91Ó°ÊÓ

When dealing with functions of multiple variables, like \( f(x, y) = 6x^2 - 2x^3 + 3y^2 + 6xy \), finding critical points involves calculating partial derivatives. Partial derivatives show how a function changes as each variable changes individually while keeping other variables constant.

Here’s how it works:

• For the function \( f(x, y) \), the partial derivative with respect to \( x \) is \( \frac{\partial f}{\partial x} \), and to find it, treat \( y \) as a constant.
• Similarly, the partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} \) treating \( x \) as a constant.

In the given example, these partial derivatives are: \( \frac{\partial f}{\partial x} = 12x - 6x^2 + 6y \) and \( \frac{\partial f}{\partial y} = 6y + 6x \).

By setting these derivatives to zero, we find potential points where the function's slope is zero in all directions, indicating possible maxima, minima, or saddle points.

Once we locate the critical points, the Second Derivative Test helps us determine their nature. This test uses the second partial derivatives of the function. It checks whether these critical points are maximum, minimum, or saddle points.

The procedure involves:

• Compute all second-order partial derivatives such as \( \frac{\partial^2 f}{\partial x^2} \), \( \frac{\partial^2 f}{\partial y^2} \), and the mixed derivative \( \frac{\partial^2 f}{\partial x \partial y} \).
• Use these to form a matrix, whose determinant helps classify each critical point.

A positive determinant \( D \) with positive \( \frac{\partial^2 f}{\partial x^2} \) indicates a local minimum; a positive \( D \) with a negative \( \frac{\partial^2 f}{\partial x^2} \) points to a local maximum. If \( D \) is negative, the point is a saddle point.

In the exercise, for points \((0,0)\) and \((1,-1)\), calculations reveal these are a local minimum and a saddle point, respectively.

The Hessian Matrix is an essential tool for understanding the concavity of functions through the second partial derivatives. It's a square matrix that incorporates these derivatives and is key in applying the Second Derivative Test.

For a function like \( f(x, y) \), the Hessian matrix \( \mathbf{H} \) is structured as follows:\[\mathbf{H} = \begin{bmatrix}\frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \\frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2}\end{bmatrix}\]In practical situations, calculating its determinant \( D \) gives valuable insight:

• If \( D > 0 \), refer to the sign of \( \frac{\partial^2 f}{\partial x^2} \) to classify a local extremum.
• If \( D < 0 \), it denotes a saddle point.

In our exercise, the Hessian matrices evaluated at \((0,0)\) and \((1,-1)\) help conclude the nature of these critical points efficiently.