91Ó°ÊÓ

Partial derivatives help us understand how a function changes as we vary each variable independently. In this exercise, we deal with the function \( f(x, y) = 3x^2 - x^3 + 12y^2 - 8y^3 + 60 \). To find the partial derivative with respect to \( x \), we treat \( y \) as a constant and differentiate with respect to \( x \) alone. This gives us \( \frac{\partial f}{\partial x} = 6x - 3x^2 \). Similarly, to find \( \frac{\partial f}{\partial y} \), we consider \( x \) as constant, leading to \( \frac{\partial f}{\partial y} = 24y - 24y^2 \). These derivatives provide the slopes of \( f(x, y) \) as \( x \) or \( y \) changes, key for identifying critical points.

The critical points, which are points where the function might have local maxima, minima, or saddle points, occur where these partial derivatives are both zero. Solving for these, we find critical points by resolving equations \( 6x - 3x^2 = 0 \) and \( 24y - 24y^2 = 0 \).

The Hessian matrix is a square matrix of second-order partial derivatives of a function. It provides crucial information about the curvature of the function at a given point, guiding us to determine the nature of critical points. For a function \( f(x, y) \), the Hessian \( H \) is calculated as:

\[ H = \begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \ \frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2} \end{bmatrix} \]

In this exercise, the Hessian matrix at each critical point is constructed using:

• \( \frac{\partial^2 f}{\partial x^2} = 6 - 6x \)
• \( \frac{\partial^2 f}{\partial y^2} = 24 - 48y \)
• \( \frac{\partial^2 f}{\partial x \partial y} = 0 \)

This matrix assists in applying the second derivative test, enabling the classification of critical points.

The second derivative test is a method to determine the nature of critical points found using the partial derivatives. For functions of two variables, it involves evaluating the determinant \( D \) of the Hessian matrix:

\[ D = \left(\frac{\partial^2 f}{\partial x^2}\right) \left(\frac{\partial^2 f}{\partial y^2}\right) - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 \]

The sign of \( D \) indicates:

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

Thus, this test allows us to classify each critical point with precision.

Local minimum points are areas in the graph where the function reaches a low value in a small neighborhood around the point. During this exercise, we use the second derivative test as a tool to identify such points. When \( D > 0 \) and \( \frac{\partial^2 f}{\partial x^2} > 0 \), the critical point is a local minimum. This indicates the graph is curving upwards, forming a trough at this point.

For example, at the critical point \((0, 0)\), after testing, \( D = 144 \) and \( \frac{\partial^2 f}{\partial x^2} = 6 > 0 \), confirming that \((0, 0)\) is indeed a local minimum. Understanding such points is crucial, especially when looking for optimal solutions in real-life scenarios.

A saddle point is a critical point where the function does not achieve a local extremum. This point resembles a saddle in its graphical representation, indicating a shift in curvature direction. When performing the second derivative test, if \( D < 0 \), the critical point is a saddle point. This occurs because the determinant of the Hessian matrix reveals different concavity directions along different sections.

In our exercise, several points such as \((0, 1)\) and \((2, 0)\) showed \( D = -144 \), classifying them as saddle points. At these points, the function shows neither a peak nor a trough, reflecting a shift in direction. Recognizing such points is key to understanding the function's behavior in multidimensional optimization tasks.