Problem 21 - Calculus Early Transcendentals

Question

Find the critical points of the following functions. Use the Second Derivative Test to determine (if possible) whether each critical point corresponds to a local maximum, local minimum, or saddle point. Confirm your results using a graphing utility. $$f(x, y)=-4 x^{2}+8 y^{2}-3$$

Step-by-Step Solution

Answer

Question: Determine the critical points and whether they are a local maximum, local minimum, or saddle point for the function f(x, y) = -4x^2 + 8y^2. Answer: The function f(x, y) has a single critical point at (0, 0) which is a saddle point.

Step 1: Compute the gradient of the function

First, we will find the gradient of the function f(x, y) by calculating the partial derivatives with respect to x and y. The partial derivatives are given by: $$ \frac{\partial f}{\partial x} = -8x \quad \text{ and } \quad \frac{\partial f}{\partial y} = 16y $$

Step 2: Find the critical points

Next, we will find the critical points by solving the following system of equations: $$ \frac{\partial f}{\partial x} = 0 \quad \text{ and } \quad \frac{\partial f}{\partial y} = 0 $$ This leads to the system: $$ -8x = 0 \\ 16y = 0 $$ Solving this system of equations, we find that the single critical point is: $$ (x, y) = (0, 0) $$

Step 3: Apply the Second Derivative Test

To apply the Second Derivative Test, we need to compute the second partial derivatives: $$ \frac{\partial^2 f}{\partial x^2} = -8 \\ \frac{\partial^2 f}{\partial y^2} = 16 \\ \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} = 0 $$ Now, we compute the determinant of the Hessian matrix, which is given by: $$ D(x, y) = \begin{vmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \\ \frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2} \end{vmatrix} $$ At the critical point (0, 0), we have: $$ D(0, 0) = \begin{vmatrix} -8 & 0 \\ 0 & 16 \end{vmatrix} = (-8)(16) - (0)(0) = (-128) $$ Since the determinant is negative, the Second Derivative Test indicates that the critical point (0, 0) is a saddle point.

Step 4: Confirm the result using a graphing utility

Using a graphing utility, we can plot the function f(x, y) and observe that, indeed, the point (0, 0) appears to be a saddle point, confirming the results obtained in our previous steps. In conclusion, the function f(x, y) has a single saddle point at (0, 0).

Key Concepts

Second Derivative Test

The Second Derivative Test is an essential tool in determining the nature of critical points found in multivariable functions. Once you have located a critical point, a point where the gradient of the function is zero, the Second Derivative Test can help decide if that point is a local maximum, local minimum, or a saddle point.

To carry out this test, you need to calculate the second partial derivatives of the function. In this case, for the function \(f(x, y) = -4x^2 + 8y^2 - 3\), the second partial derivatives are:

• \(\frac{\partial^2 f}{\partial x^2} = -8 \)
• \(\frac{\partial^2 f}{\partial y^2} = 16 \)
• \(\frac{\partial^2 f}{\partial x \partial y} = 0 \)

Now, compute the determinant of the Hessian matrix, \(D(x, y)\). At the critical point \((0, 0)\), the determinant is \((-8)(16) - (0)(0) = -128\). Because the determinant is negative, the Second Derivative Test tells us the critical point is a saddle point.

Partial Derivatives

Partial derivatives are used to understand how a function changes as each variable is varied while keeping the others fixed. For a function like \(f(x, y) = -4x^2 + 8y^2 - 3\), this involves taking derivatives with respect to \(x\) and \(y\) separately.

Let's look at this process:

• The partial derivative with respect to \(x\) is found by treating \(y\) as a constant, so \(\frac{\partial f}{\partial x} = -8x\).
• Similarly, the partial derivative with respect to \(y\) is \(\frac{\partial f}{\partial y} = 16y\).

These partial derivatives identify how a tiny change in \(x\) or \(y\) near any point corresponds to a change in the function's output. In this exercise, setting these derivatives to zero helps identify potential critical points, leading us to \((x, y) = (0, 0)\) as the sole critical point.

Hessian Matrix

The Hessian matrix is a key tool in analyzing the curvature of multivariable functions, playing a critical role in the Second Derivative Test. It is a square matrix comprising all second-order partial derivatives of the function.

For the function \(f(x, y) = -4x^2 + 8y^2 - 3\), the Hessian matrix is:\[\begin{bmatrix}\frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \\frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2}\end{bmatrix} = \begin{bmatrix}-8 & 0 \0 & 16\end{bmatrix}\]
The determinant of the Hessian matrix, \(D(x, y)\), is crucial in the Second Derivative Test. For the critical point \((0, 0)\), it is -128, indicating a saddle point due to its negative value. This matrix provides insight into how the surface of the function curves at a given point, helping us classify it as a maximum, minimum, or saddle point.