Problem 19 - 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. $$f(x, y)=4+2 x^{2}+3 y^{2}$$

Step-by-Step Solution

Answer

Answer: The critical point \((0, 0)\) is a local minimum.

Step 1: Find the partial derivatives

To find the critical points, we need to first compute the partial derivatives with respect to x and y: $$ f_x = \frac{\partial}{\partial x}(4 + 2x^2 + 3y^2) = 4x $$ $$ f_y = \frac{\partial}{\partial y}(4 + 2x^2 + 3y^2) = 6y $$

Step 2: Solve the system of equations

Now, set these partial derivatives equal to 0 and solve for x and y to find the critical points: $$ 4x = 0 \Rightarrow x = 0 $$ $$ 6y = 0 \Rightarrow y = 0 $$ The critical point is \((0, 0)\).

Step 3: Compute the Second Derivative Test discriminant

Compute the second partial derivatives: $$ f_{xx} = \frac{\partial^2}{\partial x^2}(4 + 2x^2 + 3y^2) = 4 $$ $$ f_{yy} = \frac{\partial^2}{\partial y^2}(4 + 2x^2 + 3y^2) = 6 $$ $$ f_{xy} = \frac{\partial^2}{\partial x\partial y}(4 + 2x^2 + 3y^2) = 0 $$ Compute the discriminant at the critical point using the formula: $$ D = f_{xx}f_{yy} - f_{xy}^2 $$ $$ D = (4)(6) - (0)^2 = 24 $$

Step 4: Analyze the sign of the discriminant

Since \(D > 0\) and \(f_{xx} > 0\), the Second Derivative Test tells us that the critical point \((0, 0)\) corresponds to a local minimum.

Step 5: Confirm the results using a graphing utility

The graph of the function \(f(x, y) = 4 + 2x^2 + 3y^2\) indeed shows that the function has a local minimum at the point \((0, 0)\).

Key Concepts

Partial Derivatives

When studying multivariable calculus, partial derivatives are essential for analyzing how a function changes with respect to one variable while keeping the others constant. They allow us to understand how multi-variable functions behave and identify critical points.
In our example function, \(f(x, y) = 4 + 2x^2 + 3y^2\), we found the partial derivatives by differentiating with respect to each variable separately.

• The partial derivative with respect to \(x\), denoted as \(f_x\), was found to be \(4x\).
• The partial derivative with respect to \(y\), denoted as \(f_y\), was found to be \(6y\).

Critical points occur where these partial derivatives are both zero. This helps us identify where the function may have local maxima, minima, or saddle points.
In this case, setting \(f_x = 0\) and \(f_y = 0\) leads to the critical point \((0, 0)\).

Second Derivative Test

The Second Derivative Test is a method used to classify critical points for functions of two variables. It examines the concavity of a surface to determine if a critical point is a local minimum, maximum, or saddle point.
To implement the test, we calculate the second partial derivatives:

• \(f_{xx}\) measures how \(f_x\) changes with \(x\); it was found to be \(4\).
• \(f_{yy}\) measures how \(f_y\) changes with \(y\); it was \(6\).
• \(f_{xy}\) measures the mixed partial derivative; it was \(0\).

The discriminant \(D\) is then computed using \(D = f_{xx}f_{yy} - (f_{xy})^2\). In our exercise, \(D = 24\).
Since \(D > 0\) and \(f_{xx} > 0\), this indicates that the critical point \((0, 0)\) is a local minimum. This test is crucial for understanding the nature of critical points in multivariable functions.

Local Minimum

A local minimum is a point within a function at which the function value is lower than at nearby points. It does not necessarily mean that this is the lowest value the function can take, but it is a minimal point in that specific vicinity.
In the case of the function \(f(x, y) = 4 + 2x^2 + 3y^2\), the critical point at \((0,0)\) was determined to be a local minimum. This conclusion was drawn from the Second Derivative Test.
At a local minimum, the surface will bowl slightly upwards, and the derivative values around this point will be positive. This is reflected in the positive discriminant \(D\). Understanding local minima is essential in optimization problems, where identifying the lowest points within constraints is often the goal.

Graphing Calculus Functions

Graphing calculus functions plays a vital role in visualizing and confirming theoretical mathematical conclusions, such as those drawn from solving derivatives or employing the Second Derivative Test.
The function \(f(x, y) = 4 + 2x^2 + 3y^2\) can be graphed to visually verify analytical solutions. The graph provides a 3D plot where the rise and fall of the function values can be observed.

• The point \((0, 0)\) should show a clear dip, confirming the presence of a local minimum.
• However, the visualization is limited to the region plotted, which calls for careful selection of view range and scales.

Graphing is a powerful tool that offers an intuitive understanding and confirms analytical results, bridging the gap between calculus and real-world applications.