Problem 323 - 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. $$ f(x, y)=x^{2}+4 x y+y^{2} $$

Step-by-Step Solution

Answer

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

Step 1: Find the First Partial Derivatives

To find the critical points, we first need the first partial derivatives with respect to both variables. Calculate the partial derivative with respect to \(x\): \(f_x(x, y) = \frac{\partial}{\partial x}(x^2 + 4xy + y^2) = 2x + 4y\). Now, calculate the partial derivative with respect to \(y\): \(f_y(x, y) = \frac{\partial}{\partial y}(x^2 + 4xy + y^2) = 4x + 2y\).

Step 2: Set First Partial Derivatives to Zero

To find critical points, set the first partial derivatives to zero. Solve the system of equations: \(2x + 4y = 0\) and \(4x + 2y = 0\).

Step 3: Solve the System of Equations

Solve the equations \(2x + 4y = 0\) and \(4x + 2y = 0\) simultaneously to find \(x\) and \(y\). From the first equation, simplify to get \(x = -2y\). Substitute \(x = -2y\) into the second equation: \(4(-2y) + 2y = 0\), leading to \(-8y + 2y = 0\), i.e., \(-6y = 0\). Therefore, \(y = 0\). Substitute \(y = 0\) back to find \(x = -2(0) = 0\). Thus, the critical point is \((0, 0)\).

Step 4: Find the Second Partial Derivatives

Calculate the second partial derivatives. \(f_{xx} = \frac{\partial^2 f}{\partial x^2} = 2\), \(f_{yy} = \frac{\partial^2 f}{\partial y^2} = 2\), and \(f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 4\).

Step 5: Apply the Second Derivative Test

Use the second derivative test to classify the critical point \((0, 0)\). Compute the determinant of the Hessian matrix, \(D = f_{xx}f_{yy} - (f_{xy})^2\). Substitute the second derivatives: \(D = (2)(2) - (4)^2 = 4 - 16 = -12\). 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 may achieve a maximum, minimum, or special feature in its graph. For a function of two variables, a critical point occurs when both first partial derivatives equal zero. In the context of the given function, we start by calculating the partial derivatives \(f_x(x, y)\) and \(f_y(x, y)\). To find the critical points:

• Compute the first partial derivatives for \(x\) and for \(y\).
• Set these derivatives equal to zero to form a system of equations.
• Solve the system to find the critical points, which are particular \((x, y)\) pairs.

Once determined, critical points help identify where interesting features of the surface described by the function might exist.

Partial Derivatives

Partial derivatives allow us to study how a function changes with respect to one variable, while keeping the other variables constant. For the given function, \(f(x, y) = x^2 + 4xy + y^2\), partial derivatives are calculated as follows:

• Partial derivative with respect to \(x\), \(f_x(x, y) = 2x + 4y\).
• Partial derivative with respect to \(y\), \(f_y(x, y) = 4x + 2y\).

These derivatives help to identify the direction and rate of change for each variable. They are crucial in locating the critical points of a function, which are found by setting these derivatives to zero and solving for \(x\) and \(y\). Understanding partial derivatives is essential for analyzing multivariable functions.

Hessian Matrix

The Hessian matrix is a square matrix of second-order partial derivatives for a multivariable function. It plays a crucial role in the second derivative test, which helps to determine the nature of critical points. For the function \(f(x, y) = x^2 + 4xy + y^2\), the Hessian matrix \(H\) is:\[H = \begin{bmatrix}f_{xx} & f_{xy} \f_{yx} & f_{yy}\end{bmatrix}\]Using our function's derivatives:

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

The determinant of the Hessian \(D = f_{xx}f_{yy} - (f_{xy})^2\) indicates the behavior of the function near the critical point. It can show whether it's a local maximum, minimum, or saddle point based on its value.

Saddle Point

A saddle point is a type of critical point where the surface does not have a local maximum or minimum but instead curves upward in one direction and downward in another. This creates a surface that resembles a saddle.In our example, the critical point \((0, 0)\) is classified as a saddle point by using the determinant of the Hessian matrix.

• Calculate \(D = f_{xx}f_{yy} - (f_{xy})^2\).
• For the given function, \(D = 4 - 16 = -12\).

Since the determinant \(D\) is less than zero, \((0, 0)\) does not correspond to a classic maximum or minimum. Instead, it indicates an inflection point—a saddle. This special characteristic appears frequently in multivariable calculus, highlighting the complex nature of functions involving more than one variable.