Problem 9 - Calculus Early Transcendentals

Question

Find all critical points of the following functions. $$f(x, y)=1+x^{2}+y^{2}$$

Step-by-Step Solution

Answer

Answer: The function f(x, y) has one critical point at (0, 0), and it is a local minimum.

Step 1: Find the partial derivatives

Compute the partial derivatives of the function f(x, y) with respect to x and y: For x: $$\frac{\partial f}{\partial x} = 2x$$ For y: $$\frac{\partial f}{\partial y} = 2y$$

Step 2: Set the partial derivatives equal to zero

Set the partial derivatives equal to zero and solve the resulting system of equations: $$2x = 0$$ $$2y = 0$$

Step 3: Solve the system of equations

The system of equations is quite simple, we can solve it directly: For x: $$x = 0$$ For y: $$y = 0$$ So, the critical point is \((0, 0)\).

Step 4: Classify the critical point

To classify the critical point, we can compute the second partial derivatives, and use the second partial derivative test: $$\frac{\partial^2 f}{\partial x^2} = 2$$ $$\frac{\partial^2 f}{\partial y^2} = 2$$ $$\frac{\partial^2 f}{\partial x \partial y} = 0$$ Now, we calculate the determinant of the Hessian matrix (H): $$D = \begin{vmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \\ \frac{\partial^2 f}{\partial x \partial y} & \frac{\partial^2 f}{\partial y^2} \end{vmatrix} = \begin{vmatrix} 2 & 0 \\ 0 & 2 \end{vmatrix} = (2)(2) - (0)(0) = 4$$ Since D > 0 and $$\frac{\partial^2 f}{\partial x^2} > 0$$, the critical point \((0, 0)\) is a local minimum. Therefore, the function f(x, y) has one critical point at \((0, 0)\), and it is a local minimum.

Key Concepts

Partial Derivatives

When working with a function of two variables, like the function\( f(x, y) = 1 + x^2 + y^2 \), we use partial derivatives to study how the function changes as each variable is changed individually, while keeping the other constant.
Think of it as exploring the slope of the function along the axis of either variable.

• To find the partial derivative with respect to \( x \), treat \( y \) as a constant and differentiate: \( \frac{\partial f}{\partial x} = 2x \).
• For \( y \), treat \( x \) as a constant: \( \frac{\partial f}{\partial y} = 2y \).

An essential step in finding critical points is to set these partial derivatives equal to zero, leading to equations that allow us to locate points where the function does not change along the axis when moving slightly in a neighborhood.

Second Partial Derivative Test

The second partial derivative test helps us determine the nature of critical points found from partial derivatives.
Once we have calculated the critical point—in this case, \((0, 0)\)—we use the second derivatives of the function to classify the point. This involves calculating the second partial derivatives:

• \( \frac{\partial^2 f}{\partial x^2} = 2 \)
• \( \frac{\partial^2 f}{\partial y^2} = 2 \)
• \( \frac{\partial^2 f}{\partial x \partial y} = 0 \)

These second derivatives help assess whether a point is a maximum, a minimum, or a saddle point.
Using them, we will move to evaluate the Hessian matrix.

Hessian Matrix

The Hessian matrix compiles the second partial derivatives into a small matrix for convenient assessment of critical points.
For a function \( f(x, y) \), the Hessian matrix \( H \) is structured as:\[ H = \begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \ \frac{\partial^2 f}{\partial x \partial y} & \frac{\partial^2 f}{\partial y^2} \end{bmatrix} \]In our example:\[ H = \begin{bmatrix} 2 & 0 \ 0 & 2 \end{bmatrix} \]
The sign of its determinant, \( D \), helps determine the type of critical point:- \( D = 4 \) (positive) suggests a minimum or maximum.If both the determinant \( D \) is positive and the second partial derivative in \( x \), \( \frac{\partial^2 f}{\partial x^2} \), is positive, the point is a local minimum, as seen here.

Local Minimum

In the context of multivariable calculus, a local minimum is a point where the function's value is lower than all nearby points. To identify a local minimum at a critical point, use the Hessian matrix and second derivatives.
For our function \( f(x, y) \), the Hessian determinant \( D \) is positive, and the second partial derivative \( \frac{\partial^2 f}{\partial x^2} = 2 \) is also positive. Thus, the point \((0, 0)\) qualifies as a local minimum.
This method of determining a local minimum via the Hessian is efficient for multivariable functions, ensuring conclusions about the nature of critical points rely on the function's behavior in its immediate vicinity.