Problem 324 - 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. $$f(x, y)=x^{3}+y^{3}-300 x-75 y-3$$

Step-by-Step Solution

Answer

(10, 5) and (-10, -5) are minima; (10, -5) and (-10, 5) are saddle points.

Step 1: Find the First Derivatives

First, we need to find the partial derivatives of the function to find the critical points. The first derivative with respect to x is \( f_x = \frac{\partial f}{\partial x} = 3x^2 - 300 \). The first derivative with respect to y is \( f_y = \frac{\partial f}{\partial y} = 3y^2 - 75 \).

Step 2: Solve for Critical Points

Set the first derivatives to zero to find the critical points. For \( f_x = 3x^2 - 300 = 0 \), solving gives \( x^2 = 100 \), so \( x = \pm 10 \). For \( f_y = 3y^2 - 75 = 0 \), solving gives \( y^2 = 25 \), so \( y = \pm 5 \). Therefore, the critical points are \((10, 5), (10, -5), (-10, 5), (-10, -5)\).

Step 3: Find the Second Derivatives

We need the second derivatives for the second derivative test. Compute \( f_{xx} = \frac{\partial^2 f}{\partial x^2} = 6x \) and \( f_{yy} = \frac{\partial^2 f}{\partial y^2} = 6y \). Since \( f_xy = f_yx = 0 \), these mixed partial derivatives are zero.

Step 4: Apply the Second Derivative Test

At each critical point \((x_0, y_0)\), evaluate \(D = f_{xx}(x_0, y_0) \cdot f_{yy}(x_0, y_0) - (f_{xy}(x_0, y_0))^2\). For critical points (10, 5), (-10, -5): \(D = (6x)(6y) = 360 > 0\), and \(f_{xx}(x_0, y_0) = 60\), so they are minima. For critical points (10, -5), (-10, 5): \(D = -360 < 0\), so they are saddle points.

Key Concepts

Critical Points

Critical points in calculus are where the slope of the function is zero or undefined, often indicating a potential maximum, minimum, or point of inflection. To find critical points for a multivariable function, we begin by finding the partial derivatives with respect to each variable. This involves taking the derivative of the function with respect to one variable while holding the other variables constant.
For the function given, we found the partial derivatives and set them to zero. This step is crucial as setting the first derivatives to zero identifies points where the function’s slope is flat, indicating potential extremes.

• The critical points obtained from the function are (10, 5), (10, -5), (-10, 5), and (-10, -5).

Partial Derivatives

Partial derivatives are derivatives of functions with multiple variables, taken with respect to one variable at a time. In our exercise, we have a function of two variables, \(x\) and \(y\).
The partial derivative with respect to \(x\), denoted as \(f_x\), involves differentiating the function holding \(y\) constant. Similarly, the partial derivative with respect to \(y\), \(f_y\), involves differentiating while holding \(x\) constant. The calculations provide:

• \(f_x = 3x^2 - 300\)
• \(f_y = 3y^2 - 75\)

Partial derivatives help us determine how the function changes along each direction independently.

Second Derivatives

Second derivatives are crucial for assessing the concavity of a function and are used in the second derivative test to classify critical points. For a multivariable function, we compute second partial derivatives:

• \(f_{xx}\) is found by differentiating \(f_x\) with respect to \(x\).
• \(f_{yy}\) is obtained by differentiating \(f_y\) with respect to \(y\).

Additionally, we consider mixed partial derivatives, \(f_{xy}\) and \(f_{yx}\), which are zero in our problem. Calculations give:

• \(f_{xx} = 6x\)
• \(f_{yy} = 6y\)

These derivatives form the Hessian matrix, essential for the next step, the second derivative test.

Saddle Points

Saddle points occur when a function resembles a saddle shape, curving upwards in one direction and downwards in another. At a saddle point, the second derivative test evaluates to a negative determinant, \(D < 0\).
This result indicates that the critical point is not a local extremum but rather a point where the function does not have a maximum or minimum.

• In our case, the points (10, -5) and (-10, 5) are saddle points, having determinants \(-360\).

This mixed behavior makes these points neither purely concave nor convex.

Minima

A minimum point is where the function takes on the smallest value locally. It is either a trough or a dip in the surface if considered in three dimensions. To determine whether a critical point is a minimum for our function, we examine the value of the second derivative test determinant, \(D\), and the sign of \(f_{xx}\):

• If \(D > 0\) and \(f_{xx} > 0\), as it is here, the point is a local minimum.

In the given exercise, critical points (10, 5) and (-10, -5) result in \(D = 360\) and \(f_{xx} = 60\); hence, they are minima. These calculations confirm these critical points are troughs where the function value is lower than its immediate surroundings.