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Chapter 14: Problem 22

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

Short Answer

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

Step by step solution

01

Find the First Partial Derivatives

To find critical points, we first need the first partial derivatives of the function. The function is given by \[ f(x, y) = \frac{1}{x} + xy + \frac{1}{y}. \]Compute the partial derivative with respect to \(x\): \[ f_x = -\frac{1}{x^2} + y. \]Compute the partial derivative with respect to \(y\): \[ f_y = x - \frac{1}{y^2}. \]
02

Find Critical Points

Set the first partial derivatives equal to zero to find the critical points.1. From \(f_x = 0\), \[ -\frac{1}{x^2} + y = 0 \implies y = \frac{1}{x^2}. \]2. From \(f_y = 0\),\[ x - \frac{1}{y^2} = 0 \implies x = \frac{1}{y^2}. \]This system of equations leads us to find common solutions for \(x\) and \(y\). Substitute \(y = \frac{1}{x^2}\) into \(x = \frac{1}{y^2}\) to solve for \(x\) and \(y\).
03

Solve the System of Equations

Substitute \(y = \frac{1}{x^2}\) into \(x = \frac{1}{y^2}\):\[ x = \frac{1}{(\frac{1}{x^2})^2} = x^4 \implies x - x^4 = 0. \]Factor the equation:\[ x(1-x^3) = 0. \]Therefore, the solutions are \(x = 0\) or \(x^3 = 1\). Since \(x = 0\) results in undefined values for the original function, use \(x^3 = 1\) leading to \(x = 1\).Substitute \(x = 1\) back to find \(y\):\[ y = \frac{1}{1^2} = 1. \]So the critical point is \((1, 1)\).
04

Second Derivative Test

Use the second derivative test to classify the critical points. Compute the second partial derivatives:\[ f_{xx} = \frac{2}{x^3}, \quad f_{yy} = \frac{2}{y^3}, \quad f_{xy} = 1. \]Evaluate at \((1, 1)\):\[ f_{xx}(1,1) = 2, \quad f_{yy}(1,1) = 2, \quad f_{xy}(1,1) = 1. \]Calculate determinant of the Hessian matrix, \(D\):\[ D = f_{xx}f_{yy} - (f_{xy})^2 = (2)(2) - (1)^2 = 4 - 1 = 3. \]Since \(D > 0\) and \(f_{xx}(1,1) > 0\), 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
Partial derivatives are a fundamental concept in multivariable calculus. They help us understand how a function changes as we adjust one variable while keeping others constant.
  • The partial derivative of a function with respect to a variable shows the rate of change of the function as that variable changes.
  • For a function \( f(x, y) \), the partial derivative with respect to \( x \) is denoted as \( f_x \), and with respect to \( y \) as \( f_y \).
  • In the original problem, we calculated \( f_x = -\frac{1}{x^2} + y \) and \( f_y = x - \frac{1}{y^2} \).
Calculating these derivatives is the first step to finding critical points, which tell us where local maxima, minima, or saddle points are located.
Second Derivative Test
The second derivative test helps classify the nature of critical points found using partial derivatives. It's similar to the second derivative test in single-variable calculus but takes into account multiple directions.
  • We first compute second partial derivatives: \( f_{xx} \), \( f_{yy} \), and \( f_{xy} \).
  • These derivatives provide information on the concavity of the function at the critical points.
  • Combining these, the test involves checking the values using the determinant of the Hessian matrix \( D \).
For example, at the critical point \((1, 1)\), if \( D > 0 \) and \( f_{xx} > 0 \), the point is a local minimum.
Hessian Matrix
The Hessian matrix is a square matrix of second-order partial derivatives of a function. It's essential in multivariable calculus for understanding the local curvature at critical points.
  • The Hessian matrix for a function \( f(x, y) \) is written as:\[ \begin{bmatrix} f_{xx} & f_{xy} \ f_{yx} & f_{yy} \\end{bmatrix}\]
  • In our scenario: \( f_{xx} = \frac{2}{x^3}, \; f_{yy} = \frac{2}{y^3}, \; f_{xy} = 1 \).
  • The determinant \( D \) of the Hessian matrix is used for the second derivative test: \( D = f_{xx}f_{yy} - (f_{xy})^2 \).
Evaluating the Hessian matrix at \((1, 1)\) helps confirm the nature of this critical point as a local minimum, supporting the conclusion from the second derivative test.

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Most popular questions from this chapter

You will explore functions to identify their local extrema. Use a CAS to perform the following steps: a. Plot the function over the given rectangle. b. Plot some level curves in the rectangle. c. Calculate the function's first partial derivatives and use the CAS equation solver to find the critical points. How do the critical points relate to the level curves plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer. d. Calculate the function's second partial derivatives and find the discriminant \(f_{x x} f_{y y}-f_{x y}^{2}\) . e. Using the max-min tests, classify the critical points found in part (c). Are your findings consistent with your discussion in part (c)? $$f(x, y)=x^{2}+y^{3}-3 x y, \quad-5 \leq x \leq 5, \quad-5 \leq y \leq 5$$

In Exercises \(53-60,\) sketch a typical level surface for the function. $$ f(x, y, z)=x^{2}+y^{2} $$

Minimum distance to the origin Find the point closest to the origin on the curve of intersection of the plane \(2 y+4 z=5\) and the cone \(z^{2}=4 x^{2}+4 y^{2}.\)

Parametrized Surfaces Just as you describe curves in the plane parametrically with a pair of equations \(x=f(t), y=g(t)\) defined on some parameter interval \(I,\) you can sometimes describe surfaces in space with a triple of equations \(x=f(u, v), y=g(u, v), z=h(u, v)\) defined on some parameter rectangle \(a \leq u \leq b, c \leq v \leq d .\) Many computer algebra systems permit you to plot such surfaces in parametric mode. (Parametrized surfaces are discussed in detail in Section 16.5.) Use a CAS to plot the surfaces in Exercises \(77-80 .\) Also plot several level curves in the \(x y\) -plane. $$ \begin{array}{l}{x=u \cos v, \quad y=u \sin v, \quad z=u, \quad 0 \leq u \leq 2} \\ {0 \leq v \leq 2 \pi}\end{array} $$

Find the value of \(\partial x / \partial z\) at the point \((1,-1,-3)\) if the equation $$x z+y \ln x-x^{2}+4=0$$ defines \(x\) as a function of the two independent variables \(y\) and \(z\) and the partial derivative exists.
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