Problem 17 - Calculus Early Transcendentals

Question

Find all critical points of the following functions. $$f(x, y)=x^{2}+6 x+y^{2}+8$$

Step-by-Step Solution

Answer

Answer: (-3, 0)

Step 1: Compute the partial derivatives

First, we need to compute the partial derivatives with respect to x and y. $$ \frac{\partial f}{\partial x} = 2x + 6 \\ \frac{\partial f}{\partial y} = 2y $$

Step 2: Set the partial derivatives equal to zero

Next, we will set each of the partial derivatives equal to zero and solve the resulting system of equations for x and y. $$ 2x + 6 = 0 \\ 2y = 0 $$

Step 3: Solve the system of equations

Solving the first equation for x, we get: $$ x = -3 $$ Solving the second equation for y, we get: $$ y = 0 $$

Step 4: Identify the critical points

Since the system of equations has a unique solution, we have a single critical point which is: $$ (x, y) = (-3, 0) $$ Thus, the only critical point of the function \(f(x, y) = x^2 + 6x + y^2 + 8\) is at (-3, 0).

Key Concepts

Partial Derivatives

In mathematical analysis, particularly in dealing with functions of multiple variables, we need to use partial derivatives to understand how the function behaves in relation to each variable. A partial derivative focuses on how the function changes as you tweak one variable, keeping others constant.

For the function given by \( f(x, y) = x^2 + 6x + y^2 + 8 \), computing the partial derivative with respect to \( x \) means focusing on how changes in \( x \) affect \( f \), while treating \( y \) as a constant. Here, the partial derivative with respect to \( x \) is expressed as \( \frac{\partial f}{\partial x} = 2x + 6 \).

Similarly, the partial derivative with respect to \( y \) focuses on how changes in \( y \) affect the function, holding \( x \) constant. In this exercise, the partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} = 2y \).

Using partial derivatives is crucial for finding critical points because they help determine points where the function doesn't increase or decrease in a particular direction. This sets the stage for exploring important aspects like maxima, minima, or saddle points.

System of Equations

Once partial derivatives are calculated, they can be combined to form a system of equations. This means setting each of the partial derivatives to zero and solving them simultaneously. The reason we set them to zero is that critical points, where the function does not change, occur where these derivatives are zero.

In our example, setting the partial derivatives equal to zero gives us the system of equations:

• \( 2x + 6 = 0 \)
• \( 2y = 0 \)

The task then becomes finding values of \( x \) and \( y \) that satisfy both equations. By solving the first equation, we find \( x = -3 \). From the second equation, we find \( y = 0 \).

Solving such a system effectively finds the point where the gradient, the vector of partial derivatives, is zero. This point is known as a critical point, which can indicate possible peaks or troughs in the function's surface.

Functions of Two Variables

Functions of two variables, like \( f(x, y) = x^2 + 6x + y^2 + 8 \), are an extension of single-variable functions into two dimensions. They can be visualized as surfaces in three-dimensional space, with \( x \) and \( y \) being the inputs and \( f(x, y) \) the output or height above the \( xy \)-plane.

Understanding functions of two variables involves analyzing how changes in \( x \) and \( y \) simultaneously affect the behavior of \( f \). This requires looking not just at single paths, but the whole landscape described by the function.

Critical points for these functions, therefore, correspond to points on the surface where the slope in every direction is zero. These points can represent local maxima or minima, or modes of different behavior like saddle points, where directions differ in how the function behaves.

Exploring functions of two variables is vital for areas such as optimization, where these principles can help in finding the best solutions or understanding the system's limitations.