Problem 12 - Calculus with Concepts in Calculus

Question

Find all critical points. Determine whether each critical point yields a relative maximum value, a relative minimum value, or a saddle point. $$ f(x, y)=\frac{1}{x}+\frac{1}{y}+x y $$

Step-by-Step Solution

Answer

(1, 1) is a relative minimum; (-1, 1) is a saddle point.

Step 1: Find Partial Derivatives

To find the critical points, we first need the partial derivatives of the function with respect to \( x \) and \( y \). For \( f(x, y) = \frac{1}{x} + \frac{1}{y} + xy \):\[ f_x = -\frac{1}{x^2} + y \] and \[ f_y = -\frac{1}{y^2} + x. \]

Step 2: Set Partial Derivatives to Zero

Set each partial derivative to zero and solve simultaneously: \[ -\frac{1}{x^2} + y = 0 \] and \[ -\frac{1}{y^2} + x = 0 \]. This leads to the system \[ y = \frac{1}{x^2} \] and \[ x = \frac{1}{y^2} \].

Step 3: Solve the System of Equations

Substitute \( y = \frac{1}{x^2} \) into \( x = \frac{1}{y^2} \):\[ x = \frac{1}{(\frac{1}{x^2})^2} \Rightarrow x = x^4. \] Solve \( x = x^4 \), yielding \( x = 1 \) or \( x = -1 \) as potential solutions.

Step 4: Find Corresponding \( y \) Values

For \( x = 1 \), \( y = \frac{1}{1^2} = 1 \). So, one critical point is \((1, 1)\). For \( x = -1 \), \( y = \frac{1}{(-1)^2} = 1 \). Therefore, another critical point is \((-1, 1)\).

Step 5: Use Second Derivative Test

Calculate second derivatives: \[ f_{xx} = \frac{2}{x^3}, \quad f_{yy} = \frac{2}{y^3}, \quad f_{xy} = 1. \] Use the determinant of the Hessian matrix, \( D = f_{xx}f_{yy} - (f_{xy})^2 \), to classify points.

Step 6: Evaluate the Hessian Determinant

For \( (1, 1) \), \[ f_{xx} = 2, \quad f_{yy} = 2, \quad f_{xy} = 1. \] Thus, \( D = (2)(2) - (1)^2 = 4 - 1 = 3 \), indicating a minimum because \( D > 0 \) and \( f_{xx} > 0 \). For \((-1, 1)\), \( D \) results similarly; however, this implies a saddle point due to inconsistency across \((-x, x)\).

Step 7: Conclusion of Critical Points

Therefore, \( (1, 1) \) is a relative minimum, and \( (-1, 1) \) is a saddle point.

Key Concepts

Multivariable Calculus

Multivariable calculus extends the concepts of single-variable calculus to functions of several variables. It is essential for exploring complex relationships where multiple variables interact. In this realm, we encounter functions like \( f(x, y) \), which depend on two or more independent variables. This branch of calculus enables you to examine surfaces and curves within higher-dimensional spaces.
Understanding multivariable calculus is crucial for numerous applications, like modeling real-world phenomena in physics, engineering, and economics. Calculating the rate of change and optimizing functions are prominent tasks within this framework, helping us determine maximum, minimum, or saddle points through critical points analysis. The task of analyzing the function \( f(x, y)=\frac{1}{x}+\frac{1}{y}+x y \) is a great way to delve into such concepts, as it illustrates the interaction between multiple variables and how these interactions can influence change at different points.

Partial Derivatives

Partial derivatives help us understand how a function changes as one variable is varied while the others are held constant. By focusing on one variable at a time, partial derivatives provide insight into a function's behavior from multiple angles. This is especially useful when dealing with functions of several variables, such as \( f(x, y) = \frac{1}{x} + \frac{1}{y} + xy \).
To find critical points, we calculate the partial derivatives \( f_x \) and \( f_y \). These derivatives reveal the rate of change with respect to \( x \) and \( y \) individually:

• \( f_x = -\frac{1}{x^2} + y \)
• \( f_y = -\frac{1}{y^2} + x \)

By setting these equations to zero, we can locate critical points—a vital step in understanding where a function might have extrema or saddle points.

Hessian Matrix

The Hessian matrix is a fundamental tool in multivariable calculus for determining the nature of critical points. It is a square matrix of second-order partial derivatives that provides a thorough view of a function's curvature at a specific point.
For a function \( f(x, y) \), the Hessian matrix \( H \) is given by:\[ H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{xy} & f_{yy} \end{bmatrix} \] where \( f_{xx} \), \( f_{yy} \), and \( f_{xy} \) are second-order partial derivatives.
In this exercise, they are calculated as:

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

With the Hessian matrix set up, you can use it in the second derivative test to classify critical points effectively.

Second Derivative Test

The second derivative test is a key method for evaluating critical points in multivariable calculus. It relies on the Hessian matrix to assess whether a critical point is a relative minimum, maximum, or a saddle point.
The determinant of the Hessian matrix, denoted as \( D \), is calculated by:\[ D = f_{xx}f_{yy} - (f_{xy})^2 \]This determinant helps classify the critical points based on the following criteria:

• If \( D > 0 \) and \( f_{xx} > 0 \), the point is a relative minimum.
• If \( D > 0 \) and \( f_{xx} < 0 \), the point is a relative maximum.
• If \( D < 0 \), the point is a saddle point.
• If \( D = 0 \), the test is inconclusive.

In our example, the point \((1, 1)\) is classified as a relative minimum because the Hessian's determinant is positive and indicates a positive curvature. Meanwhile, \((-1, 1)\) turns out to be a saddle point, showing a difference in behavior.