Problem 11 - Calculus and its applications

Question

Find the relative maximum and minimum values. $$ f(x, y)=4 x^{2}-y^{2} $$

Step-by-Step Solution

Answer

The function has a saddle point at \((0, 0)\), hence no relative maximum or minimum exists.

Step 1: Find the First Partial Derivatives

First, we need to find the first partial derivatives of the function with respect to each variable. For \( f(x, y) = 4x^2 - y^2 \), the partial derivative with respect to \( x \) is calculated by treating \( y \) as a constant: \( \frac{\partial f}{\partial x} = 8x \). The partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} = -2y \).

Step 2: Set Partial Derivatives to Zero

Set the first partial derivatives equal to zero to find critical points. Solve \( 8x = 0 \) which gives \( x = 0 \), and solve \( -2y = 0 \) which gives \( y = 0 \). Thus, the critical point is \( (x, y) = (0, 0) \).

Step 3: Calculate the Second Partial Derivatives

Find the second partial derivatives to apply the second derivative test. Compute \( f_{xx} = 8 \), \( f_{yy} = -2 \), and \( f_{xy} = 0 \).

Step 4: Apply the Second Derivative Test

Use the second derivative test to determine the nature of the critical point. Calculate the Hessian determinant \( D = f_{xx}f_{yy} - (f_{xy})^2 \). Thus, \( D = (8)(-2) - (0)^2 = -16 \). Since \( D < 0 \), the test indicates a saddle point.

Key Concepts

Partial Derivatives

Partial derivatives are a foundational concept in multivariable calculus. They help us study how a function changes with respect to each variable separately. For a function of two variables, like our function \( f(x, y) = 4x^2 - y^2 \), we find the partial derivatives by considering one variable at a time while keeping the other constant.
For example:
1. When calculating \( \frac{\partial f}{\partial x} \), treat \( y \) as a constant. Taking the derivative of \( 4x^2 \) in respect to \( x \) gives \( 8x \).2. Conversely, when finding \( \frac{\partial f}{\partial y} \), consider \( x \) as a constant and differentiate \( -y^2 \), yielding \( -2y \).
These partial derivatives assist in understanding how changes in \( x \) and \( y \) independently affect the function. Calculating partial derivatives is often the initial step in exploring the behavior of a function for optimization or finding critical points.

Critical Points

Critical points are where the first derivatives of a function are zero or undefined, and these provide insight into where a function could have extreme values—like minima, maxima, or saddle points. For our function, \( f(x, y) = 4x^2 - y^2 \), we set the first partial derivatives to zero to find the critical points.

• \( \frac{\partial f}{\partial x} = 8x = 0 \) gives \( x = 0 \).
• \( \frac{\partial f}{\partial y} = -2y = 0 \) gives \( y = 0 \).

Together, these yield a critical point at \((0, 0)\). This critical point acts as a potential location for a relative extremum, which could involve a maximum, minimum, or saddle point. To confirm the nature of these critical points, further investigation using the second derivative test is needed.

Second Derivative Test

The second derivative test is a method used to classify the critical points as local maxima, minima, or saddle points. After identifying the critical points, we carry out the second derivative test by computing the second partial derivatives of the function and analyzing them.
For our specific function, we determined:

• \( f_{xx} = 8 \)
• \( f_{yy} = -2 \)
• \( f_{xy} = 0 \)

These derivatives are used in computing the Hessian determinant \( D = f_{xx}f_{yy} - (f_{xy})^2 \). Calculating \( D \) yields \( -16 \). A negative \( D \) indicates a saddle point at the critical point \((0, 0)\).
The second derivative test is an effective tool in multivariable calculus for understanding the precise nature of these critical points in terms of whether they are locations of relative maxima, minima, or saddle points.