Problem 14 - Applied Calculus

Question

Find all the critical points and determine whether each is a local maximum, local minimum, a saddle point, or none of these. $$f(x, y)=x^{3}-3 x+y^{3}-3 y$$

Step-by-Step Solution

Answer

Critical points are (1, 1) local minimum, (-1, 1) saddle point, (1, -1) saddle point, (-1, -1) local maximum.

Step 1: Find the Partial Derivatives

To find the critical points, we first calculate the first partial derivatives of the function \( f(x, y) = x^3 - 3x + y^3 - 3y \). We find \( \frac{\partial f}{\partial x} = 3x^2 - 3 \) and \( \frac{\partial f}{\partial y} = 3y^2 - 3 \).

Step 2: Solve for Critical Points

Set the first partial derivatives to zero to find the critical points: \[ 3x^2 - 3 = 0 \] \[ 3y^2 - 3 = 0 \] Solving these equations gives \( x^2 = 1 \) and \( y^2 = 1 \) resulting in solutions \( x = \pm 1 \) and \( y = \pm 1 \). Thus, the critical points are: \((1, 1)\), \((-1, 1)\), \((1, -1)\), and \((-1, -1)\).

Step 3: Calculate the Second Partial Derivatives

To determine the nature of each critical point, calculate the second partial derivatives. We find: \[ \frac{\partial^2 f}{\partial x^2} = 6x \] \[ \frac{\partial^2 f}{\partial y^2} = 6y \] \[ \frac{\partial^2 f}{\partial x \partial y} = 0 \].

Step 4: Apply the Second Derivative Test

The second derivative test for functions of two variables uses the determinant of the Hessian matrix \( H \). The Hessian determinant is calculated as follows:\[ H = \left(\frac{\partial^2 f}{\partial x^2} \right)\left(\frac{\partial^2 f}{\partial y^2} \right) - \left( \frac{\partial^2 f}{\partial x \partial y}\right)^2 = (6x)(6y) - 0 = 36xy \].

Step 5: Evaluate the Hessian at Each Critical Point

Evaluate \( H \) at each critical point:- At \((1, 1)\), \( H = 36 \cdot 1 \cdot 1 = 36 \) (positive), with \( \frac{\partial^2 f}{\partial x^2} = 6 \cdot 1 = 6 > 0 \): local minimum.- At \((-1, 1)\), \( H = 36 \cdot (-1) \cdot 1 = -36 \) (negative): saddle point.- At \((1, -1)\), \( H = 36 \cdot 1 \cdot (-1) = -36 \) (negative): saddle point.- At \((-1, -1)\), \( H = 36 \cdot (-1) \cdot (-1) = 36 \) (positive), with \( \frac{\partial^2 f}{\partial x^2} = 6 \cdot (-1) = -6 < 0 \): local maximum.

Key Concepts

Partial Derivatives

Partial derivatives are essential when dealing with functions of multiple variables. In a function like \( f(x, y) = x^3 - 3x + y^3 - 3y \), we find the rate of change with respect to each variable independently. This means that:

• The partial derivative \( \frac{\partial f}{\partial x} \) provides the slope or rate of change of the function when \( y \) is constant.
• Conversely, \( \frac{\partial f}{\partial y} \) tells us how the function changes as \( x \) remains fixed.

To find the critical points of our function, we set these partial derivatives to zero, which leads us to the equations \( 3x^2 - 3 = 0 \) and \( 3y^2 - 3 = 0 \). These equations help identify the values of \( x \) and \( y \) that make the function stationary, which are potential candidates for local extrema or saddle points.

Hessian Matrix

The Hessian matrix is a square matrix that holds all the second-order partial derivatives of a function. In our problem, it plays a crucial role in determining the nature of critical points. For a function \( f(x, y) \), the Hessian matrix \( H \) is given by: \[ H = \begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \ \frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2} \end{bmatrix} \] In our example, we calculated:

• \( \frac{\partial^2 f}{\partial x^2} = 6x \)
• \( \frac{\partial^2 f}{\partial y^2} = 6y \)
• \( \frac{\partial^2 f}{\partial x \partial y} = 0 \)

The Hessian matrix assists in determining whether a point is a maximum, minimum, or saddle point by evaluating its determinant.

Second Derivative Test

The second derivative test is used for classifying critical points by analyzing the sign of the Hessian determinant. This involves the formula: \[ H = \left(\frac{\partial^2 f}{\partial x^2} \right)\left(\frac{\partial^2 f}{\partial y^2} \right) - \left( \frac{\partial^2 f}{\partial x \partial y}\right)^2 \] In this function, since \( \frac{\partial^2 f}{\partial x \partial y} = 0 \), the Hessian \( H = 36xy \). The second derivative test applies as follows:

• If \( H > 0 \) and \( \frac{\partial^2 f}{\partial x^2} > 0 \), the point is a local minimum.
• If \( H > 0 \) and \( \frac{\partial^2 f}{\partial x^2} < 0 \), the point is a local maximum.
• If \( H < 0 \), the point is a saddle point.
• If \( H = 0 \), the test is inconclusive, and further analysis is required.

This method helps delineate critical points efficiently.

Local Maximum and Minimum

After computing the second derivative test, we categorize critical points as local maxima or minima based on the results. A local maximum occurs at a point where the function value is higher than its neighboring points. Conversely, a local minimum is where the function value is smaller than its surrounding values. In our function:

• At the critical point \((1, 1)\), we find \( H = 36 \) and \( \frac{\partial^2 f}{\partial x^2} > 0 \). This confirms a local minimum.
• Conversely, at \((-1, -1)\), with \( H = 36 \) and \( \frac{\partial^2 f}{\partial x^2} < 0 \), this point is classified as a local maximum.

Identifying these points helps understand the behavior of the function at and around the critical points.

Saddle Point

A saddle point occurs where a function changes direction but isn't a local extremum, resembling a saddle shape on the graph. These points often occur where there is a value that isn't wholly greater or lesser than neighboring points. For the problem \( f(x, y) = x^3 - 3x + y^3 - 3y \):

• At the points \((-1, 1)\) and \((1, -1)\), the Hessian \( H = -36 \), indicating a saddle point since the determinant is negative.

Saddle points can provide insight into optimizing functions, highlighting where the surface curves upwards in one direction and downwards in another.