Problem 13 - Applied Calculus

Question

Find all the critical points and determine whether each is a local maximum, local minimum, or neither. $$ f(x, y)=x^{3}+y^{3}-6 y^{2}-3 x+9 $$

Step-by-Step Solution

Answer

Local minimum at (1, 4), local maximum at (-1, 0). Saddle points at (1, 0) and (-1, 4).

Step 1: Find the Partial Derivatives

First, find the first partial derivatives of the function with respect to both variables, \( x \) and \( y \). The partial derivative with respect to \( x \) is \( f_x = \frac{\partial}{\partial x}(x^3 + y^3 - 6y^2 - 3x + 9) = 3x^2 - 3 \). The partial derivative with respect to \( y \) is \( f_y = \frac{\partial}{\partial y}(x^3 + y^3 - 6y^2 - 3x + 9) = 3y^2 - 12y \).

Step 2: Set the Partial Derivatives to Zero

Set both partial derivatives equal to zero to find the critical points. This gives two equations: \( 3x^2 - 3 = 0 \) and \( 3y^2 - 12y = 0 \).

Step 3: Solve for Critical Points

Solve each equation individually. Starting with \( 3x^2 - 3 = 0 \), we have \( x^2 = 1 \), so \( x = 1 \) or \( x = -1 \). For \( 3y^2 - 12y = 0 \), factor to get \( 3y(y - 4) = 0 \). The solutions are \( y = 0 \) or \( y = 4 \). Thus, the critical points are \( (1, 0), (1, 4), (-1, 0), \) and \( (-1, 4) \).

Step 4: Use the Second Derivative Test

Find the second partial derivatives: \( f_{xx} = 6x \), \( f_{yy} = 6y - 12 \), and \( f_{xy} = 0 \). Calculate the Hessian determinant, \( H = f_{xx}f_{yy} - (f_{xy})^2 = 6x(6y - 12) \). Evaluate \( H \) and \( f_{xx} \) at each critical point.

Step 5: Determine the Nature of Each Critical Point

For the point \( (1, 0) \), \( H = 6(1)(-12) = -72 \), indicating a saddle point (neither). For \( (1, 4) \), \( H = 6 \times 12 \times 1 = 72 \) and \( f_{xx} = 6 \), implying a local minimum. For \( (-1, 0) \), \( H = 6(-1)(-12) = 72 \) and \( f_{xx} = -6 \), indicating a local maximum. For \( (-1, 4) \), \( H = -72 \), indicating another saddle point (neither).

Key Concepts

Partial Derivatives

When dealing with functions of multiple variables, understanding partial derivatives is a crucial step. In our given function, \( f(x, y) = x^3 + y^3 - 6y^2 - 3x + 9 \), we need to determine how the function changes as each variable changes, with the other held constant. For this purpose, we compute partial derivatives.

• The partial derivative with respect to \( x \), written as \( \frac{\partial f}{\partial x} \) or \( f_x \), focuses on the rate of change along the \( x \) direction while keeping \( y \) constant.
• Similarly, the partial derivative with respect to \( y \), \( \frac{\partial f}{\partial y} \) or \( f_y \), focuses on the rate of change along the \( y \) direction while keeping \( x \) constant.

By setting these partial derivatives to zero, we can locate critical points where the function's gradient is zero, indicating potential maxima, minima, or saddle points.

Second Derivative Test

The second derivative test helps us determine the nature of critical points in multivariable functions. After finding the critical points, we calculate second partial derivatives to apply this test.We look at:

• \( f_{xx} \): the second derivative with respect to \( x \).
• \( f_{yy} \): the second derivative with respect to \( y \).
• \( f_{xy} \): the mixed partial derivative first taken with respect to \( x \) and then \( y \).

These derivatives help construct the Hessian determinant \( H \), which is crucial for the second derivative test. Evaluating \( H \) and \( f_{xx} \) at each critical point informs us if the point is a local max, a local min, or a saddle point. Local minima and maxima can be identified if \( H > 0 \), depending on the sign of \( f_{xx} \). If \( H < 0 \), the point is a saddle point.

Hessian Determinant

The Hessian determinant is a mathematical tool used to evaluate critical points more thoroughly using the second derivatives of a multivariable function. It helps in understanding the curvature of the function at those points.The determinant is calculated as:\[ H = f_{xx}f_{yy} - (f_{xy})^2 \]

• When \( H > 0 \), the critical point could either be a local minimum or local maximum, depending on the sign of \( f_{xx} \).
• If \( H < 0 \), the critical point is a saddle point, which means it's neither a local max nor a local min.
• If \( H = 0 \), the test is inconclusive, and other methods must be used.

By analyzing the Hessian, we are ultimately able to classify the behavior of the function around critical points, providing insights into the function's surface shape at those places.

Local Maxima and Minima

Identifying local maxima and minima involves checking where a function has peaks and troughs in a particular region. After applying the second derivative test using the Hessian determinant, you can determine whether a critical point is a local extremum.

• A **local maximum** occurs at a point if the function value at that point is higher than all nearby points. If \( H > 0 \) and \( f_{xx} < 0 \), then we have a local maximum.
• A **local minimum** takes place at a point where the function value is less than those around it. If \( H > 0 \) and \( f_{xx} > 0 \), then it's a local minimum.
• Saddle points are neither maxima nor minima but points where the function changes direction.

Local maxima and minima are valuable in optimization, helping identify the best or worst values that a function can achieve within a given region.