Problem 17 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. \(f(x, y)=e^{x^{2}+y^{2}}\)

Step-by-Step Solution

Answer

The relative extrema of the function \(f(x, y) = e^{x^2 + y^2}\) is a local minimum at the point (0, 0) with a value of 1.

Step 1: Calculate First Partial Derivatives

First, calculate the first partial derivatives of the given function with respect to x and y: \[f_x = \dfrac{\partial f}{\partial x} = 2xe^{x^2 + y^2}\] \[f_y = \dfrac{\partial f}{\partial y} = 2ye^{x^2 + y^2}\]

Step 2: Find the Critical Points

Set the first partial derivatives to 0 and solve for x and y: \[f_x = 2xe^{x^2 + y^2} = 0\] \[f_y = 2ye^{x^2 + y^2} = 0\] Both equations involve the exponential function \(e^{x^2 + y^2}\), which is never equal to 0. Therefore, the only way to satisfy the equations is for x or y to be 0: For \(f_x = 0\), we have x = 0, For \(f_y = 0\), we have y = 0. So, the only critical point is (0, 0).

Step 3: Calculate Second Partial Derivatives

Calculate the second partial derivatives with respect to x, y: \[f_{xx} = \dfrac{\partial^2f}{\partial x^2} = 2e^{x^2 + y^2} + 4x^2e^{x^2 + y^2}\] \[f_{yy} = \dfrac{\partial^2f}{\partial y^2} = 2e^{x^2 + y^2} + 4y^2e^{x^2 + y^2}\] \[f_{xy} = \dfrac{\partial^2f}{\partial x\partial y} = 4xye^{x^2 + y^2}\]

Step 4: Second-derivative Test

To classify the critical point (0, 0), we will use the second-derivative test by calculating the determinant D at the critical point: \[D(x, y) = f_{xx}f_{yy} - f_{xy}^2\] Evaluate the second partial derivatives at the critical point (0, 0): \[f_{xx}(0, 0) = 2\] \[f_{yy}(0, 0) = 2\] \[f_{xy}(0, 0) = 0\] Now, calculate the determinant D(0, 0): \[D(0,0) = (2)(2) - (0)^2 = 4\] Since \(D > 0\) and \(f_{xx}(0, 0) > 0\), the critical point (0, 0) is a local minimum.

Step 5: Determine Relative Extrema

Since the critical point (0, 0) is a local minimum, the relative extrema of the function is: Local minimum at the point (0, 0), with a corresponding function value of \(f(0, 0)=e^0=1\). So, the relative extrema of the function \(f(x, y) = e^{x^2 + y^2}\) is a local minimum at point (0, 0) with the value of 1.

Key Concepts

Critical Points

In multivariable calculus, critical points of a function are potentially important spots where the function changes its behavior. For a function of two variables, critical points are found when the first partial derivatives with respect to both variables are zero. For the given function, \(f(x, y)=e^{x^2+y^2}\), these are points where the slopes, as computed through the first derivatives \(f_x\) and \(f_y\), flatten out.

Here, the first partial derivatives are:

• \(f_x = 2xe^{x^2 + y^2}\)
• \(f_y = 2ye^{x^2 + y^2}\)

As the exponential function is never zero, for each expression to equal zero, both \(x\) and \(y\) must be zero. This results in a critical point at (0, 0). These critical points help us further analyze the function's behavior using methods such as the second derivative test.

Second Derivative Test

The second derivative test in multi-variable calculus helps us classify the critical points as local minima, maxima, or saddle points. This test involves evaluating second partial derivatives and using them to compute the Hessian determinant. The determinant \(D\) is calculated as \(D = f_{xx}f_{yy} - f_{xy}^2\), which is then used to determine the point's nature.

For the function \(f(x, y)=e^{x^2+y^2}\), the relevant second partial derivatives are:

• \(f_{xx} = 2e^{x^2 + y^2} + 4x^2e^{x^2 + y^2}\)
• \(f_{yy} = 2e^{x^2 + y^2} + 4y^2e^{x^2 + y^2}\)
• \(f_{xy} = 4xye^{x^2 + y^2}\)

At the critical point (0, 0), evaluating these gives:

• \(f_{xx}(0, 0) = 2\)
• \(f_{yy}(0, 0) = 2\)
• \(f_{xy}(0, 0) = 0\)

The determinant is thus \(D = (2)(2) - (0)^2 = 4\). Since \(D > 0\) and \(f_{xx}(0, 0) > 0\), the critical point (0, 0) is identified as a local minimum.

Partial Derivatives

Partial derivatives are derivatives of functions with more than one variable, and they denote the function's change with respect to one variable while keeping the others constant. They play a crucial role in identifying critical points by allowing us to understand how the function behaves in different directions.

For the function \(f(x, y) = e^{x^2 + y^2}\), the first partial derivatives are computed as:

• \(f_x = \frac{\partial f}{\partial x} = 2xe^{x^2 + y^2}\)
• \(f_y = \frac{\partial f}{\partial y} = 2ye^{x^2 + y^2}\)

These equations help in determining regions where the function doesn’t change, leading to the discovery of the critical points. In the context of the second derivative test, the second partial derivatives \(f_{xx}, f_{yy},\) and \(f_{xy}\) are calculated, providing further insights into the curvature of the function's surface around each critical point. These derivatives allow us to compute the Hessian determinant, essential for classifying the critical points effectively.

Relative Extrema

In multivariable calculus, relative extrema refer to places where a function takes on local minimum or maximum values. Identifying these extrema is crucial for understanding the behavior of functions in terms of producing the highest or lowest values locally, i.e., within a vicinity.

Here, we've determined that the critical point (0, 0) is a local minimum. This means around this point, the function \(f(x, y)=e^{x^2+y^2}\) reaches its lowest value compared to nearby points. The function value at the local minimum is computed as \(f(0, 0) = e^0 = 1\), indicating that the smallest output value of the function in the local neighborhood around (0, 0) is 1.

• This classification helps in various applications, like optimization, where one seeks to maximize or minimize outputs.
• Recognizing relative extrema is also valuable in fields ranging from economics to engineering where system behaviors might rely on output levels.

Thus, through detailed examination using critical points and the second derivative test, one can effectively identify and understand relative extrema.