Problem 4 - Calculus for Biology and Medicine

Question

The functions are defined for all \((x, y) \in R^{2} .\) Find all candidates for local extrema, and use the Hessian matrix to determine the type (maximum, minimum, or saddle point). $$ f(x, y)=-x y-2 y^{2} $$

Step-by-Step Solution

Answer

The function has a saddle point at \((0, 0)\).

Step 1: Find the First Partial Derivatives

To find the candidates for local extrema, we need the first partial derivatives of the function \( f(x, y) = -xy - 2y^2 \). These are:- \( f_x(x, y) = \frac{d}{dx}(-xy - 2y^2) = -y \)- \( f_y(x, y) = \frac{d}{dy}(-xy - 2y^2) = -x - 4y \).

Step 2: Solve for Critical Points

Set the first partial derivatives equal to zero to find the critical points.- Solve \( f_x(x, y) = -y = 0 \) which gives \( y = 0 \).- Solve \( f_y(x, y) = -x - 4y = 0 \). Substitute \( y = 0 \) to get \( -x = 0 \), so \( x = 0 \).The critical point is \((x, y) = (0, 0)\).

Step 3: Compute the Second Partial Derivatives

The second partial derivatives are needed to construct the Hessian matrix.- \( f_{xx} = \frac{d}{dx}(-y) = 0 \).- \( f_{yy} = \frac{d}{dy}(-x - 4y) = -4 \).- \( f_{xy} = \frac{d}{dy}(-y) = -1 \).- \( f_{yx} = \frac{d}{dx}(-x - 4y) = -1 \).

Step 4: Construct the Hessian Matrix

The Hessian matrix \( H \) is given by:\[H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{yx} & f_{yy} \end{bmatrix} = \begin{bmatrix} 0 & -1 \ -1 & -4 \end{bmatrix}\]

Step 5: Evaluate the Determinant of the Hessian

The determinant of the Hessian \( H \) is calculated as follows:\[\det(H) = (0)(-4) - (-1)(-1) = 0 - 1 = -1\]Since the determinant is negative, \((0, 0)\) is a saddle point.

Key Concepts

Hessian matrix

The Hessian matrix is a square matrix made up of second-order partial derivatives. It is crucial in multivariable calculus to study the local curvature of functions. For a function of two variables, the Hessian is a 2x2 matrix given by:

• First, compute the second-order partial derivatives of the function.
• Arrange these derivatives in the matrix format as \[H = \begin{bmatrix} f_{xx} & f_{xy} \f_{yx} & f_{yy} \end{bmatrix}\]
• The diagonal elements are the second partial derivatives with respect to each variable, while the off-diagonal elements are the mixed partial derivatives.

In the provided exercise, designing the Hessian helped in determining the nature of the critical point. Evaluating the Hessian's determinant reveals the behavior by assessing if the point is a maximum, minimum, or saddle point.

local extrema

Local extrema refer to the local minimum or maximum points of a function. These points are potential candidates for the highest or lowest value in a region near the point. To find local extrema, follow this approach:

• Locate critical points where the first partial derivatives are zero.
• Use the Hessian matrix to determine the type of extremum. The determinant here helps decide if it's a local maximum, minimum, or saddle point.

In practical terms, if at a critical point the Hessian's determinant is positive, you consider further:

• If the second derivative with respect to one variable is positive, you have a local minimum.
• If negative, it's a local maximum.
• A negative determinant indicates a saddle point.

In the given problem, the critical point was (0, 0), evaluated using the Hessian's determinant to identify a saddle point instead of a local extremum.

partial derivatives

Partial derivatives measure how a function changes as one of its variables changes, holding others constant. This technique is vital in functions of several variables, where each partial derivative essentially slices through the multi-dimensional graph in a particular direction.Here's how to compute:

• Identify each relevant variable of the function.
• Find the derivative with respect to one variable, treating others as constants.

In the original exercise, the partial derivatives for the function\[f(x, y) = -xy - 2y^2\]were determined as:

• \( f_x(x, y) = -y \)
• \( f_y(x, y) = -x - 4y \)

Setting these derivatives to zero is the first step in locating the function's critical points.

critical points

Critical points in a multivariable function occur where all first partial derivatives are zero or undefined. They are essential in locating potential extrema.Steps to find critical points involve:

• Calculate the first partial derivatives.
• Set each of these equal to zero, solving the resulting system of equations.
• The solutions give you points where the function's rate of change is zero.

These points are crucial in determining the nature of the function at that point. The Hessian matrix, coupled with these coordinates, decides whether each point is a local extremum or a saddle point.In the problem given, solving:

• \( f_x(x, y) = -y = 0 \) yielded \( y = 0 \)
• \( f_y(x, y) = -x - 4y = 0 \) with \( y = 0 \) led to \( x = 0 \)

Thus, the critical point found was \((0, 0)\). Using further steps such as the Hessian matrix analysis, we concluded it was a saddle point.