Problem 319 - Calculus 3

Question

For the following exercises, use the second derivative test to identify any critical points and determine whether each critical point is a maximum, minimum, saddle point, or none of these. $$ f(x, y)=x^{2} y^{2} $$

Step-by-Step Solution

Answer

The critical point is \((0, 0)\) and the test is inconclusive.

Step 1: Find the first-order partial derivatives

To apply the second derivative test, we first need to find the first-order partial derivatives of the function \( f(x, y) = x^2 y^2 \). The partial derivative of \( f \) with respect to \( x \) is:\[ f_x = \frac{\partial}{\partial x}(x^2 y^2) = 2xy^2 \]The partial derivative of \( f \) with respect to \( y \) is:\[ f_y = \frac{\partial}{\partial y}(x^2 y^2) = 2yx^2 \]

Step 2: Find critical points

Critical points occur where the first-order partial derivatives are zero. Set \( f_x = 2xy^2 = 0 \) and \( f_y = 2yx^2 = 0 \) to find critical points.For \( f_x = 0 \): either \( x = 0 \) or \( y = 0 \).For \( f_y = 0 \): either \( y = 0 \) or \( x = 0 \).Thus, the critical points are when \( x = 0 \) or \( y = 0 \). The critical point is \( (0,0) \).

Step 3: Find the second-order partial derivatives

We now find the second-order partial derivatives of \( f \).\[ f_{xx} = \frac{\partial^2}{\partial x^2}(x^2 y^2) = 2y^2 \]\[ f_{yy} = \frac{\partial^2}{\partial y^2}(x^2 y^2) = 2x^2 \]\[ f_{xy} = \frac{\partial^2}{\partial x \partial y}(x^2 y^2) = 4xy \]

Step 4: Apply the second derivative test

The second derivative test involves the Hessian determinant, given by\[ D = f_{xx}(x_0, y_0)f_{yy}(x_0, y_0) - (f_{xy}(x_0, y_0))^2 \].At the critical point \( (0,0) \):\[ f_{xx} = 2\cdot0^2 = 0, \ f_{yy} = 2\cdot0^2 = 0, \ f_{xy} = 4\cdot0\cdot0 = 0 \]So, \[ D = 0\cdot 0 - (0)^2 = 0 \].When \( D = 0 \), the second derivative test is inconclusive.

Key Concepts

Partial Derivatives

Partial derivatives are used to find the rate at which a function changes as one of its variables is altered, keeping the other variables constant.
In the exercise, we're dealing with a function of two variables, \( f(x, y) = x^2 y^2 \).
To start, we calculate the first partial derivatives:

• The partial derivative with respect to \( x \) is found by differentiating the function while treating \( y \) as a constant. This gives us \( f_x = 2xy^2 \).
• The partial derivative with respect to \( y \) is found by differentiating with \( x \) held constant, resulting in \( f_y = 2yx^2 \).

These derivatives help in identifying the behavior of the function's surface and are crucial in finding the critical points.

Critical Points

Critical points are locations on the function's surface where the slope is zero in all directions, indicating potential maxima, minima, or saddle points.
To find these, we set our partial derivatives to zero:

• \( f_x = 0 \) means \( 2xy^2 = 0 \). This can be true if \( x = 0 \) or \( y = 0 \).
• Similarly, \( f_y = 0 \) implies \( 2yx^2 = 0 \), also leading to \( y = 0 \) or \( x = 0 \).

Thus, solving these equations results in the critical point at \( (0, 0) \).
Identifying critical points serves as a foundation for further analysis since they can indicate where changes in the nature of the function might occur.

Hessian Determinant

The Hessian determinant helps determine the nature of a critical point.
It involves the second-order partial derivatives and is part of the second derivative test.
The formula given in the exercise is:\[ D = f_{xx} f_{yy} - (f_{xy})^2 \]For our function, the second-order partial derivatives are:

• \( f_{xx} = 2y^2 \)
• \( f_{yy} = 2x^2 \)
• \( f_{xy} = 4xy \)

At the critical point \((0, 0)\), calculating these gives us all derivatives as zero, resulting in \( D = 0 \).
At this point, the second derivative test is inconclusive, meaning we cannot readily determine if \((0, 0)\) is a maximum, minimum, or a saddle point from this determinant alone.

Saddle Point

A saddle point refers to a point on a function's surface where the surface curves upwards in one direction and downwards in another.
It resembles a saddle and does not represent a local maximum or minimum.
In our case, despite our Hessian determinant returning zero, indicating an inconclusive test, understanding saddle points helps frame hypotheses.
If the determinants had differed, we'd get insight into the surface's curvature.
In contexts where \( D > 0 \) and mixed derivatives contribute negatively, saddle points become apparent, contrasting with the intuitions for maxima and minima.
Thus, while \( (0, 0) \) remains uncertain without further tools, recognizing potential saddle points enriches our understanding of a function's multifaceted surface.