91Ó°ÊÓ

To start with, partial derivatives are derivatives where we hold one of the variables constant. For a function of two variables, such as the given function \( f(x, y) = y e^{x} - 3x - y + 5 \), we compute the partial derivatives with respect to \(x\) and \(y\).

The partial derivative with respect to \( x \) of our function is calculated by treating \( y \) as a constant, giving us:
\( \frac{∂f}{∂x} = y e^{x} - 3 \).

For the partial derivative with respect to \( y \), we treat \( x \) as a constant, resulting in: \( \frac{∂f}{∂y} = e^{x} - 1 \).

Evaluating these at the point \((0, 3)\), we substitute \( x = 0 \) and \( y = 3 \) into the partial derivative equations:
\( \frac{∂f}{∂x}(0, 3) = 3 e^{0} - 3 = 0 \)
\( \frac{∂f}{∂y}(0, 3) = e^{0} - 1 = 0 \)

This tells us that both partial derivatives are zero at the point \((0, 3)\), which is our starting point for using the second-derivative test.

The Hessian matrix is a square matrix of second-order partial derivatives of a scalar-valued function. For our function \(f(x, y)\), the Hessian matrix is particularly important in determining the concavity of the function.

To construct the Hessian matrix, we calculate the second partial derivatives:

• \( \frac{∂^2 f}{∂x^2} = y e^{x} \)
• \( \frac{∂^2 f}{∂y^2} = 0 \)
• \( \frac{∂^2 f}{∂x ∂y} = e^{x} \)
• \( \frac{∂^2 f}{∂y ∂x} = e^{x} \)

Evaluating these at the point \((0, 3)\):

• \( \frac{∂^2 f}{∂x^2}(0, 3) = 3 e^{0} = 3 \)
• \( \frac{∂^2 f}{∂y^2}(0, 3) = 0 \)
• \( \frac{∂^2 f}{∂x ∂y}(0, 3) = e^{0} = 1 \)
• \( \frac{∂^2 f}{∂y ∂x}(0, 3) = e^{0} = 1 \)

The Hessian matrix at this point is then:
\[ H = \begin{bmatrix} 3 & 1 \ 1 & 0 \ \ end{bmatrix} \]

To determine the nature of the critical point, we calculate the determinant of the Hessian matrix:
\[ D = (3)(0) - (1)(1) = -1 \]
A negative determinant suggests that the point is neither a local maximum nor a minimum, but rather a saddle point.

Local extrema refer to points where a function reaches a local maximum or minimum. Determining local extrema involves not only finding where the first partial derivatives are zero but also using the second-derivative test.

In our case, after finding that both \( \frac{∂f}{∂x} = 0 \) and \( \frac{∂f}{∂y} = 0 \) at \( (0, 3) \), we apply the second-derivative test. As the determinant of the Hessian matrix \( D \) is negative, we confirm that \( (0, 3) \) is a saddle point. This means that the function does not have a local maximum or minimum at this point. Instead, the function has both a local minimum and maximum behavior in different directions around \( (0, 3) \).

Understanding these concepts is crucial for analyzing the behavior of functions in multivariable calculus and can be very useful in optimization problems and in the study of surface behavior in higher dimensions.