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Chapter 14: Problem 28

Find all the local maxima, local minima, and saddle points of the functions in Exercises \(1-30\) . $$ f(x, y)=\frac{1}{x}+x y+\frac{1}{y} $$

Short Answer

Expert verified
The function has a local minimum at (1,1).

Step by step solution

01

Find the Partial Derivatives

To identify critical points, first find the partial derivatives of the function. Compute \( f_x \) and \( f_y \). For \( f(x, y)=\frac{1}{x}+xy+\frac{1}{y} \): \[ f_x = \frac{-1}{x^2} + y \] \[ f_y = x - \frac{1}{y^2} \]
02

Set Partial Derivatives Equal to Zero

Find the critical points by setting the partial derivatives equal to zero. \[ \frac{-1}{x^2} + y = 0 \Rightarrow y = \frac{1}{x^2} \] \[ x - \frac{1}{y^2} = 0 \Rightarrow x = \frac{1}{y^2} \]
03

Solve the System of Equations

Substitute \( y = \frac{1}{x^2} \) into \( x = \frac{1}{y^2} \), giving: \[ x = \frac{1}{\left(\frac{1}{x^2}\right)^2} = x^4 \] This simplifies to \( x^4 = x \), resulting in \( x(x^3 - 1) = 0 \). So, \( x = 0 \) or \( x = 1 \).
04

Determine Corresponding y-values

For \( x = 0 \), \( y = \frac{1}{x^2} \) is undefined (excluding \( x = 0 \) from valid critical points).For \( x = 1 \), substitute back: \[ y = \frac{1}{1^2} = 1 \] Thus, \((x, y) = (1, 1)\) is a critical point.
05

Use the Second Derivative Test

To classify the critical points, compute the second derivatives: \[ f_{xx} = \frac{2}{x^3}, \quad f_{yy} = \frac{2}{y^3}, \quad f_{xy} = 1 \] Evaluate these at \((1,1)\): \[ f_{xx}(1,1) = 2, \quad f_{yy}(1,1) = 2, \quad f_{xy}(1,1) = 1 \] Calculate the determinant of the Hessian: \[ D = f_{xx}f_{yy} - \left(f_{xy}\right)^2 = (2)(2) - (1)^2 = 3 \]
06

Classify the Critical Points

Using the second derivative test for the Hessian determinant \( D = 3 \) (\( D > 0 \)), and \( f_{xx} > 0 \), the point \((1,1)\) is a local minimum.

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Key Concepts

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

Partial Derivatives
In calculus, when dealing with functions of several variables, partial derivatives are a key tool. They let us understand how a function changes as we vary one variable while keeping others constant. For the function given, \( f(x, y)=\frac{1}{x}+xy+\frac{1}{y} \), we find the partial derivatives to locate critical points.
  • \( f_x \), the partial derivative with respect to \( x \), is calculated by differentiating \( f \) while treating \( y \) as a constant: \( f_x = \frac{-1}{x^2} + y \).
  • Similarly, \( f_y \), the partial derivative with respect to \( y \), is obtained by treating \( x \) as a constant: \( f_y = x - \frac{1}{y^2} \).
Setting these partial derivatives to zero helps us find the critical points. The equations \( \frac{-1}{x^2} + y = 0 \) and \( x - \frac{1}{y^2} = 0 \) must be solved simultaneously to find these points.
Second Derivative Test
Once you find the critical points using the partial derivatives, you need to determine their nature. The second derivative test is a way to classify critical points as local maxima, minima, or saddle points.
First, calculate the second-order partial derivatives:
  • \( f_{xx} = \frac{2}{x^3} \)
  • \( f_{yy} = \frac{2}{y^3} \)
  • \( f_{xy} = 1 \)
These derivatives collectively help us form the Hessian matrix, which is key to the test. Evaluating these at \((1, 1)\) gives \( f_{xx}(1,1) = 2 \), \( f_{yy}(1,1) = 2 \), and \( f_{xy}(1,1) = 1 \). These values are necessary for calculating the Hessian determinant.
Hessian Determinant
The Hessian determinant aids in the classification of critical points, an essential step of the second derivative test. The determinant \( D \) is calculated as follows:
  • \( D = f_{xx}f_{yy} - (f_{xy})^2 \)
For our function at the critical point \((1,1)\), we find \[ D = (2)(2) - (1)^2 = 3 \].
  • If \( D > 0 \) and \( f_{xx} > 0 \), it indicates a local minimum.
  • If \( D > 0 \) and \( f_{xx} < 0 \), it indicates a local maximum.
  • If \( D < 0 \), the point is a saddle point.
In our example, since \( D = 3 > 0 \) and \( f_{xx} = 2 > 0 \), the critical point \((1,1)\) is confirmed to be a local minimum.

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Most popular questions from this chapter

In Exercises \(51-56,\) find the limit of \(f\) as \((x, y) \rightarrow(0,0)\) or show that the limit does not exist. $$ f(x, y)=\frac{y^{2}}{x^{2}+y^{2}} $$

To find the extreme values of a function \(f(x, y)\) on a curve \(x=x(t), y=y(t),\) we treat \(f\) as a function of the single variable \(t\) and use the Chain Rule to find where \(d f / d t\) is zero. As in any other single-variable case, the extreme values of \(f\) are then found among the values at the a. critical points (points where \(d f / d t\) is zero or fails to exist), and b. endpoints of the parameter domain. Find the absolute maximum and minimum values of the following functions on the given curves. Functions: a. \(f(x, y)=x^{2}+y^{2} \quad\) b. \(g(x, y)=1 /\left(x^{2}+y^{2}\right)\) Curves: i. The line \(x=t, \quad y=2-2 t\) ii. The line segment \(x=t, \quad y=2-2 t, \quad 0 \leq t \leq 1\)

Find the linearizations \(L(x, y, z)\) of the functions in Exercises \(37-42\) at the given points. $$ \begin{array}{l}{f(x, y, z)=x^{2}+y^{2}+z^{2} \text { at }} \\\ {\begin{array}{lll}{\text { a. }(1,1,1)} & {\text { b. }(0,1,0)} & {\text { c. }(1,0,0)}\end{array}}\end{array} $$

In Exercises \(31-36,\) find the linearization \(L(x, y)\) of the function \(f(x, y)\) at \(P_{0} .\) Then find an upper bound for the magnitude \(|E|\) of the error in the approximation \(f(x, y) \approx L(x, y)\) over the rectangle \(R\) . $$ \begin{array}{l}{f(x, y)=1+y+x \cos y \text { at } P_{0}(0,0)} \\ {R :|x| \leq 0.2, \quad|y| \leq 0.2} \\ {\text { (Use }|\cos y| \leq 1 \text { and }|\sin y| \leq 1 \text { in estimating } E . )}\end{array} $$

In Exercises \(43-46,\) 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 \(R\) . $$ \begin{array}{l}{f(x, y, z)=x^{2}+x y+y z+(1 / 4) z^{2} \quad \text { at } \quad P_{0}(1,1,2)} \\ {R : \quad|x-1| \leq 0.01, \quad|y-1| \leq 0.01, \quad|z-2| \leq 0.08}\end{array} $$
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