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Chapter 14: Problem 25

Find all the local maxima, local minima, and saddle points of the functions in Exercises \(1-30\) . $$ f(x, y)=4 x y-x^{4}-y^{4} $$

Short Answer

Expert verified
Local maxima at (1, 1) and (-1, -1); saddle point at (0, 0).

Step by step solution

01

Find the Critical Points

To find critical points, we first need to take the partial derivatives of the function and set them equal to zero. Given function: \[ f(x, y) = 4xy - x^4 - y^4 \]1. Find \( f_x \): \[ f_x = \frac{\partial}{\partial x}(4xy - x^4 - y^4) = 4y - 4x^3 \] 2. Find \( f_y \): \[ f_y = \frac{\partial}{\partial y}(4xy - x^4 - y^4) = 4x - 4y^3 \]Set \( f_x = 0 \) and \( f_y = 0 \):\[ 4y - 4x^3 = 0 \]\[ 4x - 4y^3 = 0 \] This simplifies to two equations:\[ y = x^3 \]\[ x = y^3 \]
02

Solve Simultaneous Equations

Now solve the system of equations:1. Substitute \( y = x^3 \) into \( x = y^3 \):\[ x = (x^3)^3 = x^9 \]2. Simplify this equation:\[ x - x^9 = 0 \]\[ x(1 - x^8) = 0 \]3. Solutions to this equation are \( x = 0 \) or \( 1 - x^8 = 0 \).Solving \( 1 - x^8 = 0 \) gives:\[ x^8 = 1 \]\[ x = \pm 1 \] (since \( x \) and \( -x \) must be real). The corresponding values for \( y \) are:- If \( x = 0 \), then \( y = 0^3 = 0 \).- If \( x = 1 \), then \( y = 1^3 = 1 \).- If \( x = -1 \), then \( y = (-1)^3 = -1 \). Thus, potential critical points are \((0,0)\), \((1,1)\), and \((-1,-1)\).
03

Classify the Critical Points

To classify the critical points, we need to find the second derivatives and evaluate the Hessian determinant:1. Calculate second derivatives:\[ f_{xx} = \frac{\partial}{\partial x}(4y - 4x^3) = -12x^2 \]\[ f_{yy} = \frac{\partial}{\partial y}(4x - 4y^3) = -12y^2 \]\[ f_{xy} = \frac{\partial}{\partial y}(4y - 4x^3) = 4 \]\[ f_{yx} = 4 \]2. Calculate Hessian determinant:\[ H = f_{xx}f_{yy} - (f_{xy})^2 = (-12x^2)(-12y^2) - 16 \]\[ H = 144x^2y^2 - 16 \]3. Evaluate the determinant at each critical point:- For \( (0, 0) \): \[ H(0, 0) = 144(0)(0) - 16 = -16 \] 鈥� This is a saddle point.- For \( (1, 1) \) and \( (-1, -1) \): \[ H = 144(1)(1) - 16 = 128 \] \( f_{xx}(1, 1) = -12 \), which is less than 0, indicating a local maximum.Thus, \((1, 1)\) is a local maximum and \((-1, -1)\) is also a local maximum as they are symmetric critical points.

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Key Concepts

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

Critical Points
In multivariable calculus, critical points are places where the partial derivatives of a function are zero. These points are important because they are where local maxima, minima, or saddle points might exist. For the function \(f(x, y) = 4xy - x^4 - y^4\), the process to determine critical points involves finding its partial derivatives with respect to \(x\) and \(y\), then setting these partial derivatives equal to zero.
  • Compute the partial derivative with respect to \(x\), denoted as \(f_x\).
  • Compute the partial derivative with respect to \(y\), denoted as \(f_y\).
  • Set \(f_x = 0\) and \(f_y = 0\) to find the critical points.
This gives us a system of equations that helps us find potential critical points. In this exercise, solving the system results in the critical points: \((0,0)\), \((1,1)\), and \((-1,-1)\). These critical points require further evaluation to determine their type.
Second Derivatives
To effectively classify critical points, we need to explore the second derivatives of the function. Second derivatives provide insight into the behavior of the function around its critical points. The second partial derivatives are:
  • \(f_{xx}\): the partial derivative of \(f_x\) with respect to \(x\).
  • \(f_{yy}\): the partial derivative of \(f_y\) with respect to \(y\).
  • \(f_{xy}\) or \(f_{yx}\): the mixed partial derivatives, which are usually equal for well-behaved functions.
For our function, these are computed as: \[ f_{xx} = -12x^2, \, f_{yy} = -12y^2, \, f_{xy} = f_{yx} = 4 \] This second derivative information is crucial for understanding how the function curves or saddles at different points.
Hessian Determinant
The Hessian determinant is a scalar value derived from the second derivatives of a function, and it helps classify the type of critical points. For a function \(f(x, y)\), its Hessian matrix is composed of second partial derivatives: \[ H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{yx} & f_{yy} \end{bmatrix} \] The Hessian determinant \(H\) is calculated as: \[ H = f_{xx}f_{yy} - (f_{xy})^2 \] For our function, \(H = 144x^2y^2 - 16\). A positive determinant at a critical point indicates a local minimum or maximum, while a negative determinant suggests a saddle point.
  • Negative \(H\) at \((0,0)\) indicates it's a saddle point.
  • Positive \(H\) at \((1,1)\) and \((-1,-1)\) suggests these points may be maxima if \(f_{xx} < 0\).
Thus, we use the Hessian determinant to effectively categorize our critical points.
Saddle Points
Saddle points are critical points where the function does not have a local extreme value. Instead, the function increases in one direction and decreases in another. Essentially, a saddle point resembles a valley in one direction and a hill in another. In mathematical terms, saddle points occur when the Hessian determinant is negative. For our exercise, the point \((0,0)\) is identified as a saddle point since the Hessian determinant \(H = -16\) at this point. Here, the value of \(H\) being negative alerts us to the changing nature of the function's surface around \((0,0)\). The identification of saddle points is crucial in optimization problems because they are not optimal solutions, but knowing their locations helps in understanding the overall behavior of the function's surface.

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Most popular questions from this chapter

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 \(R\) . $$ \begin{array}{l}{f(x, y)=x y^{2}+y \cos (x-1) \text { at } P_{0}(1,2)} \\ {R :|x-1| \leq 0.1, \quad|y-2| \leq 0.1}\end{array} $$

In Exercises \(65-70,\) 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 critical plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer. $$ f(x, y)=x^{2}+y^{3}-3 x y, \quad-5 \leq x \leq 5, \quad-5 \leq y \leq 5 $$

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 \(f(x, y, z)=x y+y z\) subject to the constraints \(x^{2}+y^{2}-2=0\) and \(x^{2}+z^{2}-2=0\)

In Exercises \(51-56,\) find the limit of \(f\) as \((x, y) \rightarrow(0,0)\) or show that the limit does not exist. $$ f(x, y)=\cos \left(\frac{x^{3}-y^{3}}{x^{2}+y^{2}}\right) $$

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