Problem 338 - 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. $$f(x, y)=2 x^{2}+2 x y+y^{2}+2 x-3$$

Step-by-Step Solution

Answer

The function has a local minimum at the point \((-1, 1)\).

Step 1: Find Partial Derivatives

First, start by finding the partial derivatives of the function \( f(x, y) = 2x^2 + 2xy + y^2 + 2x - 3 \). The partial derivative with respect to \( x \) is: \( f_x = \frac{\partial}{\partial x}(2x^2 + 2xy + y^2 + 2x - 3) = 4x + 2y + 2 \).The partial derivative with respect to \( y \) is: \( f_y = \frac{\partial}{\partial y}(2x^2 + 2xy + y^2 + 2x - 3) = 2x + 2y \).

Step 2: Find Critical Points

Set both partial derivatives equal to zero to find the critical points.Setting \( f_x = 0 \): \( 4x + 2y + 2 = 0 \)Setting \( f_y = 0 \): \( 2x + 2y = 0 \)Solve these system of linear equations. From \( 2x + 2y = 0 \), we have \( x + y = 0 \) or \( y = -x \). Substitute \( y = -x \) in \( 4x + 2y + 2 = 0 \):\( 4x + 2(-x) + 2 = 0 \)\( 4x - 2x + 2 = 0 \)\( 2x = -2 \)\( x = -1 \).Then \( y = -(-1) = 1 \).So, the critical point is \( (-1, 1) \).

Step 3: Compute Second Partial Derivatives

Compute the second partial derivatives to apply the second derivative test.The second partial derivatives are:- \( f_{xx} = \frac{\partial^2 f}{\partial x^2} = 4 \)- \( f_{yy} = \frac{\partial^2 f}{\partial y^2} = 2 \)- \( f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 2 \)

Step 4: Find the Determinant of the Hessian Matrix

Evaluate the determinant of the Hessian matrix at the critical point \( (-1, 1) \).The determinant of the Hessian matrix \( H \) is given by:\[ D = f_{xx}f_{yy} - (f_{xy})^2 \]Substitute the values:\[ D = 4 \times 2 - (2)^2 = 8 - 4 = 4 \]

Step 5: Classify the Critical Point

Use the second derivative test to classify the critical point \((x, y) = (-1, 1)\):- Since \( D = 4 > 0 \) and \( f_{xx} = 4 > 0 \), the function \( f \) has a local minimum at \((-1, 1)\).

Key Concepts

Critical Points

In multivariable calculus, critical points are key to understanding the behavior of a function. To find critical points for a function like the given one, we first find its partial derivatives and set them equal to zero. These equations help us pinpoint where the function's slope is zero in any direction, suggesting possible maxima, minima, or saddle points.

For the function \( f(x, y) = 2x^2 + 2xy + y^2 + 2x - 3 \), we calculated the partial derivatives with respect to both \( x \) and \( y \):

• \( f_x = 4x + 2y + 2 \)
• \( f_y = 2x + 2y \)

To find the critical points, solve the system where both partial derivatives equal zero. This solution method helps us determine where the tangent plane to the surface defined by the function is horizontal. The critical point for this function was found at \((-1, 1)\).

Partial Derivatives

Partial derivatives are fundamental in understanding how a multivariable function changes with respect to each variable independently. They are similar to taking a derivative for a single-variable function, but we do it while keeping the other variables constant.

For the function \( f(x, y) = 2x^2 + 2xy + y^2 + 2x - 3 \):

• The partial derivative with respect to \( x \) is \( f_x = 4x + 2y + 2 \).
• The partial derivative with respect to \( y \) is \( f_y = 2x + 2y \).

By setting these partial derivatives equal to zero, we can determine the locations at which the function does not increase or decrease, aiding in finding the critical points. This process is crucial for utilizing the second derivative test afterwards.

Hessian Matrix

The Hessian Matrix is an essential tool when working with functions of multiple variables. It encapsulates all the second partial derivatives of a function and provides key information about the curvature of the function's graph.

The Hessian for a two-variable function \( f(x, y) \) looks like this:

• \( f_{xx} \) - second partial derivative with respect to \( x \)
• \( f_{yy} \) - second partial derivative with respect to \( y \)
• \( f_{xy} \) and \( f_{yx} \) - mixed partial derivatives

For our function, these were found as:

• \( f_{xx} = 4 \)
• \( f_{yy} = 2 \)
• \( f_{xy} = f_{yx} = 2 \)

To determine the type of critical point, we compute the determinant of the Hessian. A positive determinant suggests a local extremum, which we determine further by checking the sign of \( f_{xx} \) for minimum or maximum.

Local Minimum

When seeking to classify critical points, the second derivative test provides us with a roadmap. It relies on the Hessian matrix to assess the local behavior of the function. Specifically, it helps determine if the critical point is a local minimum, maximum, or saddle point.

For a function \( f(x, y) \), the second derivative test observes:

• If the determinant \( D = f_{xx}f_{yy} - (f_{xy})^2 > 0 \) and \( f_{xx} > 0 \), the function has a local minimum at that point.
• If \( D > 0 \) and \( f_{xx} < 0 \), there is a local maximum.
• If \( D < 0 \), the point is a saddle point, indicating mixed behavior akin to concave up in one direction and concave down in the other.

In our exercise, \( D = 4 \) and \( f_{xx} = 4 \), both positive, thus verifying a local minimum exists at \((-1, 1)\). This result fits logically within the function's overall shape.