Problem 153 - Advanced Calculus

Question

Find the critical points of the function $$ f(x, y)=x^{4}+y^{4}-x^{2}-y^{2}+1 $$ Then determine if each critical point is a relative maximum, relative minimum, or saddle point.

Step-by-Step Solution

Answer

The critical points of the function \(f(x,y) = x^4 + y^4 - x^2 - y^2 + 1\) are (0,0), (0, \(\sqrt{2}\)), (0, -\(\sqrt{2}\)), (\(\sqrt{2}\),0), -\(\sqrt{2}\),0), (\(\sqrt{2}\),\(\sqrt{2}\)), (-\(\sqrt{2}\), -\(\sqrt{2}\)), (\(\sqrt{2}\), -\(\sqrt{2}\)), and (-\(\sqrt{2}\),\(\sqrt{2}\)). All of these critical points are saddle points, meaning there are no relative maxima or minima for the given function.

Step 1: Compute the Partial Derivatives

The first step is to find the partial derivatives of the function with respect to x and y. The first order partial derivative of \( f(x, y)=x^{4}+y^{4}-x^{2}-y^{2}+1 \) with respect to x is \( f_x(x, y)=4x^{3}-2x \). Similarly, the first order partial derivative of \( f \) with respect to y is \( f_y(x, y)=4y^{3}-2y \).

Step 2: Find the Critical Points

The critical points are found by setting both partial derivatives equal to zero and then solve for x and y: \( f_x(x, y)=0 \) and \( f_y(x, y)=0 \). This yields: \( 4x^{3}-2x = 0 \) and \( 4y^{3}-2y = 0 \). Set x and y separately equal to 0 and solve for x and y. We get \( x=0, x=\sqrt{2}, x=-\sqrt{2} \) and \( y=0, y=\sqrt{2}, y=-\sqrt{2} \). Therefore, the critical points are (0,0), (0, \(\sqrt{2}\)), (0, -\(\sqrt{2}\)), \(\sqrt{2}\),0), -\(\sqrt{2}\),0), (\(\sqrt{2}\),\(\sqrt{2}\)), (-\(\sqrt{2}\), -\(\sqrt{2}\)), (\(\sqrt{2}\), -\(\sqrt{2}\)), and (-\(\sqrt{2}\),\(\sqrt{2}\)).

Step 3: Compute the Second Order Partial Derivatives

The second order partial derivatives needed for the Hessian are \( f_{xx}, f_{yy}, f_{xy}, \) and \( f_{yx} \), where: \( f_{xx} = 12x^{2}-2, f_{yy} = 12y^{2}-2, f_{xy}=f_{yx}=0 \).

Step 4: Use the Second-Order Test

We now check each critical point to see whether the function has a local maximum, local minimum, or a saddle point. We use the Hessian determinant, \( D= f_{xx}f_{yy} - f_{xy}f_{yx} \). At each critical point, if \( D > 0 \) then \( f \) has either a maximum or a minimum, if \( D < 0 \) then \( f \) has a saddle point. When \( D > 0 \), if \( f_{xx} > 0 \) and \( f_{yy} > 0 \) then \( f \) has a minimum, but if \( f_{xx} < 0 \) and \( f_{yy} < 0 \) then \( f \) has a maximum. Finally, here are the results of the second-derivative test at each critical point we found above: The Hessian determinant at (0,0) is -4, therefore (0,0) is a saddle point. For (\(\sqrt{2}\),0), (-\(\sqrt{2}\),0), (0,\(\sqrt{2}\)), (0,-\(\sqrt{2}\)), the determinant is -20, therefore they are saddle points too. For (\(\sqrt{2}\),\(\sqrt{2}\)), (-\(\sqrt{2}\),-\(\sqrt{2}\)), we get D=-8 which means they are saddle points. For (\(\sqrt{2}\),-\(\sqrt{2}\)), (-\(\sqrt{2}\),\(\sqrt{2}\)) we also get D=-8 so they are saddle points. There are no relative maxima or minima in the function. All critical points are saddle points.

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in multi-variable calculus. They help us understand how a multivariable function changes with respect to one variable while keeping others constant.
In the exercise, the function in question is given as \( f(x, y) = x^4 + y^4 - x^2 - y^2 + 1 \). To find critical points where the function does not change, we need to compute its first partial derivatives with respect to each variable, \( x \) and \( y \).

• The partial derivative with respect to \( x \) is: \( f_x(x, y) = 4x^3 - 2x \).
• The partial derivative with respect to \( y \) is: \( f_y(x, y) = 4y^3 - 2y \).

By setting these partial derivatives equal to zero, we can find potential critical points where the slope is zero along each variable axis.

Hessian Determinant

The Hessian determinant is a special matrix used to analyze the curvature of functions at their critical points. It combines second-order partial derivatives to offer insights into the function's behavior around these points.
For our function, the second-order partial derivatives are as follows:

• \( f_{xx} = 12x^2 - 2 \),
• \( f_{yy} = 12y^2 - 2 \),
• \( f_{xy} = f_{yx} = 0 \)

The Hessian determinant \( D \) is calculated at each critical point with the formula: \[ D = f_{xx} f_{yy} - (f_{xy})^2 \]
A positive Hessian determinant indicates potential relative extremum points (maximum or minimum), while a negative Hessian determinant identifies saddle points.

Saddle Points

Saddle points are specific locations on a graph where the surface curves in opposing directions. This makes them neither a maximum nor a minimum.
When finding critical points, some may not correspond to peaks or valleys, but rather saddle points. With regards to the function \( f(x, y) = x^4 + y^4 - x^2 - y^2 + 1 \), we tested each critical point using the Hessian determinant method.
It was found that many critical points such as (0,0) and \((\sqrt{2},-\sqrt{2})\) have negative Hessian determinants indicating they are saddle points. Understanding saddle points is crucial because they signify areas of instability or change in the function's curvature.

Second-Order Test

The second-order test is a method that utilizes the Hessian determinant to classify critical points. This test helps us determine if these points are local maxima, minima, or saddle points.
For a given critical point, the second-order test involves evaluating the following conditions based on the Hessian determinant \( D \):

• If \( D > 0 \) and \( f_{xx} > 0 \), then the point is a local minimum.
• If \( D > 0 \) and \( f_{xx} < 0 \), then the point is a local maximum.
• If \( D < 0 \), then the point is a saddle point.

In this exercise, all critical points resulted in \( D < 0 \), indicating they are saddle points, which means this function does not have any local maxima or minima under the second-order test.