Problem 327 - 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. $$f(x, y)=3 x^{2}-2 x y+y^{2}-8 y$$

Step-by-Step Solution

Answer

The critical point \((2, 6)\) is a local minimum.

Step 1: Find the First Derivatives

To use the second derivative test, we first need to find the first derivatives of \( f(x, y) \). Calculate \( f_x(x, y) \) and \( f_y(x, y) \).The partial derivative with respect to \( x \):\[ f_x(x, y) = \frac{\partial}{\partial x}(3x^2 - 2xy + y^2 - 8y) = 6x - 2y. \]The partial derivative with respect to \( y \):\[ f_y(x, y) = \frac{\partial}{\partial y}(3x^2 - 2xy + y^2 - 8y) = -2x + 2y - 8. \]

Step 2: Find Critical Points

Critical points occur where both first derivatives are zero. Solve the system of equations:1. \( 6x - 2y = 0 \)2. \( -2x + 2y - 8 = 0 \)First equation gives: \( y = 3x \).Substitute \( y = 3x \) into the second equation:\[ -2x + 2(3x) - 8 = 0 \]\[ -2x + 6x - 8 = 0 \]\[ 4x = 8 \]\[ x = 2 \].Substitute \( x = 2 \) into \( y = 3x \) to find \( y \):\[ y = 3(2) = 6 \].Thus, the critical point is \((x, y) = (2, 6)\).

Step 3: Compute the Second Derivatives

We need the second derivatives to apply the second derivative test. Compute the second derivatives:1. \( f_{xx}(x, y) = \frac{\partial}{\partial x}(6x - 2y) = 6 \).2. \( f_{yy}(x, y) = \frac{\partial}{\partial y}(-2x + 2y - 8) = 2 \).3. \( f_{xy}(x, y) = \frac{\partial}{\partial y}(6x - 2y) = -2 \).

Step 4: Apply the Second Derivative Test

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

Key Concepts

Critical Points

Critical points in a function of two variables, like our given function, are points where both partial derivatives are zero. This is where the slope of the function is flat in all directions, which can imply potential maxima, minima, or saddle points.

To find critical points, you first compute the partial derivatives of the function with respect to each variable. For the function \( f(x, y) = 3x^2 - 2xy + y^2 - 8y \), the partial derivatives are:

• \( f_x(x, y) = 6x - 2y \)
• \( f_y(x, y) = -2x + 2y - 8 \)

You then set these derivatives equal to zero to find solutions for \( x \) and \( y \). These solutions are your critical points. For this function, solving the equations yields the critical point at \( (2, 6) \).

Partial Derivatives

Partial derivatives are crucial when dealing with functions of multiple variables. They represent the rate of change of the function with respect to one variable while keeping the other variables constant.

In our current exercise, we calculate the partial derivatives with respect to \( x \) and \( y \) to find the direction and rate of change along each axis separately. Compute the derivatives like so:

• The partial derivative with respect to \( x \) is \( f_x(x, y) = 6x - 2y \).
• The partial derivative with respect to \( y \) is \( f_y(x, y) = -2x + 2y - 8 \).

Understanding these derivatives is key to finding the critical points and further analyzing the function's behavior at these points.

Determining Local Extremes

Once critical points are identified, the next step is to determine whether these points are local maxima, minima, or saddle points. This is done using the second derivative test in multivariable calculus.

First, you need to find the second derivatives of the function:

• \( f_{xx}(x, y) = 6 \)
• \( f_{yy}(x, y) = 2 \)
• \( f_{xy}(x, y) = -2 \)

Next, calculate the determinant of the Hessian matrix, \( D \), at the critical point \((2, 6)\): \[ D = f_{xx}(2, 6)f_{yy}(2, 6) - (f_{xy}(2, 6))^2 = 12 - 4 = 8 \]Since \( D > 0 \) and \( f_{xx} > 0 \), the critical point \((2, 6)\) is a local minimum. This test helps us understand the behavior of the function in the vicinity of the critical points.

Multivariable Calculus

Multivariable calculus extends calculus to functions with more than one variable. It includes concepts like partial derivatives, gradient vectors, and multiple integrals, enabling deeper analysis of complex functions.

In the context of this exercise, we emphasized the application of partial derivatives to locate and classify critical points. Using these derivatives, along with the second derivative test, is fundamental to understanding the topology of a function's surface.

Studying multivariable calculus allows us to explore functions in broader contexts, such as those represented in physical phenomena and engineering problems, where multiple inputs and outputs coexist.