Problem 322 - 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. $$ f(x, y)=8 x y(x+y)+7 $$

Step-by-Step Solution

Answer

The critical point (0, 0) is a saddle point.

Step 1: Find the First Derivatives

To find the critical points, we first need to compute the first partial derivatives of the function. The given function is \[ f(x, y) = 8xy(x+y) + 7. \]The first partial derivative with respect to \( x \) is: \( f_x(x, y) = \frac{\partial}{\partial x}[8xy(x+y) + 7] \) which simplifies to \( f_x = 16xy + 8y^2 \).The first partial derivative with respect to \( y \) is: \( f_y(x, y) = \frac{\partial}{\partial y}[8xy(x+y) + 7] \) which simplifies to \( f_y = 16xy + 8x^2 \).

Step 2: Solve for Critical Points

To find the critical points, set the first partial derivatives equal to zero:\[ f_x = 16xy + 8y^2 = 0, \]\[ f_y = 16xy + 8x^2 = 0. \]From these equations, factor out common terms to find solutions. For both equations, divide by 8:\[ 2xy + y^2 = 0 \quad \Rightarrow \quad y(2x + y) = 0, \]\[ 2xy + x^2 = 0 \quad \Rightarrow \quad x(2y + x) = 0. \]The solutions to these are:1. \( y = 0 \) and \( x = 0 \). Thus a critical point at \( (0, 0) \).2. \( 2x + y = 0 \Rightarrow y = -2x \) and \( 2y + x = 0 \Rightarrow x = -2y \). Solve \( y = -2x \) and \( x = -2y \): - Substitute \( y = -2x \) into \( x = -2y \), leading to \( x = -2(-2x) \rightarrow x = 4x \), which implies \( x = 0 \). - Substitute \( x = -2y \) back, giving \( y = 0 \).Thus the critical point is also \( (0, 0) \).

Step 3: Find the Second Derivatives

Calculate the second partial derivatives to use the second derivative test. Start from first derivatives:\[ f_{xx}(x, y) = \frac{\partial}{\partial x}(16xy + 8y^2) = 16y, \]\[ f_{yy}(x, y) = \frac{\partial}{\partial y}(16xy + 8x^2) = 16x, \]\[ f_{xy}(x, y) = \frac{\partial}{\partial y}(16xy + 8y^2) = \frac{\partial}{\partial x}(16xy + 8x^2) = 16. \]

Step 4: Apply Second Derivative Test

The second derivative test involves computing the Hessian determinant, \( D \), defined as:\[ D = f_{xx}(0, 0)f_{yy}(0, 0) - [f_{xy}(0, 0)]^2. \]Substitute the values at the critical point \((0, 0)\):\[ f_{xx}(0, 0) = 16(0) = 0, \]\[ f_{yy}(0, 0) = 16(0) = 0, \]\[ f_{xy}(0, 0) = 16. \]Now compute \( D \):\[ D = (0)(0) - (16)^2 = -256. \]Since \( D < 0 \), the critical point \( (0, 0) \) is a saddle point.

Key Concepts

Critical Points

Critical points are essential in determining where a function's behavior changes, such as in finding maxima, minima, or saddle points. For a function of two variables, like \( f(x, y) = 8xy(x+y) + 7 \), critical points occur where both partial derivatives are equal to zero. This means:

• The partial derivative with respect to \( x \), \( f_x \), must be zero.
• The partial derivative with respect to \( y \), \( f_y \), must also be zero.

To find these critical points, you calculate \( f_x \) and \( f_y \) as shown in the steps, and then solve the equations \( f_x = 0 \) and \( f_y = 0 \). In the exercise, both equations simplify to give the critical point at \( (0, 0) \). Once located, these points are assessed further using the second derivative test.

Hessian Determinant

The Hessian determinant, often noted as \( D \), is computed using second partial derivatives to classify the nature of the critical points. The determinant is based on three second partial derivatives:

• \( f_{xx} \) is the second derivative with respect to \( x \),
• \( f_{yy} \) is the second derivative with respect to \( y \),
• \( f_{xy} \) (or \( f_{yx} \)) is the mixed derivative, which is the same due to symmetry provided by Clairaut's theorem.

The formula for the Hessian determinant is: \[ D = f_{xx}(x, y)f_{yy}(x, y) - [f_{xy}(x, y)]^2 \] In our example, at the critical point \( (0, 0) \), the calculation provided \( D = -256 \), which helps us determine that the point is a saddle.

Partial Derivatives

Partial derivatives are essential in multivariable calculus as they allow us to understand how a function changes with respect to one variable while keeping others constant. For the function \( f(x, y) = 8xy(x+y) + 7 \), partial derivatives are calculated for both \( x \) and \( y \).

• \( f_x(x, y) \) represents how \( f \) changes as \( x \) changes, with \( y \) constant, and is \( 16xy + 8y^2 \).
• \( f_y(x, y) \) represents how \( f \) changes as \( y \) changes, with \( x \) constant, and is \( 16xy + 8x^2 \).

Solving \( f_x = 0 \) and \( f_y = 0 \) simultaneously helps identify critical points. They are pivotal to the second derivative test, which assesses the point's nature.

Saddle Point

A saddle point refers to a kind of critical point where the function does not exhibit a local maximum or minimum but instead shows a change in curvature (concavity). In the second derivative test, once we find a critical point, we determine its nature by checking the Hessian determinant, \( D \).

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

In our example, since the determinant \( D = -256 \) is negative at the critical point \( (0, 0) \), this indicates that we have a saddle point. Understanding saddle points is vital as it shows that despite being a critical point, the function's curvature changes differently along different directions.