Problem 331 - 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. $$ f(x, y)=x^{2}+2 y^{2}-x^{2} y $$

Step-by-Step Solution

Answer

The critical points are (0,0) a local minimum, (2,1) and (-2,1) saddle points.

Step 1: Find First Partial Derivatives

Compute the first partial derivatives \( f_x \) and \( f_y \).\[ f_x = \frac{\partial}{\partial x} (x^2 + 2y^2 - x^2y) = 2x - 2xy \]\[ f_y = \frac{\partial}{\partial y} (x^2 + 2y^2 - x^2y) = 4y - x^2 \]

Step 2: Find Critical Points

Set the first partial derivatives equal to zero and solve for \(x\) and \(y\).For \( f_x = 0 \):\[ 2x - 2xy = 0 \]\[ 2x(1 - y) = 0 \]This gives solutions \( x = 0 \) or \( y = 1 \).For \( f_y = 0 \):\[ 4y - x^2 = 0 \]\[ x^2 = 4y \]Substituting in the two cases:- When \( x = 0 \), \( 4y = 0 \) gives \( y = 0 \). Thus, one critical point is \((0,0)\).- When \( y = 1 \), \( x^2 = 4 \times 1 \) gives \( x = \pm 2 \). Thus, two critical points are \((2, 1)\) and \((-2, 1)\).

Step 3: Find Second Partial Derivatives

Compute the second partial derivatives for the second derivative test.\[ f_{xx} = \frac{\partial}{\partial x}(2x - 2xy) = 2 - 2y \]\[ f_{yy} = \frac{\partial}{\partial y}(4y - x^2) = 4 \]\[ f_{xy} = \frac{\partial}{\partial y}(2x - 2xy) = -2x \]\[ f_{yx} = \frac{\partial}{\partial x}(4y - x^2) = -2x \]

Step 4: Apply Second Derivative Test

Evaluate the determinant of the Hessian matrix \( D = f_{xx}f_{yy} - (f_{xy})^2 \) at each critical point.For \((0,0)\):\[ f_{xx} = 2, \; f_{yy} = 4, \; f_{xy} = 0 \]\[ D = (2)(4) - (0)^2 = 8 \]Since \( D > 0 \) and \( f_{xx} > 0 \), \((0,0)\) is a local minimum.For \((2,1)\) and \((-2,1)\):\[ f_{xx} = 2 - 2 = 0, \; f_{yy} = 4, \; f_{xy} = -4 \text{ for both points} \]\[ D = (0)(4) - (-4)^2 = -16 \]Since \( D < 0 \), \((2,1)\) and \((-2,1)\) are saddle points.

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept when dealing with functions of multiple variables, such as functions of the form \( f(x, y) \). In simple terms, a partial derivative measures how a function changes as one of its variables is varied while keeping the other variables constant.
For example, if we have a function \( f(x, y) = x^2 + 2y^2 - x^2y \), we compute the partial derivative of \( f \) with respect to \( x \), denoted as \( f_x \), by treating \( y \) as a constant. This gives us \( f_x = 2x - 2xy \).
Similarly, the partial derivative of \( f \) with respect to \( y \), denoted as \( f_y \), involves treating \( x \) as a constant, resulting in \( f_y = 4y - x^2 \). These derivatives help us understand the slope of the function in different directions. Understanding partial derivatives is crucial for locating critical points of multivariable functions, which leads to our next topic.

Critical Points

Critical points in the context of multivariable calculus are points where the gradient of the function (a vector of all first partial derivatives) is zero, meaning that the slope of the function is flat in all directions. For our function \( f(x, y) = x^2 + 2y^2 - x^2y \), we find critical points by solving the system of equations \( f_x = 0 \) and \( f_y = 0 \).
This requires us to simultaneously solve the equations:

• \( 2x - 2xy = 0 \) which simplifies to \( 2x(1 - y) = 0 \), giving the solutions \( x = 0 \) or \( y = 1 \).
• \( 4y - x^2 = 0 \) which gives \( x^2 = 4y \).

By taking different cases, we find the critical points, such as \( (0, 0) \), \( (2, 1) \), and \( (-2, 1) \). Each of these points might be a point of interest for maximum, minimum, or saddle point, but further testing is required to determine their nature. This brings us to using the Hessian matrix.

Hessian Matrix

The Hessian matrix is a square matrix of second-order partial derivatives of a scalar-valued function, or scalar field. For functions \( f(x, y) \), the Hessian is used in the second derivative test to determine the nature of critical points.
For our function \( f(x, y) \), the Hessian matrix \( H \) is composed of the second partial derivatives:

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

The determinant of the Hessian, \( D = f_{xx}f_{yy} - (f_{xy})^2 \), is key to the second derivative test. If \( D > 0 \) at a point and \( f_{xx} > 0 \), the point is a local minimum; if \( f_{xx} < 0 \), it's a local maximum. If \( D < 0 \), the point is a saddle point.
For our critical points, calculations show \( (0,0) \) as a local minimum, and points \( (2,1) \) and \( (-2,1) \) as saddle points, due to their respective Hessian determinants and second derivatives.