Problem 320 - 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. $$f(x, y)=x^{2}-6 x+y^{2}+4 y-8$$

Step-by-Step Solution

Answer

The critical point (3, -2) is a local minimum.

Step 1: Calculate the First Derivatives

To find the critical points of the function \(f(x, y) = x^2 - 6x + y^2 + 4y - 8\), we first calculate the partial derivatives with respect to \(x\) and \(y\). The partial derivative with respect to \(x\) is:\[ f_x = \frac{\partial f}{\partial x} = 2x - 6 \]The partial derivative with respect to \(y\) is:\[ f_y = \frac{\partial f}{\partial y} = 2y + 4 \]

Step 2: Find the Critical Points

Set the first derivatives equal to zero to find the critical points. Set \(f_x = 0\): \[ 2x - 6 = 0 \]Solve for \(x\): \[ x = 3 \]Set \(f_y = 0\): \[ 2y + 4 = 0 \]Solve for \(y\): \[ y = -2 \]Thus, the critical point is \((3, -2)\).

Step 3: Calculate the Second Derivatives

To apply the second derivative test, calculate the second partial derivatives:\( f_{xx} = \frac{\partial^2 f}{\partial x^2} = 2 \)\( f_{yy} = \frac{\partial^2 f}{\partial y^2} = 2 \)\( f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 0 \)

Step 4: Apply the Second Derivative Test

The second derivative test classifies the critical point by computing the determinant of the Hessian matrix, \( D \):\[ D = f_{xx}f_{yy} - (f_{xy})^2 = (2)(2) - (0)^2 = 4 \]Since \(D > 0\) and \(f_{xx} > 0\), the critical point \((3, -2)\) is a local minimum.

Key Concepts

Critical Points

Critical points are essential in analyzing the behavior of functions. They are points in the domain where the function's partial derivatives are zero or undefined. In the context of multivariable functions like \( f(x, y) = x^2 - 6x + y^2 + 4y - 8 \), finding critical points involves several steps:

• First, compute the partial derivatives of the function with respect to each variable.
• Then, set these derivatives equal to zero.
• Solve the resulting system of equations to find values of \( x \) and \( y \) that satisfy both equations.

In our example, solving **\( 2x - 6 = 0 \)** and **\( 2y + 4 = 0 \)** gives the critical point \((3, -2)\). The importance of critical points lies in their potential to be extrema (maxima or minima) or saddle points of the function.

Partial Derivatives

When dealing with functions of several variables, partial derivatives are a powerful tool. They help us understand how a function changes when we vary one variable while keeping others constant. For instance, in the function \( f(x, y) = x^2 - 6x + y^2 + 4y - 8 \), the partial derivatives are:

• **\( \frac{\partial f}{\partial x} = 2x - 6 \)**, which measures how changes in \( x \) affect \( f \).
• **\( \frac{\partial f}{\partial y} = 2y + 4 \)**, which measures how changes in \( y \) affect \( f \).

These derivatives are used to find critical points by setting them equal to zero. Essentially, you are finding where the surface of the function is flat in each direction. This allows you to ascertain locations where the function might have special properties, like turning, leveling off, or reaching a maximum or minimum.

Hessian Matrix

The Hessian matrix is a square matrix of second-order mixed partial derivatives of a scalar-valued function. Think of it as a mathematical tool to explore the nature of critical points. In our multivariable function example, the Hessian matrix is:\[H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{xy} & f_{yy} \end{bmatrix} = \begin{bmatrix} 2 & 0 \ 0 & 2 \end{bmatrix}\]To assess a critical point using the Hessian matrix, compute the determinant, denoted as \( D \):

• \( D = f_{xx}f_{yy} - (f_{xy})^2 \)

For our function, \( D = 4 \), which indicates a definite shape of the surface around the critical point. The sign of \( D \) and the second derivatives will categorize the critical point as a maximum, minimum, or saddle point.

Local Minimum

Determining whether a critical point is a local minimum requires using the second derivative test. For our function, the second derivative test gives us conditions based on the Hessian matrix determinant \( D \) and the second partial derivative \( f_{xx} \):

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

At the critical point \((3, -2)\), we found \( D = 4 \) and \( f_{xx} = 2 \), so it is a local minimum. This means in a neighborhood around \((3, -2)\), the function \( f(x, y) \) reaches a lower value locally compared to nearby points.