Problem 27 - 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. $$f(x, y)=2 x y e^{-x^{2}-y^{2}}$$

Step-by-Step Solution

Answer

Question: Identify the critical points of the function \(f(x,y) = 2xye^{-x^2 - y^2}\), classify them as a local maximum, local minimum, or saddle point, and confirm the types using a graphing utility. Answer: The critical points of \(f(x,y)\) are \((0,0),\ (\pm\frac{1}{\sqrt{2}},0),\ (0,\pm\frac{1}{\sqrt{2}})\). The second derivative test indicates that \((\pm\frac{1}{\sqrt{2}},0)\) and \((0,\pm\frac{1}{\sqrt{2}})\) are saddle points. The test is inconclusive for \((0,0)\), so we must use a graphing utility which confirms that there is neither a local maximum nor local minimum at \((0,0)\).

Step 1: Find the partial derivatives of the function

First, we need to find the partial derivatives with respect to x and y. Here's the function again: $$f(x,y) = 2xye^{-x^2 - y^2}$$ Let's find the partial derivatives: $$\frac{\partial f}{\partial x} = 2ye^{-x^2 - y^2}(1 - 2x^2)$$ $$\frac{\partial f}{\partial y} = 2xe^{-x^2 - y^2}(1 - 2y^2)$$

Step 2: Identify the critical points

Now we need to find the points where both partial derivatives are 0: $$2ye^{-x^2 - y^2}(1 - 2x^2) = 0$$ $$2xe^{-x^2 - y^2}(1 - 2y^2) = 0$$ Solving these equations yields the critical points: $$(0,0),\quad (\pm\frac{1}{\sqrt{2}},0), \quad (0,\pm\frac{1}{\sqrt{2}})$$

Step 3: Apply the second derivative test

Calculate the second partial derivatives of the function: $$\frac{\partial^2 f}{\partial x^2} = 4x^2y^2e^{-x^2 - y^2}(2x^2 - 3)$$ $$\frac{\partial^2 f}{\partial y^2} = 4x^2y^2e^{-x^2 - y^2}(2y^2 - 3)$$ $$\frac{\partial^2 f}{\partial x \partial y} = 4x^2y^2e^{-x^2 - y^2}$$ Evaluate the second partial derivatives at each critical point and calculate the determinant D of the Hessian matrix: $$D(x,y) = \frac{\partial^2 f}{\partial x^2} \frac{\partial^2 f}{\partial y^2} - (\frac{\partial^2 f}{\partial x \partial y})^2$$ At \((0,0)\): $$D(0,0) = 0$$ As the determinant is 0, the test is inconclusive, and we cannot determine if this point is a local maximum, local minimum, or saddle point using the second derivative test. We need to graph the function to determine its type. At \((\pm\frac{1}{\sqrt{2}},0)\) and \((0,\pm\frac{1}{\sqrt{2}})\): $$D(\pm\frac{1}{\sqrt{2}},0) = D(0,\pm\frac{1}{\sqrt{2}}) = -4e^{-1} < 0$$ As the determinant is negative, \((\pm\frac{1}{\sqrt{2}},0)\), and \((0,\pm\frac{1}{\sqrt{2}})\) are saddle points.

Step 4: Confirm the results using a graphing utility

Using a graphing utility, plot the function \(f(x,y)\). This should confirm that \((\pm\frac{1}{\sqrt{2}},0)\) and \((0,\pm\frac{1}{\sqrt{2}})\) are saddle points and that the function does not have a local maximum or minimum at \((0,0)\).

Key Concepts

Partial Derivatives

To analyze a function of two variables, like \( f(x, y) = 2xy e^{-x^2-y^2} \), we often start by finding partial derivatives. A partial derivative represents the rate of change of the function with respect to one variable while keeping the other variable constant. This is akin to looking at the slope along a specific direction in a three-dimensional space. For our function, the partial derivative with respect to \( x \) is \( \frac{\partial f}{\partial x} = 2y e^{-x^2-y^2}(1 - 2x^2) \). Similarly, the partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} = 2x e^{-x^2-y^2}(1 - 2y^2) \).
These expressions help us find critical points by setting each derivative to zero and solving the resulting system of equations. These critical points are where the function could have a local maximum, minimum, or saddle point.

Second Derivative Test

Once we have partial derivatives and critical points, the next step is to use the Second Derivative Test. This test helps us determine the nature of those critical points – whether they are maxima, minima, or saddle points. For a function \( f(x, y) \), we start by calculating the second partial derivatives:

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

We then evaluate these derivatives at each critical point. The key is to calculate the determinant of the Hessian matrix using \( D = \frac{\partial^2 f}{\partial x^2} \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 \). The value of \( D \) helps us to ascertain the nature of the critical points:

• If \( D > 0 \), and \( \frac{\partial^2 f}{\partial x^2} > 0 \), it's a local minimum.
• If \( D > 0 \), and \( \frac{\partial^2 f}{\partial x^2} < 0 \), it's a local maximum.
• If \( D < 0 \), the point is a saddle point.
• If \( D = 0 \), the test is inconclusive.

Hessian Matrix

The Hessian Matrix is a square matrix of second-order partial derivatives of a scalar-valued function. It's essential for performing the Second Derivative Test. For a function \( f \), the Hessian matrix is defined as: \[\begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \\frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2}\end{bmatrix}\]
In our example, the Hessian matrix plays a central role: it's used to calculate the determinant \( D \). Calculating determinant \( D \) at different points helps us identify the nature of critical points as either local maxima, minima, or saddle points. The behavior and curvature of the function’s surface depend significantly on the values within this matrix.

Saddle Points

Saddle points are special critical points where the function behaves curiously. At a saddle point, the function doesn't define a peak or a trough. Instead, it exhibits a mix, curving upwards in one direction and downwards in another.
For the given function, critical points \((\pm\frac{1}{\sqrt{2}}, 0)\) and \((0, \pm\frac{1}{\sqrt{2}})\) are saddle points. We determined these because, at these points, the Hessian determinant \( D \) was found to be negative, confirming their saddle nature. Saddle points are vital in understanding the behavior of a multi-variable function and often indicate changes in the function’s behavior around those points. They're named so because they resemble the saddle of a horse, with one axis arching upwards and the other downwards.