Problem 11 - APEX Calculus

Question

Find the critical points of the given function. Use the Second Derivative Test to determine if each critical point corresponds to a relative maximum, minimum, or saddle point. $$ f(x, y)=x^{2} y^{2} $$

Step-by-Step Solution

Answer

The critical point (0,0) is inconclusive by the Second Derivative Test.

Step 1: Compute the First Derivatives

First, we need to find the partial derivatives of the function \( f(x, y) = x^2 y^2 \) with respect to \( x \) and \( y \). These are:For \( x \):\[ f_x(x, y) = \frac{\partial}{\partial x}(x^2 y^2) = 2xy^2 \]For \( y \):\[ f_y(x, y) = \frac{\partial}{\partial y}(x^2 y^2) = 2x^2y \]

Step 2: Find Critical Points

Critical points occur where both first partial derivatives are equal to zero. Therefore, we solve the system of equations:\[ 2xy^2 = 0 \]\[ 2x^2y = 0 \]From the first equation, \( 2xy^2 = 0 \), it follows that either \( x = 0 \) or \( y = 0 \).From the second equation, \( 2x^2y = 0 \), it follows that either \( x = 0 \) or \( y = 0 \).The critical points are thus \( (0, 0) \).

Step 3: Compute the Second Derivatives

Next, we need the second partial derivatives for the Second Derivative Test:\[ f_{xx} = \frac{\partial^2}{\partial x^2}(x^2 y^2) = \frac{\partial}{\partial x}(2xy^2) = 2y^2 \]\[ f_{yy} = \frac{\partial^2}{\partial y^2}(x^2 y^2) = \frac{\partial}{\partial y}(2x^2 y) = 2x^2 \]\[ f_{xy} = \frac{\partial^2}{\partial y \partial x}(x^2 y^2) = \frac{\partial}{\partial y}(2xy^2) = 4xy \]

Step 4: Apply the Second Derivative Test

The Second Derivative Test involves evaluating the determinant of the Hessian matrix at each critical point. The Hessian matrix is given by:\[ H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{xy} & f_{yy} \end{bmatrix} = \begin{bmatrix} 2y^2 & 4xy \ 4xy & 2x^2 \end{bmatrix} \]The determinant of the Hessian is:\[ D = f_{xx}f_{yy} - (f_{xy})^2 = (2y^2)(2x^2) - (4xy)^2 = 4x^2y^2 - 16x^2y^2 = -12x^2y^2 \]At the critical point \( (0, 0) \):\[ D = -12(0)^2(0)^2 = 0 \]As \( D = 0 \), the test is inconclusive at this critical point.

Key Concepts

Partial Derivatives

When working with functions of multiple variables, like the function \( f(x, y) = x^2 y^2 \), understanding partial derivatives is key to analyzing how the function behaves in different directions. Partial derivatives essentially measure how the function changes as you vary one of the variables while keeping the other constant. For instance:

• The partial derivative with respect to \( x \), denoted as \( f_x \), evaluates how the function changes as \( x \) varies while \( y \) remains fixed. For our function, \( f_x(x, y) = 2xy^2 \).
• The partial derivative with respect to \( y \), denoted as \( f_y \), evaluates how the function changes as \( y \) varies while \( x \) remains constant. Here, \( f_y(x, y) = 2x^2y \).

Computing these derivatives helps identify where the slope of the function is zero, pinpointing potential critical points.

Second Derivative Test

The Second Derivative Test serves as a tool to classify critical points found using first derivatives. After calculating the partial derivatives and finding where they equal zero, providing potential critical points, this test helps us decide if a point is a local maximum, minimum, or saddle point.The test involves examining the second derivatives:

• \( f_{xx} \): The second partial derivative with respect to \( x \) twice.
• \( f_{yy} \): The second partial derivative with respect to \( y \) twice.
• \( f_{xy} \): The mixed second derivative taking \( x \) then \( y \).

These derivatives are used to form the Hessian matrix, which is crucial in applying this test. The determinant of this matrix, \( D \), determines the nature of each critical point. If \( D > 0 \) and \( f_{xx} > 0 \), there's a local minimum. If \( D > 0 \) and \( f_{xx} < 0 \), there's a local maximum. If \( D < 0 \), the point is a saddle point. If \( D = 0 \), the test is inconclusive, as seen at the point \( (0, 0) \) for our function.

Hessian Matrix

The Hessian matrix is a square matrix formed from the second-order partial derivatives of a function. This matrix plays a vital role in determining the curvature of the surface the function defines and is a critical part of the Second Derivative Test.For a function of two variables \( f(x, y) \), the Hessian matrix \( H \) is represented as:\[H = \begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2} \end{bmatrix}\]In our case, this becomes:\[H = \begin{bmatrix} 2y^2 & 4xy 4xy & 2x^2 \end{bmatrix}\]Calculating the determinant of this Hessian matrix \( D = f_{xx}f_{yy} - (f_{xy})^2 \) provides information necessary about the critical points. The value of \( D \) at these points helps you identify their nature, although for \( (0, 0) \) in our function, the determinant being zero suggests that further analysis may be required beyond the Second Derivative Test alone.