Problem 24 - Calculus and its applications

Question

Find the relative maximum and minimum values as well as any saddle points. $$ f(x, y)=x y+\frac{2}{x}+\frac{4}{y} $$

Step-by-Step Solution

Answer

The function has a local minimum at the point (2, 0.5).

Step 1: Find First Partial Derivatives

To find critical points, we first need the partial derivatives of \( f(x,y) \). Calculate \( f_x = \frac{\partial f}{\partial x} \) and \( f_y = \frac{\partial f}{\partial y} \). These are given by:\[f_x = y - \frac{2}{x^2}, \quad f_y = x - \frac{4}{y^2}\]

Step 2: Find Critical Points

Critical points occur where \( f_x = 0 \) and \( f_y = 0 \). Solve these equations simultaneously:1. Set \( y - \frac{2}{x^2} = 0 \) so \( y = \frac{2}{x^2} \).2. Set \( x - \frac{4}{y^2} = 0 \) so \( y^2 = \frac{4}{x} \). Solve both to find \( x = 2 \) and \( y = 0.5 \). Thus, the critical point is \((2, 0.5)\).

Step 3: Find Second Partial Derivatives

Next, find the second partial derivatives to classify the critical point using the Hessian determinant method.\[f_{xx} = \frac{4}{x^3}, \quad f_{yy} = \frac{8}{y^3}, \quad f_{xy} = 1\]

Step 4: Evaluate Hessian Determinant

Hessian \( H \) is calculated as follows:\[ H = f_{xx} f_{yy} - (f_{xy})^2 \]Evaluating at \( (2, 0.5) \):\[ f_{xx} = \frac{4}{8} = 0.5, \quad f_{yy} = \frac{8}{0.125} = 64, \quad f_{xy} = 1\]\[ H = (0.5)(64) - 1^2 = 32 - 1 = 31\]

Step 5: Classify the Critical Point

Since \( H > 0 \) and \( f_{xx} > 0 \) at the critical point \( (2, 0.5) \), the Hessian indicates that this point is a local minimum.

Key Concepts

Partial Derivatives

To understand functions of several variables, partial derivatives play a crucial role. Partial derivatives help us measure how a function changes as each variable is varied independently. When dealing with multivariable calculus, we often have functions such as \( f(x, y) \), which depends on both \( x \) and \( y \).

Partial derivatives, denoted \( f_x \) and \( f_y \), represent the derivative of the function concerning one variable while keeping the other variable constant.

• For example, \( f_x = \frac{\partial f}{\partial x} \) gives us the rate of change of \( f \) with respect to \( x \), holding \( y \) constant.
• Similarly, \( f_y = \frac{\partial f}{\partial y} \) is the rate of change of \( f \) with respect to \( y \), keeping \( x \) constant.

For the function \( f(x, y) = xy + \frac{2}{x} + \frac{4}{y} \), the partial derivatives are:
\[ f_x = y - \frac{2}{x^2}, \quad f_y = x - \frac{4}{y^2} \]These expressions will be set to zero in the next steps to find the critical points.

Critical Points

In multivariable calculus, critical points are the coordinates where the function has the potential to change from increasing to decreasing or vice-versa. To locate these points, we solve the system of equations obtained by setting all first partial derivatives to zero.

Given \( f_x = y - \frac{2}{x^2} \) and \( f_y = x - \frac{4}{y^2} \), we find the critical points by:

• Solving \( y - \frac{2}{x^2} = 0 \) results in \( y = \frac{2}{x^2} \).
• Solving \( x - \frac{4}{y^2} = 0 \) results in \( y^2 = \frac{4}{x} \).

Solving these results provides the critical point \((x, y) = (2, 0.5)\). These are the coordinates where we need to check the nature of extremas.

Hessian Determinant

The Hessian determinant is a crucial tool in multivariable calculus for classifying critical points. It is constructed using the second partial derivatives of a function. The Hessian \( H \) for a function \( f \) with two variables \( x \) and \( y \) is given by:
\[ H = f_{xx} f_{yy} - (f_{xy})^2 \]For the function \( f(x, y) \), the second partial derivatives are:

• \( f_{xx} = \frac{4}{x^3} \)
• \( f_{yy} = \frac{8}{y^3} \)
• \( f_{xy} = 1 \)

Evaluating at the critical point \((2, 0.5)\), we compute \( f_{xx} = 0.5 \), \( f_{yy} = 64 \), and \( f_{xy} = 1 \). Substituting these values yields:
\[ H = (0.5)(64) - (1)^2 = 32 - 1 = 31 \]
A positive Hessian determinant indicates that the critical point is either a local minimum or maximum, depending on the sign of \( f_{xx} \).

Local Extrema

Local extrema are points on the graph of a function where the function achieves locally maximum or minimum values. To classify whether a critical point is a local minimum, a local maximum, or a saddle point, we can use the Hessian determinant and the second partial derivative \( f_{xx} \).

Here's how it works:

• If the Hessian \( H > 0 \) and \( f_{xx} > 0 \), the critical point is a local minimum.
• If \( H > 0 \) and \( f_{xx} < 0 \), the critical point is a local maximum.
• If \( H < 0 \), the critical point is a saddle point.

For our problem, at \( (2, 0.5) \), since \( H = 31 > 0 \) and \( f_{xx} = 0.5 > 0 \), \( (2, 0.5) \) is a local minimum. Saddle point indications occur when the function curves upward in one direction and downward in another, resulting in \( H < 0 \), which does not happen here.