/*! 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 8 Find all the local maxima, local... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find all the local maxima, local minima, and saddle points of the functions. $$f(x, y)=x^{2}-2 x y+2 y^{2}-2 x+2 y+1$$

Short Answer

Expert verified
The function has a local minimum at \((1,0)\); no local maxima or saddle points exist.

Step by step solution

01

Find the Partial Derivatives

To determine critical points, first compute the first partial derivatives of the function. The function is \(f(x, y) = x^2 - 2xy + 2y^2 - 2x + 2y + 1\). The partial derivatives are:\[ f_x = \frac{\partial f}{\partial x} = 2x - 2y - 2 \]\[ f_y = \frac{\partial f}{\partial y} = -2x + 4y + 2 \]
02

Solve for Critical Points

Set each of the partial derivatives to zero and solve the system of equations:1. \( 2x - 2y - 2 = 0 \)2. \( -2x + 4y + 2 = 0 \)From (1), simplify to find \( x = y + 1 \). Substitute this into (2):\[ -2(y+1) + 4y + 2 = 0 \]\[ -2y - 2 + 4y + 2 = 0 \]\[ 2y = 0 \]\[ y = 0 \]Substitute \( y = 0 \) back into \( x = y + 1 \):\( x = 1 \).The critical point is \((1, 0)\).
03

Compute the Second Partial Derivatives

To classify the critical point, calculate the second partial derivatives:\[ f_{xx} = \frac{\partial^2 f}{\partial x^2} = 2 \]\[ f_{yy} = \frac{\partial^2 f}{\partial y^2} = 4 \]\[ f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = -2 \]
04

Compute the Hessian Determinant

The Hessian determinant \( H \) is given by:\[ H = f_{xx}f_{yy} - (f_{xy})^2 \]\[ H = 2 \times 4 - (-2)^2 = 8 - 4 = 4 \]
05

Classify the Critical Point Using the Hessian

Use the Hessian determinant to classify the critical point at \((1, 0)\):- Since \(H > 0\) and \(f_{xx} > 0\), the critical point is a local minimum.The function has a local minimum at \((1, 0)\). There are no other critical points, and hence, no local maxima or saddle points.

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 91Ó°ÊÓ!

Key Concepts

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

Partial Derivatives
Understanding the concept of partial derivatives is crucial in multivariable calculus. They allow us to examine how a function changes as we alter each individual variable, holding the others constant.
Partial derivatives are similar to ordinary derivatives but are applied to functions with two or more variables. In our example, the function is dependent on two variables, \(x\) and \(y\). Thus, the partial derivatives \(f_x\) and \(f_y\) tell us how the function changes with respect to \(x\) and \(y\) respectively. By calculating these, we gain insight into how the function's surface tilts in each direction.
When we set these derivatives to zero, we find points that could represent peaks, valleys, or saddle points in the function's surface – points where the function does not slope in any direction.
Critical Points
Critical points are essential in calculus optimization since they represent spots where the function may achieve a local maximum, local minimum or a saddle point.
To find critical points, we solve the system of equations formed by setting each first partial derivative to zero.
In the provided example, by solving the system, we determine that there is a critical point at \((1, 0)\). These points are important because they potentially identify peaks or troughs on the function surface. Properly identifying and classifying critical points is the backbone of optimizing functions in calculus.
Hessian Determinant
The Hessian determinant is a powerful tool used to classify critical points. This calculation helps us understand the nature of these points – whether they are local minima, local maxima, or saddle points.
The Hessian matrix is made from the second partial derivatives of a function and is used to compute the Hessian determinant, \(H\). For a function of two variables, like ours, \(H\) is calculated as \[H = f_{xx}f_{yy} - (f_{xy})^2\] where \(f_{xx}\), \(f_{yy}\), and \(f_{xy}\) are the second partial derivatives.
In our case, \(H = 4\), which indicates a positive determinant. With \(f_{xx} > 0\), we conclude that the critical point \((1, 0)\) is a local minimum. Understanding the Hessian determinant's role helps clarify why certain critical points are classified in particular ways.

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

If a function \(f(x, y)\) has continuous second partial derivatives throughout an open region \(R,\) must the first-order partial derivatives of \(f\) be continuous on \(R ?\) Give reasons for your answer.

For what values of the constant \(k\) does the Second Derivative Test guarantee that \(f(x, y)=x^{2}+k x y+y^{2}\) will have a saddle point at (0,0)\(?\) A local minimum at (0,0)\(?\) For what values of \(k\) is the Second Derivative Test inconclusive? Give reasons for your answers.

Minimize the function \(f(x, y, z)=x^{2}+y^{2}+z^{2}\) subject to the constraints \(x+2 y+3 z=6\) and \(x+3 y+9 z=9.\)

Find the linearization \(L(x, y, z)\) of the function \(f(x, y, z)\) at \(P_{0} .\) Then find an upper bound for the magnitude of the error \(E\) in the approximation \(f(x, y, z) \approx L(x, y, z)\) over the region \(R\) \(f(x, y, z)=\sqrt{2} \cos x \sin (y+z)\) at \(P_{0}(0,0, \pi / 4)\) R: \(|x| \leq 0.01, \quad|y| \leq 0.01, \quad|z-\pi / 4| \leq 0.01\)

The discriminant \(f_{x x} f_{y y}-f_{x y}^{2}\) is zero at the origin for each of the following functions, so the Second Derivative Test fails there. Determine whether the function has a maximum, a minimum, or neither at the origin by imagining what the surface \(z=f(x, y)\) looks like. Describe your reasoning in each case. $$\text { a. } f(x, y)=x^{2} y^{2}$$ $$\text { b. } f(x, y)=1-x^{2} y^{2}$$ $$\text { c. }f(x, y)=x y^{2}$$ $$\text { d. } f(x, y)=x^{3} y^{2}$$ $$\text { e. } f(x, y)=x^{3} y^{3}$$ $$f(x, y)=x^{4} y^{4}$$

See all solutions

Recommended explanations on Math Textbooks

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.