Chapter 4: Problem 331
For the following exercises, use the second derivative test to identify any critical points and determine whether each critical point is a maximum, minimum, saddle point, or none of these. $$f(x, y)=x^{2}+2 y^{2}-x^{2} y$$
Short Answer
Expert verified
Critical points are (0,0) - minimum, (2,1) and (-2,1) are saddle points.
Step by step solution
01
Find first derivatives
We need to find the first partial derivatives of the function \( f(x, y) = x^2 + 2y^2 - x^2y \). Compute the partial derivative with respect to \( x \) first: \[ f_x(x, y) = \frac{\partial}{\partial x}(x^2 + 2y^2 - x^2y) = 2x - 2xy \]Now compute the partial derivative with respect to \( y \): \[ f_y(x, y) = \frac{\partial}{\partial y}(x^2 + 2y^2 - x^2y) = 4y - x^2 \]
02
Find critical points
To find the critical points, set the first derivatives equal to zero and solve the system of equations:\( 2x - 2xy = 0 \) and \( 4y - x^2 = 0 \).From \( 2x - 2xy = 0 \), we get \( x(1-y) = 0 \). Thus, \( x = 0 \) or \( y = 1 \).If \( x = 0 \), substitute into \( 4y - x^2 = 0 \) to get \( 4y = 0 \). Hence, \( y = 0 \). This gives us the critical point \( (0, 0) \).If \( y = 1 \), substitute into \( 4y - x^2 = 0 \) to get \( 4(1) - x^2 = 0 \). Hence, \( x^2 = 4 \) which gives \( x = 2 \) or \( x = -2 \). Thus, the critical points are \( (2, 1) \) and \( (-2, 1) \).
03
Compute second derivatives
Compute the second partial derivatives needed for the second derivative test:\[ f_{xx} = \frac{\partial}{\partial x}(2x - 2xy) = 2 - 2y \]\[ f_{yy} = \frac{\partial}{\partial y}(4y - x^2) = 4 \]\[ f_{xy} = \frac{\partial}{\partial y}(2x - 2xy) = -2x \]
04
Apply the second derivative test
Apply the second derivative test with function \( D(x, y) = f_{xx}f_{yy} - (f_{xy})^2 \). **Evaluate** this for each critical point:- For \( (0, 0) \): \[ D(0, 0) = (2)(4) - (0)^2 = 8 \] \( f_{xx}(0, 0) = 2 > 0 \), so \( (0, 0) \) is a local minimum.- For \( (2, 1) \): \[ D(2, 1) = (0)(4) - (-4)^2 = -16 \] Since \( D(2, 1) < 0 \), \( (2, 1) \) is a saddle point.- For \( (-2, 1) \): \[ D(-2, 1) = (0)(4) - (4)^2 = -16 \] Since \( D(-2, 1) < 0 \), \( (-2, 1) \) is a saddle point.
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.
Second Derivative Test
The Second Derivative Test is a method used in calculus and differential equations to classify critical points of a differentiable function. It helps determine whether a critical point is a local minimum, local maximum, or a saddle point by examining the second partial derivatives.
To perform the Second Derivative Test for functions of two variables, compute the following:
This test is particularly useful because it provides a simple way to evaluate critical points without having to sketch the graph of the function.
To perform the Second Derivative Test for functions of two variables, compute the following:
- The second partial derivatives: \( f_{xx} \), \( f_{yy} \), and \( f_{xy} \).
- Calculate the discriminant: \[ D(x, y) = f_{xx}f_{yy} - (f_{xy})^2 \]
This test is particularly useful because it provides a simple way to evaluate critical points without having to sketch the graph of the function.
Partial Derivatives
Partial derivatives are a core concept in multivariable calculus. They express how a function changes as one of the variables changes, while the other variables are held constant. For a function \( f(x, y) \), the partial derivatives with respect to \( x \) and \( y \) are denoted \( f_x \) and \( f_y \), respectively.
• **Finding Partial Derivatives**: To find \( f_x \), differentiate \( f(x, y) \) with respect to \( x \), treating \( y \) as a constant. Similarly, derive \( f_y \) by differentiating with respect to \( y \), holding \( x \) constant.
In our given function \( f(x, y) = x^2 + 2y^2 - x^2y \), the partial derivatives were:
• \( f_x = 2x - 2xy \)
• \( f_y = 4y - x^2 \)
Partial derivatives are used in identifying critical points and evaluating the behavior of functions of multiple variables.
• **Finding Partial Derivatives**: To find \( f_x \), differentiate \( f(x, y) \) with respect to \( x \), treating \( y \) as a constant. Similarly, derive \( f_y \) by differentiating with respect to \( y \), holding \( x \) constant.
In our given function \( f(x, y) = x^2 + 2y^2 - x^2y \), the partial derivatives were:
• \( f_x = 2x - 2xy \)
• \( f_y = 4y - x^2 \)
Partial derivatives are used in identifying critical points and evaluating the behavior of functions of multiple variables.
Local Minimum
A local minimum is a point on the graph of a function where the function value is lower than at any nearby points. For a point to be classified as a local minimum in the context of the Second Derivative Test, the discriminant \( D(x, y) \) must be positive, and \( f_{xx} \) must also be positive.
In our worked example, the critical point \( (0, 0) \) was identified as a local minimum. Here's why:
• The discriminant was calculated as \( D(0, 0) = 8 \), which is greater than zero, indicating that it is definitively a minimum rather than a saddle point or maximum.
• The second derivative \( f_{xx}(0, 0) = 2 \) is positive, confirming a local minimum at this point.
Local minima are important in optimization, helping us find points that represent the smallest values of the function within a specific region.
In our worked example, the critical point \( (0, 0) \) was identified as a local minimum. Here's why:
• The discriminant was calculated as \( D(0, 0) = 8 \), which is greater than zero, indicating that it is definitively a minimum rather than a saddle point or maximum.
• The second derivative \( f_{xx}(0, 0) = 2 \) is positive, confirming a local minimum at this point.
Local minima are important in optimization, helping us find points that represent the smallest values of the function within a specific region.
Saddle Point
A saddle point on a surface is a critical point that is neither a local maximum nor a local minimum. Instead, it represents a point where the function moves upwards in one direction and downwards in another, resembling a horse saddle.
The Second Derivative Test identifies a saddle point when the discriminant \( D(x, y) \) is negative. This indicates that the shape of the function around the point is such that it does not exhibit a clear maximum or minimum.
In the provided example, the critical points \( (2, 1) \) and \( (-2, 1) \) were identified as saddle points:
The Second Derivative Test identifies a saddle point when the discriminant \( D(x, y) \) is negative. This indicates that the shape of the function around the point is such that it does not exhibit a clear maximum or minimum.
In the provided example, the critical points \( (2, 1) \) and \( (-2, 1) \) were identified as saddle points:
- The value of \( D(x, y) \) for both these points was calculated as \(-16\), which is less than zero.
- This indicated that around these points, the function balanced between increasing and decreasing, characteristic of a saddle point.