Problem 12 - Calculus Early Transcendentals

Question

Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. $$f(x, y)=x y+\frac{1}{x}+\frac{1}{y}$$

Step-by-Step Solution

Answer

Local minimum at (1, 1); local maximum at (-1, -1). No saddle points.

Step 1: Find the partial derivatives

To find the local extreme points, we first need to compute the partial derivatives of the function \(f(x, y) = xy + \frac{1}{x} + \frac{1}{y}\). The partial derivative with respect to \(x\) is \(f_x = y - \frac{1}{x^2}\), and with respect to \(y\), it is \(f_y = x - \frac{1}{y^2}\).

Step 2: Set partial derivatives to zero

To find critical points, we set these partial derivatives equal to zero. So, we solve \(y - \frac{1}{x^2} = 0\) and \(x - \frac{1}{y^2} = 0\) to find the values of \(x\) and \(y\) that make both derivatives zero.

Step 3: Solve for critical points

From \(y = \frac{1}{x^2}\), substitute into the second equation: \(x = x^4\). Solving \(x = x^4\) gives the solutions \(x = 1\) and \(x = -1\). For \(x = 1\), \(y = 1\), and for \(x = -1\), \(y = -1\). Thus, the critical points are \((1, 1)\) and \((-1, -1)\).

Step 4: Compute second derivatives

Compute the second partial derivatives: \(f_{xx} = \frac{2}{x^3}\), \(f_{yy} = \frac{2}{y^3}\), and \(f_{xy} = 1\). These second derivatives will be used in testing the critical points.

Step 5: Use the second derivative test

Apply the second derivative test at each critical point using the determinant \(D = f_{xx} f_{yy} - (f_{xy})^2\). Evaluate at \((1, 1)\) and \((-1, -1)\). At \((1, 1)\), \(D = (2)(2) - (1)^2 = 3\). Since \(D > 0\) and \(f_{xx} > 0\), \((1, 1)\) is a local minimum. At \((-1, -1)\), \(D = (-2)(-2) - (1)^2 = 3\). Since \(D > 0\) and \(f_{xx} < 0\), \((-1, -1)\) is a local maximum.

Step 6: Conclusion

The function has a local minimum at \((1, 1)\) and a local maximum at \((-1, -1)\). There are no saddle points since all critical points are either local maxima or minima.

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in the calculus of several variables. When dealing with functions of more than one variable, a partial derivative measures how the function changes as one particular variable changes while holding the other variables constant.
For the function given in the exercise, \(f(x, y) = xy + \frac{1}{x} + \frac{1}{y}\), we compute the partial derivatives with respect to \(x\) and \(y\):

• Partial derivative with respect to \(x\) (\(f_x\)): This indicates how the function \(f\) changes as \(x\) varies, which is given by \(f_x = y - \frac{1}{x^2}\).
• Partial derivative with respect to \(y\) (\(f_y\)): This shows how the function \(f\) changes as \(y\) varies, i.e., \(f_y = x - \frac{1}{y^2}\).

Calculating these partial derivatives is crucial for finding where potential local extrema or saddle points might occur.

Critical Points

Critical points occur where the partial derivatives of a function are zero or undefined. These points are candidates for local maxima, minima, or saddle points.
By setting \(f_x = 0\) and \(f_y = 0\), we are tasked with solving the system of equations:

• \(y - \frac{1}{x^2} = 0\)
• \(x - \frac{1}{y^2} = 0\)

These equations provide the potential critical points, which in the exercise were found to be \((1, 1)\) and \((-1, -1)\). This step is essential as these points will be used further to determine the nature of the extremum through additional tests like the Second Derivative Test.

Second Derivative Test

The Second Derivative Test is a method used to classify critical points as local maxima, minima, or saddle points. It involves calculating the second partial derivatives of the function and evaluating a determinant formed by these derivatives.
For the function in the exercise, the second derivatives are:

• \(f_{xx} = \frac{2}{x^3}\)
• \(f_{yy} = \frac{2}{y^3}\)
• \(f_{xy} = 1\)

The determinant \(D\) used in this test is given by:\[ D = f_{xx} f_{yy} - (f_{xy})^2 \]Evaluating \(D\) at the critical points helps decide:

• If \(D > 0\) and \(f_{xx} > 0\), it's a local minimum.
• If \(D > 0\) and \(f_{xx} < 0\), it's a local maximum.
• If \(D < 0\), it's a saddle point.

In the exercise, applying this test indicated \((1, 1)\) as a local minimum and \((-1, -1)\) as a local maximum.

Local Maximum and Minimum Values

Local maximum and minimum values are specific points in a function's domain where the function reaches a peak (maximum) or a trough (minimum), respectively, relative to nearby points.
In analyzing functions of two variables, once the critical points have been located and classified using the Second Derivative Test, one can determine these local extrema:

• A local maximum occurs at a point if the function's value there is higher than nearby values within a small neighborhood.
• A local minimum is where the function's value is lower than neighboring values.

In the solution, the critical point \((1, 1)\) was determined to be a local minimum, while \((-1, -1)\) was a local maximum, indicating the function's behavior in these regions. Understanding these points helps in visualizing the overall shape and behavior of the function in its domain.