Problem 24 - 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. $$f(x, y)=x y e^{-x-y}$$

Step-by-Step Solution

Answer

Question: Determine the nature of the critical points of the function \(f(x, y) = xy e^{-x-y}\) using the Second Derivative Test. Answer: The critical points of the function are \((0,0)\) and \((\frac{1}{2}, \frac{1}{2})\). The Second Derivative Test is inconclusive for the critical point \((0,0)\), so we cannot determine its nature. The critical point \((\frac{1}{2}, \frac{1}{2})\) is a saddle point.

Step 1: Find first-order partial derivatives

To find the critical points, we need to calculate the first-order partial derivatives respect to x and y and set them to 0: $$ \frac{\partial f}{\partial x} = y e^{-x-y}(1 - x), \quad \frac{\partial f}{\partial y} = x e^{-x-y}(1 - y) $$

Step 2: Set partial derivatives equal to 0

Now, set the first-order partial derivatives equal to 0 to get our critical points: $$ y e^{-x-y}(1 - x) = 0 \quad (1) $$ $$ x e^{-x-y}(1 - y) = 0 \quad (2) $$

Step 3: Find critical points

From equation (1), we have two possibilities: 1. \(y=0\) 2. \(1-x=y\) Similarly, from equation (2), we have two possibilities as well: 1. \(x=0\) 2. \(1-y=x\) Now we can see that two critical points emerge when we match these possibilities: 1. \(x=0\) and \(y=0\), leading to the critical point \((0, 0)\). 2. \(1-x=y\) and \(1-y=x\), leading to the critical point \((\frac{1}{2}, \frac{1}{2})\). So, we have two critical points: \((0, 0)\) and \((\frac{1}{2}, \frac{1}{2})\).

Step 4: Apply Second Derivative Test

Now we will calculate the second-order partial derivatives and form the Hessian matrix to apply the Second Derivative Test: $$ \frac{\partial^2 f}{\partial x^2} = e^{-x-y} y^2 (x-2), \quad \frac{\partial^2 f}{\partial y^2} = e^{-x-y} x^2 (y-2), \quad \frac{\partial^2 f}{\partial x \partial y} = e^{-x-y}(1-x-y) $$ The Hessian matrix for this function is: $$ H = \begin{bmatrix} e^{-x-y} y^2 (x-2) & e^{-x-y}(1-x-y) \\ e^{-x-y}(1-x-y) & e^{-x-y} x^2 (y-2) \end{bmatrix} $$ Now, we evaluate the determinant of Hessian matrix at each critical point: 1. At \((0, 0)\), the determinant of Hessian matrix is 0, so the Second Derivative Test is inconclusive. 2. At \((\frac{1}{2}, \frac{1}{2})\), the determinant of Hessian matrix is negative, which indicates that this point is a saddle point.

Step 5: Confirm results using graphing utility

Using a graphing utility, we can plot the function \(f(x, y)=x y e^{-x-y}\) and observe the critical points. We can confirm that \((\frac{1}{2}, \frac{1}{2})\) is a saddle point, but the graphing utility will not be able to tell us the nature of the critical point \((0, 0)\) since the Second Derivative Test was inconclusive. To summarize: 1. \((0, 0)\) is a critical point, but the Second Derivative Test is inconclusive, so we cannot determine its nature. 2. \((\frac{1}{2}, \frac{1}{2})\) is a critical point and a saddle point, as shown by the Second Derivative Test and confirmed by the graphing utility.

Key Concepts

Second Derivative Test

The Second Derivative Test is a method used to classify critical points of a function of two variables. Once critical points are identified, the test helps determine if these points are local maxima, local minima, or saddle points. To perform this test, we inspect the second-order partial derivatives of the function and form what's known as the Hessian matrix. By analyzing its determinant, we can infer the nature of the critical points.

If the determinant of the Hessian matrix at a critical point is positive, then:

• If the second partial derivative with respect to x twice, \(\frac{\partial^2 f}{\partial x^2}\), is greater than 0, the critical point is a local minimum.
• If \(\frac{\partial^2 f}{\partial x^2}\) is less than 0, the critical point is a local maximum.

If the determinant is negative, the point is a saddle point.

If the determinant is zero, the test is inconclusive, and other methods or tests may be needed to classify the critical point.

Partial Derivatives

Partial derivatives are derivatives of functions of multiple variables taken with respect to one variable at a time, while treating all other variables as constants. In such multivariable functions, like the exercise function \(f(x, y) = xy e^{-x-y}\), the partial derivative with respect to \(x\), denoted \(\frac{\partial f}{\partial x}\), shows how \(f\) changes as \(x\) changes while keeping \(y\) constant.

Similarly, \(\frac{\partial f}{\partial y}\) indicates the change in \(f\) concerning \(y\) while \(x\) is constant.

Partial derivatives are crucial for finding critical points. By setting these derivatives equal to 0, we can identify potential maxima, minima, or saddle points. In our example, we calculate both \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\) and solve the resulting equations to find all critical points \((x, y)\).

Hessian Matrix

The Hessian matrix is a square matrix of all second-order partial derivatives of a function. For a function \(f(x, y)\), the Hessian matrix \(H\) looks like this:
\[H = \begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \\frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2} \end{bmatrix}\]

It quantitatively describes the local curvature of the function. The elements on the main diagonal are the second partial derivatives of the function with respect to each variable, while the off-diagonal elements are mixed partial derivatives.

The determinant of this matrix plays a vital role in the Second Derivative Test. For example, evaluating this determinant at a critical point helps determine whether it is a local extremum or a saddle point. The signs and values of these derivatives yield insight into the function's behavior near critical points.

Saddle Point

A saddle point is a type of critical point where the function does not achieve a local maximum or minimum. Instead, the function increases in one direction and decreases in another, resembling a saddle shape. In two-dimensional functions, saddle points occur when the determinant of the Hessian matrix at a critical point is negative.

In our exercise with the function \(f(x, y) = xy e^{-x-y}\), at the critical point \((\frac{1}{2}, \frac{1}{2})\), the Hessian determinant was found to be negative, thus confirming it as a saddle point. Understanding this concept is crucial when analyzing functions in multivariable calculus because it highlights regions where the function's behavior changes rapidly.

Remember, saddle points can be confusing at first because they don't fit into the categories of maxima or minima but instead represent a complex interaction in multiple variable functions.