/*! 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 2 If \(f(x, y)=x^{2}+y^{2}\) and \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 2

If \(f(x, y)=x^{2}+y^{2}\) and \(g(x, y)=2 x+3 y-4=0,\) write the Lagrange multiplier conditions that must be satisfied by a point that maximizes or minimizes \(f\) subject to \(g(x, y)=0\).

Short Answer

Expert verified
Answer: The Lagrange multiplier conditions are \(2x - 2\lambda = 0\), \(2y - 3\lambda = 0\), and \(-2x - 3y + 4 = 0\).

Step by step solution

01

Introduce the Lagrange function

The Lagrange function \(L(x, y, \lambda)\) is given by $$ L(x, y, \lambda) = f(x, y) - \lambda g(x, y) = x^2 + y^2 - \lambda(2x + 3y - 4). $$
02

Find the partial derivatives of the Lagrange function

Calculate the partial derivatives of the Lagrange function with respect to \(x, y\), and \(\lambda\). $$ \frac{\partial L}{\partial x} = 2x - 2\lambda, \;\; \frac{\partial L}{\partial y} = 2y - 3\lambda, \;\; \frac{\partial L}{\partial \lambda} = -2x - 3y + 4. $$
03

Set these derivatives equal to zero

To maximize or minimize the function \(f(x, y)\) subject to the constraint \(g(x, y) = 0\), set the partial derivatives of the Lagrange function equal to zero. $$ 2x - 2\lambda = 0,\;\; 2y - 3\lambda = 0, \;\; -2x - 3y + 4 = 0. $$ These are the Lagrange multiplier conditions that must be satisfied by a point that maximizes or minimizes \(f(x, y)\) subject to \(g(x, y) = 0\).

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!

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

Show that the plane \(a x+b y+c z=d\) and the line \(\mathbf{r}(t)=\mathbf{r}_{0}+\mathbf{v} t,\) not in the plane, have no points of intersection if and only if \(\mathbf{v} \cdot\langle a, b, c\rangle=0 .\) Give a geometric explanation of the result.

Consider the following functions \(f.\) a. Is \(f\) continuous at (0,0)\(?\) b. Is \(f\) differentiable at (0,0)\(?\) c. If possible, evaluate \(f_{x}(0,0)\) and \(f_{y}(0,0)\) d. Determine whether \(f_{x}\) and \(f_{y}\) are continuous at \((0,0).\) e. Explain why Theorems 5 and 6 are consistent with the results in parts \((a)-(d).\) $$f(x, y)=\sqrt{|x y|}$$

Limits at (0,0) may be easier to evaluate by converting to polar coordinates. Remember that the same limit must be obtained as \(r \rightarrow 0\) along all paths to (0,0) Evaluate the following limits or state that they do not exist. $$\lim _{(x, y) \rightarrow(0,0)} \frac{(x-y)^{2}}{x^{2}+x y+y^{2}}$$

a. Consider the function \(w=f(x, y, z)\). List all possible second partial derivatives that could be computed. b. Let \(f(x, y, z)=x^{2} y+2 x z^{2}-3 y^{2} z\) and determine which second partial derivatives are equal. c. How many second partial derivatives does \(p=g(w, x, y, z)\) have?

Match equations a-f with surfaces A-F. a. \(y-z^{2}=0\) b. \(2 x+3 y-z=5\) c. \(4 x^{2}+\frac{y^{2}}{9}+z^{2}=1\) d. \(x^{2}+\frac{y^{2}}{9}-z^{2}=1\) e. \(x^{2}+\frac{y^{2}}{9}=z^{2}\) f. \(y=|x|\)
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Theoretical and Mathematical Physics

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.