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Chapter 7: Problem 23

Find the relative maximum and minimum values and the saddle points. $$ f(x, y)=2 y^{2}+x^{2}-x^{2} y $$

Short Answer

Expert verified
(0,0) is a local minimum. (2,1) and (-2,1) are saddle points.

Step by step solution

01

- Find the Partial Derivatives

First, find the first-order partial derivatives of the function with respect to both variables. \[ f_x = \frac{\partial f}{\partial x} = 2x - 2xy \]\[ f_y = \frac{\partial f}{\partial y} = 4y - x^2 \]
02

- Set Partial Derivatives to Zero

Set each partial derivative to zero and solve for critical points: \[ 2x - 2xy = 0 \]\[ 4y - x^2 = 0 \]
03

- Solve for Critical Points

Solve the system of equations found in Step 2: From \( 2x = 2xy \), factored gives \( x(1-y) = 0 \), thus \( x = 0 \) or \( y = 1 \).If \( x = 0 \), substitute into \( 4y - x^2 = 0 \), giving \( 4y = 0 \Rightarrow y = 0 \). This point is \( (0,0) \).If \( y = 1 \), substitute into \( 4y - x^2 = 0 \), giving \( 4(1) - x^2 = 0 \Rightarrow x^2 = 4 \Rightarrow x = 2 \ or \ x = -2 \). These points are \( (2,1) \) and \( (-2,1) \).
04

- Determine Nature of Critical Points

To classify the critical points, use the second partial derivatives:\[ f_{xx} = \frac{\partial^2 f}{\partial x^2} = 2 - 2y \]\[ f_{yy} = \frac{\partial^2 f}{\partial y^2} = 4 \]\[ f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = -2x \]
05

- Evaluate the Determinant of the Hessian Matrix

Calculate the Hessian matrix determinant, denoted by \( D \):\[ D = f_{xx} f_{yy} - (f_{xy})^2 \]For each critical point, calculate \( D \) and use it to classify the points.
06

- Evaluate at Specific Points (0,0)

For \( (0,0) \):\[ f_{xx}(0,0) = 2 \]\[ f_{yy}(0,0) = 4 \]\[ f_{xy}(0,0) = 0 \]\[ D = (2)(4) - (0)^2 = 8 > 0 \Rightarrow f_{xx} > 0 \text{, local minimum} \]
07

- Evaluate at Specific Points (2,1) and (-2,1)

For \( (2,1) \) and \( (-2,1) \):\[ f_{xx}(2,1) = f_{xx}(-2,1) = 0 \]\[ D = (0)(4) - (-4)^2 = -16 < 0 \Rightarrow \text{saddle points} \]
08

- Summarize the Results

Using the above classifications, summarize the critical points and their nature: \[ (0,0): \text{local minimum} \]\[ (2,1): \text{saddle point} \]\[ (-2,1): \text{saddle point} \]

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Key Concepts

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

Partial Derivatives
To find the relative extrema of a function with two variables, we first need to determine its partial derivatives. These derivatives give us the rate of change of the function with respect to each variable. For our function \( f(x, y) = 2y^2 + x^2 - x^2 y \), the partial derivatives are calculated as follows:
  • The partial derivative with respect to \( x \) is denoted \( f_x \) and is given by \( f_x = \frac{\partial f}{\partial x} = 2x - 2xy \).
  • The partial derivative with respect to \( y \) is denoted \( f_y \) and is given by \( f_y = \frac{\partial f}{\partial y} = 4y - x^2 \).
These partial derivatives are essential for identifying critical points, which are needed to find relative extrema.
Critical Points
Critical points are points where the first-order partial derivatives of a function are both zero or undefined. They are potential candidates for relative maxima, minima, or saddle points. To find these points for our function:
  • Set \( f_x = 0 \):\ \( 2x - 2xy = 0 \rightarrow x(1 - y) = 0 \) giving \( x = 0 \) or \( y = 1 \).
  • Set \( f_y = 0 \): \( 4y - x^2 = 0 \rightarrow x^2 = 4y \).
By solving simultaneously, we get the critical points \( (0,0) \), \( (2,1) \), and \( (-2,1) \). Determining their nature involves examining the second partial derivatives and the Hessian matrix.
Hessian Matrix
The Hessian matrix provides a way to classify critical points by analyzing the second-order partial derivatives. For our function, the second partial derivatives are:
  • \( f_{xx} = \frac{\partial^2 f}{\partial x^2} = 2 - 2y \)
  • \( f_{yy} = \frac{\partial^2 f}{\partial y^2} = 4 \)
  • \( f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = -2x \)
The Hessian matrix \( H \) is then given by:
\[ H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{xy} & f_{yy} \end{bmatrix} \]
The determinant of the Hessian, \( D \), is calculated as \( D = f_{xx} f_{yy} - (f_{xy})^2 \). Evaluating \( D \) at each critical point helps us determine if we have a local minimum, maximum, or saddle point.
Saddle Points
Saddle points are critical points where the surface of the function changes direction, and they do not correspond to relative minima or maxima. To identify saddle points, we check the determinant of the Hessian matrix:
  • For \( (2,1) \) and \( (-2,1) \), \( f_{xx} = 0 \), \( f_{yy} = 4 \), \( f_{xy} = -4 \).
  • Calculating the determinant \( D \): \( D = (0)(4) - (-4)^2 = -16 < 0 \), indicating saddle points.
Thus, the points \( (2,1) \) and \( (-2,1) \) are saddle points since the Hessian determinant \( D \) is negative at these points.
Local Minimum
A local minimum is a point where the function has a lower value than at nearby points. To determine if a critical point is a local minimum, we use the Hessian matrix:
  • For \( (0,0) \), \( f_{xx} = 2 \), \( f_{yy} = 4 \), \( f_{xy} = 0 \).
  • Calculating \( D \): \( D = (2)(4) - (0)^2 = 8 > 0 \), and since \( f_{xx} > 0 \), this indicates a local minimum.
Thus, the point \( (0,0) \) is indeed a local minimum of the function.

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