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Chapter 14: Problem 6

Find all the local maxima, local minima, and saddle points of the functions in Exercises \(1-30\) . $$ f(x, y)=y^{2}+x y-2 x-2 y+2 $$

Short Answer

Expert verified
The function has a saddle point at \((-2, 2)\). There are no local maxima or minima.

Step by step solution

01

Find the First Partial Derivatives

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

Set First Partial Derivatives to Zero

To find the critical points, set the first partial derivatives equal to zero:1. \( f_x(x, y) = y - 2 = 0 \) implies \( y = 2 \).2. \( f_y(x, y) = 2y + x - 2 = 0 \) substituting \( y = 2 \) gives: \[ 2(2) + x - 2 = 0 \Rightarrow x + 4 - 2 = 0 \Rightarrow x = -2 \]
03

Evaluate the Second Partial Derivatives

Calculate the second partial derivatives to use in the second derivative test:- \( f_{xx} = \frac{\partial^2}{\partial x^2}(f) = 0 \)- \( f_{yy} = \frac{\partial^2}{\partial y^2}(f) = 2 \)- \( f_{xy} = \frac{\partial^2}{\partial x \partial y}(f) = 1 \)
04

Compute the Determinant of the Hessian Matrix

The Hessian matrix \( H \) for the function is:\[H = \begin{bmatrix}f_{xx} & f_{xy} \f_{xy} & f_{yy}\end{bmatrix} = \begin{bmatrix}0 & 1 \1 & 2\end{bmatrix}\]The determinant of \( H \) is:\[ \text{det}(H) = (0)(2) - (1)(1) = -1 \]
05

Analyze the Second Derivative Test Result

Since the determinant of the Hessian matrix \( \text{det}(H) = -1 \), which is less than zero, the point \((-2, 2)\) is a saddle point. Saddle points occur where the determinant of the Hessian is negative, indicating mixed types of curvatures in different directions.

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Key Concepts

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

Partial Derivatives
Differentiating a function partially means finding the rate at which the function changes with respect to one variable while keeping the other variables constant. These derivatives help locate critical points, where the function's slope could be flat or zero.
In our exercise, we dealt with the function \( f(x, y) = y^2 + xy - 2x - 2y + 2 \).
This required us to compute the partial derivatives with respect to \( x \) and \( y \):
  • For \( x \), \( f_x(x, y) = y - 2 \)
  • For \( y \), \( f_y(x, y) = 2y + x - 2 \)
By setting these derivatives to zero, \( f_x(x, y) = 0 \) and \( f_y(x, y) = 0 \), you find the critical points.
This is where the function might have local extrema or saddle points.
Second Derivative Test
The second derivative test helps classify critical points found using the first partial derivatives. This test involves taking the second partial derivatives of the function, thereby getting the Hessian matrix.
The idea is to evaluate the nature of the critical points based on these second derivatives.
In this context, the second derivative test offers us detailed insights:
  • If the determinant of the Hessian is positive and the second derivative with respect to \( x \), \( f_{xx} \), is positive at that point, the point is a local minimum.
  • If the determinant is positive and \( f_{xx} \) is negative, the point is a local maximum.
  • If the determinant is negative, you have found a saddle point.
In our case, the determinant was negative, proving it's a saddle point.
Hessian Matrix
The Hessian matrix is crucial for the second derivative test. It's a square matrix composed of the second partial derivatives of a multivariable function. For the function in the example, the Hessian matrix is:\[ H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{xy} & f_{yy} \end{bmatrix} = \begin{bmatrix} 0 & 1 \ 1 & 2 \end{bmatrix} \]Here:
  • \( f_{xx} \) is the second derivative with respect to \( x \), which is 0 in this problem.
  • \( f_{yy} \) is the second derivative with respect to \( y \), which turned out to be 2.
  • \( f_{xy} \) and \( f_{yx} \) are the mixed partial derivatives equal to 1.
The determinant of this Hessian matrix is \(-1\).
This determinant guides us in using the second derivative test to successfully identify the nature of critical points.
Saddle Point
A saddle point in mathematics refers to a location on the graph of a function where the function looks like a saddle; sheer in some directions and curved in others.
Saddle points occur because of mixed curvatures, meaning they are not extrema but still critical.
In our specific exercise, the point \((-2, 2) \) was identified as a saddle point.
This resulted from the Hessian determinant being negative, \(-1\), which is a strong indicator of a saddle point.
Saddle points are unique as they mark the crossing of different types of curvatures in a function, thereby having significant importance in understanding the behavior of a function in multi-dimensions.

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Most popular questions from this chapter

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 \(x=t, \quad y=2-2 t, \quad 0 \leq t \leq 1\)

Use a CAS to perform the following steps implementing the method of Lagrange multipliers for finding constrained extrema: a. Form the function \(h=f-\lambda_{1} g_{1}-\lambda_{2} g_{2},\) where \(f\) is the function to optimize subject to the constraints \(g_{1}=0\) and \(g_{2}=0 .\) b. Determine all the first partial derivatives of \(h\) , including the partials with respect to \(\lambda_{1}\) and \(\lambda_{2},\) and set them equal to \(0 .\) c. Solve the system of equations found in part (b) for all the unknowns, including \(\lambda_{1}\) and \(\lambda_{2} .\) d. Evaluate \(f\) at each of the solution points found in part (c) and select the extreme value subject to the constraints asked for in the exercise. Minimize \(f(x, y, z)=x^{2}+y^{2}+z^{2}\) subject to the constraints \(x^{2}-x y+y^{2}-z^{2}-1=0\) and \(x^{2}+y^{2}-1=0\)

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. \(f(x, y)=x+y\) c. \(h(x, y)=2 x^{2}+y^{2}\) Curves: i. The semicircle \(x^{2}+y^{2}=4, \quad y \geq 0\) ii. The quarter circle \(x^{2}+y^{2}=4, \quad x \geq 0, \quad y \geq 0\) Use the parametric equations \(x=2 \cos t, y=2 \sin t\)

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

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