Problem 10 - APEX Calculus

Question

Find the critical points of the given function. Use the Second Derivative Test to determine if each critical point corresponds to a relative maximum, minimum, or saddle point. $$ f(x, y)=\frac{1}{3} x^{3}-x+\frac{1}{3} y^{3}-4 y $$

Step-by-Step Solution

Answer

Critical points are (1,2) - min, (1,-2) - saddle, (-1,2) - saddle, (-1,-2) - max.

Step 1: Find the First Partial Derivatives

To find the critical points, we first calculate the first partial derivatives of the function with respect to both variables:\[ f_x = \frac{\partial}{\partial x}\left(\frac{1}{3}x^3 - x + \frac{1}{3}y^3 - 4y\right) = x^2 - 1 \]\[ f_y = \frac{\partial}{\partial y}\left(\frac{1}{3}x^3 - x + \frac{1}{3}y^3 - 4y\right) = y^2 - 4 \]

Step 2: Solve for Critical Points

Set the first partial derivatives equal to zero to find the critical points:\[ f_x = 0 \Rightarrow x^2 - 1 = 0 \]\[ x = \pm 1 \]\[ f_y = 0 \Rightarrow y^2 - 4 = 0 \]\[ y = \pm 2 \]This gives us four critical points: \((1, 2), (1, -2), (-1, 2), (-1, -2)\).

Step 3: Find the Second Partial Derivatives

Calculate the second partial derivatives:\[ f_{xx} = \frac{\partial^2}{\partial x^2}\left(\frac{1}{3}x^3 - x + \frac{1}{3}y^3 - 4y\right) = 2x \]\[ f_{yy} = \frac{\partial^2}{\partial y^2}\left(\frac{1}{3}x^3 - x + \frac{1}{3}y^3 - 4y\right) = 2y \]\[ f_{xy} = \frac{\partial^2}{\partial y \partial x}\left(\frac{1}{3}x^3 - x + \frac{1}{3}y^3 - 4y\right) = 0 \]

Step 4: Use the Second Derivative Test

Apply the second derivative test using the determinant of the Hessian matrix:\[ D(x, y) = f_{xx} f_{yy} - (f_{xy})^2 \]For each critical point:1. **For \((1, 2)\):** \[ D(1, 2) = (2 \cdot 1)(2 \cdot 2) - 0^2 = 4 > 0 \] Since \(f_{xx} = 2 > 0\), it's a relative minimum.2. **For \((1, -2)\):** \[ D(1, -2) = (2 \cdot 1)(2 \cdot (-2)) - 0^2 = -4 < 0 \] It's a saddle point.3. **For \((-1, 2)\):** \[ D(-1, 2) = (2 \cdot (-1))(2 \cdot 2) - 0^2 = -4 < 0 \] It's a saddle point.4. **For \((-1, -2)\):** \[ D(-1, -2) = (2 \cdot (-1))(2 \cdot (-2)) - 0^2 = 4 > 0 \] Since \(f_{xx} = -2 < 0\), it's a relative maximum.

Key Concepts

Partial Derivatives

Partial derivatives are a vital concept when dealing with functions of several variables. They represent the rate of change of a function with respect to one variable, holding the others constant.
To find the partial derivatives of the function \( f(x, y) = \frac{1}{3}x^3 - x + \frac{1}{3}y^3 - 4y \), we compute:

• \( f_x \), the partial derivative with respect to \( x \), is calculated by differentiating the function treating \( y \) as a constant. This gives \( f_x = x^2 - 1 \).
• Similarly, \( f_y \), the partial derivative with respect to \( y \), is derived by differentiating the function while holding \( x \) constant. This gives \( f_y = y^2 - 4 \).

Finding these derivatives allows us to determine critical points, where these partial derivatives equal zero.

Second Derivative Test

The Second Derivative Test helps identify the nature of critical points. After computing the second partial derivatives, we use them to determine whether each critical point is a relative maximum, minimum, or a saddle point.
The second derivative test involves evaluating the following determinant, known as the Hessian determinant \( D(x, y) \):

• \[ D(x, y) = f_{xx} f_{yy} - (f_{xy})^2 \]

Where \( f_{xx} \) and \( f_{yy} \) are the second partial derivatives with respect to \( x \) and \( y \) respectively, and \( f_{xy} \) is the mixed partial derivative.

• If \( D(x, y) > 0 \) and \( f_{xx} > 0 \), the function has a relative minimum at that point.
• If \( D(x, y) > 0 \) and \( f_{xx} < 0 \), the function has a relative maximum.
• If \( D(x, y) < 0 \), the point is a saddle point.

This test provides an efficient way to classify critical points.

Hessian Matrix

The Hessian Matrix is a square matrix of second-order partial derivatives of a scalar-valued function, or scalar field. It provides critical insights into the curvature of the function. For functions of two variables like \( f(x, y) \), the Hessian matrix is given by:

• \[H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{yx} & f_{yy} \end{bmatrix} \]

In our exercise, the Hessian matrix helps in applying the Second Derivative Test. We find:

• \( f_{xx} = 2x \)
• \( f_{yy} = 2y \)
• \( f_{xy} = 0 \)

These components are used to compute the Hessian determinant \( D(x, y) \), which tells us about the nature of the critical points. The sign and value of \( D(x, y) \) guide us in analyzing the behaviour of the function at these points.

Relative Maximum and Minimum

Relative maxima and minima are important concepts in calculus, describing the highest or lowest values a function can achieve relative to surrounding points.
In our example, after applying the Second Derivative Test, we analyzed four critical points: \((1, 2), (1, -2), (-1, 2)\), and \((-1, -2)\). This analysis yields:

• \((1, 2)\): With \( D(1, 2) = 4 > 0 \) and \( f_{xx} = 2 > 0 \), this point is a relative minimum, meaning it is lower than neighboring points.
• \((1, -2)\): Here, \( D(1, -2) = -4 < 0 \), indicating a saddle point, where the point is not a maximum nor a minimum.
• \((-1, 2)\): Similarly, \( D(-1, 2) = -4 < 0 \), confirming another saddle point.
• \((-1, -2)\): With \( D(-1, -2) = 4 > 0 \) and \( f_{xx} = -2 < 0 \), this point is a relative maximum, higher than points in its immediate vicinity.

This process demonstrates how calculus reveals the function's topology.