Problem 12 - Calculus and its applications

Question

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

Step-by-Step Solution

Answer

The function has no relative maximum or minimum values; \((0,0)\) is a saddle point.

Step 1: Compute Partial Derivatives

To find the critical points, we need to first calculate the partial derivatives of the function. Compute the partial derivative of the function with respect to \(x\): \( f_x = \frac{\partial}{\partial x}(x^2 - y^2) = 2x \).Now compute the partial derivative with respect to \(y\): \( f_y = \frac{\partial}{\partial y}(x^2 - y^2) = -2y \).

Step 2: Solve for Critical Points

Set each partial derivative equal to zero to find the critical points:\( f_x = 2x = 0 \) gives \( x = 0 \).\( f_y = -2y = 0 \) gives \( y = 0 \).This indicates a critical point at \((x, y) = (0, 0)\).

Step 3: Second Derivative Test

To determine the nature of the critical point, use the second derivative test for functions of two variables:Calculate the second partial derivatives:\( f_{xx} = \frac{\partial^2 f}{\partial x^2} = 2 \),\( f_{yy} = \frac{\partial^2 f}{\partial y^2} = -2 \),\( f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 0 \).Compute the determinant of the Hessian matrix:\( D = f_{xx}f_{yy} - (f_{xy})^2 = (2)(-2) - (0)^2 = -4 \).

Step 4: Analyze the Determinant

Since \( D < 0 \), the critical point \((0, 0)\) is a saddle point. As a saddle point, it is neither a relative maximum nor a relative minimum. Such points indicate that the surface curves upward in one direction and downward in another direction at that point.

Key Concepts

Critical Points

In multivariable calculus, critical points are essential for analyzing the behavior of a function. They indicate where the gradient (or slope) is zero or undefined, and they can reveal potential maximum, minimum, or saddle points. In our exercise, we begin by identifying these key points of the function by setting the partial derivatives equal to zero.

For the function \(f(x, y) = x^2 - y^2\), partial derivatives were calculated:

• Partial derivative with respect to \(x\): \(f_x = 2x\)
• Partial derivative with respect to \(y\): \(f_y = -2y\)

Setting \(f_x = 0\) gives \(x = 0\), and \(f_y = 0\) gives \(y = 0\). This results in a critical point at \((0, 0)\). Critical points are crucial as they are potential candidates for local maximum, minimum, or saddle points. However, further tests, such as the second derivative test, are needed to classify them accurately.

Partial Derivatives

Partial derivatives play a fundamental role in multivariable calculus by showing how a function changes with respect to one variable while keeping the others constant. They tell us the rate at which the function's value is changing in the direction of each axis independently.

In the context of our exercise, we calculated the partial derivatives of \(f(x, y) = x^2 - y^2\):

• For \(x\): \(f_x = \frac{\partial}{\partial x}(x^2 - y^2) = 2x\)
• For \(y\): \(f_y = \frac{\partial}{\partial y}(x^2 - y^2) = -2y\)

These derivatives indicate the slopes of the tangent planes along the \(x\) and \(y\) directions. Solving these equations reveals where the slopes are zero, which are the critical points of the function. Understanding partial derivatives is essential for exploring the behavior of functions with several variables.

Second Derivative Test

The second derivative test in multivariable calculus helps determine the nature of critical points. After finding a critical point, we need to perform this test to classify whether the critical point is a local maximum, local minimum, or saddle point.

For our function, we compute the second partial derivatives:

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

We then calculate the determinant of the Hessian matrix \(D = f_{xx}f_{yy} - (f_{xy})^2 = (2)(-2) - 0^2 = -4\). With \(D < 0\), the critical point \((0, 0)\) is a saddle point.

Saddle points occur when the function curves upward in one direction and downward in another, and as such, are neither maximum nor minimum points. Understanding this test aids us in fully characterizing the behavior of a function around its critical points.