and minima of multivariable functions鈥攚hich is exactly what we do when locating critical points.

The beauty of multivariable calculus is in how it allows us to analyze and describe the world around us, which is inherently multidimensional. For instance, the function f(x, y) in our exercise could represent a physical phenomenon where two variables influence the outcome. Understanding how to handle these variables separately and together is at the heart of the subject.

Identifying critical points in multivariable calculus not only helps in graphing these functions but also in solving optimization problems in fields like economics, engineering, and the sciences.

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

Find the points (if they exist) at which the following planes and curves intersect. $$8 x+y+z=60 ; \quad \mathbf{r}(t)=\left\langle t, t^{2}, 3 t^{2}\right\rangle, \text { for }-\infty

Identify and briefly describe the surfaces defined by the following equations. $$z^{2}+4 y^{2}-x^{2}=1$$

Consider the following equations of quadric surfaces. a. Find the intercepts with the three coordinate axes, when they exist. b. Find the equations of the x y-, x z^{-}, \text {and } y z-\text {traces, when they exist. c. Sketch a graph of the surface. $$-\frac{x^{2}}{6}-24 y^{2}+\frac{z^{2}}{24}-6=0$$

Given a differentiable function \(w=f(x, y, z),\) the goal is to find its maximum and minimum values subject to the constraints \(g(x, y, z)=0\) and \(h(x, y, z)=0\) where \(g\) and \(h\) are also differentiable. a. Imagine a level surface of the function \(f\) and the constraint surfaces \(g(x, y, z)=0\) and \(h(x, y, z)=0 .\) Note that \(g\) and \(h\) intersect (in general) in a curve \(C\) on which maximum and minimum values of \(f\) must be found. Explain why \(\nabla g\) and \(\nabla h\) are orthogonal to their respective surfaces. b. Explain why \(\nabla f\) lies in the plane formed by \(\nabla g\) and \(\nabla h\) at a point of \(C\) where \(f\) has a maximum or minimum value. c. Explain why part (b) implies that \(\nabla f=\lambda \nabla g+\mu \nabla h\) at a point of \(C\) where \(f\) has a maximum or minimum value, where \(\lambda\) and \(\mu\) (the Lagrange multipliers) are real numbers. d. Conclude from part (c) that the equations that must be solved for maximum or minimum values of \(f\) subject to two constraints are \(\nabla f=\lambda \nabla g+\mu \nabla h, g(x, y, z)=0\) and \(h(x, y, z)=0\).

A classical equation of mathematics is Laplace's equation, which arises in both theory and applications. It governs ideal fluid flow, electrostatic potentials, and the steadystate distribution of heat in a conducting medium. In two dimensions, Laplace's equation is $$\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0.$$ Show that the following functions are harmonic; that is, they satisfy Laplace's equation. $$u(x, y)=e^{a x} \cos a y, \text { for any real number } a$$
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

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.