Problem 329 - Calculus Volume 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. $$f(x, y)=y^{2}+x y+3 y+2 x+3$$

Step-by-Step Solution

Answer

The critical point \((1, -2)\) is a saddle point.

Step 1: Find the First Partial Derivatives

To identify critical points, we first find the partial derivatives of the function with respect to \(x\) and \(y\). The partial derivative of \( f(x, y) \) with respect to \( x \) is \( f_x = y + 2 \). The partial derivative with respect to \( y \) is \( f_y = 2y + x + 3 \).

Step 2: Set Partial Derivatives to Zero

Critical points occur where both partial derivatives are zero. Set \( f_x = 0 \) and \( f_y = 0 \). This gives us the equations: \( y + 2 = 0 \) and \( 2y + x + 3 = 0 \).

Step 3: Solve the System of Equations

Solving the equation \( y + 2 = 0 \) gives \( y = -2 \). Substitute \( y = -2 \) into \( 2y + x + 3 = 0 \), resulting in \( 2(-2) + x + 3 = 0 \), or \( x - 1 = 0 \), giving \( x = 1 \). Thus, the critical point is \((1, -2)\).

Step 4: Compute the Second Partial Derivatives

Calculate the second partial derivatives: \( f_{xx} = 0 \), \( f_{yy} = 2 \), and \( f_{xy} = 1 \). These are needed for the second derivative test.

Step 5: Apply the Second Derivative Test

For the function \( f(x, y) \), determine the nature of the critical point using the second derivative test. Compute the determinant \( D = f_{xx}f_{yy} - (f_{xy})^2 \). Thus, \( D = 0 \times 2 - (1)^2 = -1 \).

Step 6: Interpret the Result

Since \( D < 0 \), the point \((1, -2)\) is a saddle point according to the second derivative test.

Key Concepts

Critical Points

Critical points are where a function's partial derivatives equal zero.
These points are where potential local maxima, minima, or saddle points could occur.
To find these points in a multivariable function, like our given function \( f(x, y) = y^2 + xy + 3y + 2x + 3 \), we first calculate the first partial derivatives with respect to each variable. Once these are set to zero, we solve the resulting system of equations tolocate the critical points.

• Partial derivative with respect to \(x\): \( f_x = y + 2 \)
• Partial derivative with respect to \(y\): \( f_y = 2y + x + 3 \)

By solving \(f_x = 0\) and \(f_y = 0\), we find our critical point is (1, -2).
This point lays the groundwork for further analysis.

Partial Derivatives

Partial derivatives are used to determine how a function changes as one variable changes, while keeping other variables constant. In a multivariable function, each variable can have its own partial derivative.
These derivatives allow us to study the function's behavior at critical points.

• The partial derivative with respect to \( x \), noted as \( f_x \), shows the rate of change of the function as \( x \) changes.
• The partial derivative with respect to \( y \), noted as \( f_y \), shows the rate of change as \( y \) changes.

In our example, finding \( f_x \) and \( f_y \) and setting them to zero helps us locate the critical point.
Understanding partial derivatives is essential for further steps like computing the second derivative for determining the nature of these critical points.

Saddle Point

A saddle point occurs in a function when the point is neither a local maximum nor a local minimum.
Instead, the point is a mix of both; in one direction it behaves like a maximum and in another, like a minimum.
When using the second derivative test, if the determinant \( D \) is negative, as we found \( D = -1 \),
the critical point is classified as a saddle point.
In this case, the critical point (1, -2) exhibits this saddle-like behavior.
Saddle points are crucial in analyzing landscapes of functions because they indicate points of instability or change in curvature, essential in optimization problems.

Second Partial Derivatives

Second partial derivatives involve taking the partial derivative of a partial derivative, providing deeper insight into the behavior of a function.
These derivatives are vital in the second derivative test that determines the nature of critical points. The necessary derivatives are:

• \( f_{xx} \): the second partial derivative with respect to \( x \), gives information on concavity in the \( x \)-direction.
• \( f_{yy} \): the second partial derivative with respect to \( y \), shows concavity in the \( y \)-direction.
• \( f_{xy} \): the mixed partial derivative, checks how the function changes in a diagonal direction.

These derivatives are used in calculating \( D = f_{xx}f_{yy} - (f_{xy})^2 \),
which indicates the curvature at the critical point.
Knowing the second partial derivatives allows us to use the second derivative test, providing essential insights into points on a surface.