/*! 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 6 Examine the function for relativ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 6

Examine the function for relative extrema and saddle points. $$ f(x, y)=9-(x-3)^{2}-(y+2)^{2} $$

Short Answer

Expert verified
The critical point of the function \(f(x, y)=9-(x-3)^{2}-(y+2)^{2}\) is found to be at (3,-2). After evaluating the determinant of the Hessian matrix, it is determined that this is a relative maximum.

Step by step solution

01

Find Partial Derivatives

To find the critical points, first find the partial derivatives of the function with respect to \(x\) and \(y\). \[\frac{\partial f}{\partial x}=-2(x-3),\] \[ \frac{\partial f}{\partial y}=-2(y+2).\]
02

Set Partial Derivatives Equal to Zero

Then, set these equal to zero to solve for \(x\) and \(y\) respectively. This will give you coordinates of the critical points. \[-2(x-3) = 0 \rightarrow x = 3,\] \[ -2(y+2) = 0 \rightarrow y = -2.\] Thus, the critical point is at (3, -2).
03

Evaluate the Second Partial Derivatives

After finding the critical point, find the second partial derivatives:\[\frac{\partial^2 f}{\partial x^2} = -2,\] \[ \frac{\partial^2 f}{\partial y^2} = -2,\] and the mixed second partial derivative: \[\frac{\partial^2 f}{\partial y \partial x} = 0.\]
04

Evaluate the Determinant of the Hessian Matrix

Now, find the determinant of the Hessian matrix. It is defined as \[ H = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial y \partial x}\right)^2.\] Plug in the values to get that \(H = (-2)(-2) - (0)^2 = 4.\] If the determinant is positive, the point will be an extremum, so our point (3,-2) is a relative extremum.
05

Checking whether it is a relative maximum or minimum

Examine the sign of the second-order partial derivatives. If both are negative, then our point is a local maximum. If both are positive, then it is a local minimum. Since both second partial derivatives are negative, our point is a relative maximum.

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

Sketch the region \(R\) whose area is given by the double integral. Then change the order of integration and show that both orders yield the same area. $$ \int_{0}^{4} \int_{0}^{\sqrt{x}} d y d x $$

Use the regression capabilities of a graphing utility or a spreadsheet to find the least squares regression quadratic for the given points. Then plot the points and graph the least squares regression quadratic. $$ (0,0),(2,2),(3,6),(4,12) $$

Evaluate the double integral. $$ \int_{0}^{4} \int_{0}^{x} \frac{2}{x^{2}+1} d y d x $$

Use the regression capabilities of \(a\) graphing utility or a spreadsheet to find any model that best fits the data points. $$ \begin{aligned} &(1,13), \quad(2,16.5),(4,24),(5,28),(8,39),(11,50.25) \\ &(17,72),(20,85) \end{aligned} $$

Sketch the region of integration and evaluate the double integral. $$ \int_{0}^{6} \int_{y / 2}^{3}(x+y) d x d y $$
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Statistics

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

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.