Problem 328 - 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. $$ f(x, y)=3 x^{2}+2 x y+y^{2} $$

Step-by-Step Solution

Answer

The critical point is (0, 0), and it is a local minimum.

Step 1: Find First Partial Derivatives

To locate the critical points, we need to find where the first partial derivatives are equal to zero. Compute the first partial derivatives of the function:\( f_x(x,y) = \frac{\partial}{\partial x}(3x^2 + 2xy + y^2) = 6x + 2y \)\( f_y(x,y) = \frac{\partial}{\partial y}(3x^2 + 2xy + y^2) = 2x + 2y \).

Step 2: Set First Partial Derivatives to Zero

Set the first partial derivatives to zero to find critical points:\( 6x + 2y = 0 \)\( 2x + 2y = 0 \).

Step 3: Solve the System of Equations

Solve the system of equations from the previous step:From \( 2x + 2y = 0 \), we have \( x + y = 0 \), which simplifies to \( y = -x \).Substitute \( y = -x \) into \( 6x + 2y = 0 \):\( 6x + 2(-x) = 0 \) gives \( 6x - 2x = 0 \), so \( 4x = 0 \), thus \( x = 0 \).Since \( y = -x \), \( y = 0 \) and we find the critical point is \( (0, 0) \).

Step 4: Find Second Partial Derivatives

Compute the second partial derivatives for the second derivative test:\( f_{xx}(x,y) = \frac{\partial^2}{\partial x^2}(3x^2 + 2xy + y^2) = 6 \)\( f_{yy}(x,y) = \frac{\partial^2}{\partial y^2}(3x^2 + 2xy + y^2) = 2 \)\( f_{xy}(x,y) = \frac{\partial^2}{\partial x \partial y}(3x^2 + 2xy + y^2) = 2 \).

Step 5: Apply Second Derivative Test

The second derivative test uses the discriminant \( D = f_{xx}f_{yy} - (f_{xy})^2 \):\( D = (6)(2) - (2)^2 = 12 - 4 = 8 \).Evaluate at the critical point \((0,0)\):Since \( D > 0 \) and \( f_{xx}(0, 0) = 6 > 0 \), the critical point \( (0, 0) \) is a local minimum.

Key Concepts

Critical Points

Critical points in a multivariable function are similar to points in a single-variable function where the function does not increase or decrease in any particular direction. These are the points of interest where we seek to determine whether a local maximum, local minimum, or other type of behavior occurs.
To locate critical points, we solve for where the first partial derivatives of the function are zero. This implies that the function's rate of change, or gradient, is zero at these points.

• Objective: Set the first partial derivatives of the function to zero.
• Outcome: Identify potential critical points by solving these equations.

In our example function, we determined that the first partial derivatives lead us to the simultaneous equations:

• \( 6x + 2y = 0 \)
• \( 2x + 2y = 0 \)

Solving these gives us the critical point \((0, 0)\), which becomes the focus for further classification using the second derivative test.

Partial Derivatives

Partial derivatives are essential tools in multivariable calculus, representing the rate of change of a function with respect to one variable while keeping the others constant. They function similarly to the derivative in single-variable calculus but account for multiple dimensions.
For a function of two variables, such as \( f(x, y) \), the partial derivatives \( f_x(x, y) \) and \( f_y(x, y) \) describe the slope of the surface at any given point in the x and y directions, respectively.

• First Partial Derivatives: Used to locate critical points by finding where the rate of change in both directions is zero.
• Second Partial Derivatives: Used in the second derivative test to help classify critical points.

In the exercise, the second partial derivatives \( f_{xx}(x, y), f_{yy}(x, y), \) and \( f_{xy}(x, y) \) were calculated to apply the second derivative test effectively.

Local Minimum

A local minimum in multivariable calculus is a point where the function has a value smaller than all nearby points. At this point, the function curves upwards in all directions, resembling a small valley.
The second derivative test helps to identify such a point by examining the sign and value of the discriminant \( D \) and the second partial derivatives.

• The Discriminant \( D = f_{xx}f_{yy} - (f_{xy})^2 \).
• If \( D > 0 \) and \( f_{xx} > 0 \), the point is a local minimum.

For the specific critical point \((0, 0)\) in our function, the second derivative test revealed:

• \( D = 8 > 0 \)
• \( f_{xx} = 6 > 0 \)

This confirmed that \((0, 0)\) is indeed a local minimum, highlighting how second derivatives can classify the nature of a critical point.

Saddle Points

Saddle points are like the strange cousins of local extrema in multivariable calculus. These points are neither local maxima nor local minima. Instead, the surface curves up in one direction and down in another, forming a saddle-like shape.
The second derivative test helps identify a saddle point. At such points, the discriminant \( D \) plays a crucial role.

• If \( D < 0 \): The critical point is a saddle point.
• Behavior: A mixed increase and decrease around the point.

While our exercise did not eventually result in a saddle point, understanding this concept is vital for complete mastery of the second derivative test, as other functions might have saddle points instead of a local minimum or maximum.