Problem 5 - 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). $$ f(x, y)=-2 x^{2}+y^{2}-6 y $$

Step-by-Step Solution

Answer

The stationary point (0, 3) is a saddle point.

Step 1: Calculate First Partial Derivatives

To find the stationary points, calculate the first partial derivatives of the function with respect to each variable.Given function: \( f(x, y) = -2x^2 + y^2 - 6y \)- Partial derivative w.r.t. \( x \): \( f_x = \frac{\partial f}{\partial x} = -4x \)- Partial derivative w.r.t. \( y \): \( f_y = \frac{\partial f}{\partial y} = 2y - 6 \)

Step 2: Set Partial Derivatives to Zero

To find critical points, set each partial derivative to zero and solve for \( x \) and \( y \).From \( f_x = -4x = 0 \), we get \( x = 0 \).From \( f_y = 2y - 6 = 0 \), we get \( y = 3 \).The stationary point is \((x, y) = (0, 3)\).

Step 3: Compute the Second Partial Derivatives

Calculate the second partial derivatives which are needed for the Hessian matrix.- Second partial derivative w.r.t. \( x \): \( f_{xx} = \frac{\partial^2 f}{\partial x^2} = -4 \)- Second partial derivative w.r.t. \( y \): \( f_{yy} = \frac{\partial^2 f}{\partial y^2} = 2 \)- Mixed partial derivatives: \( f_{xy} = f_{yx} = \frac{\partial^2 f}{\partial x \partial y} = 0 \)

Step 4: Form the Hessian Matrix

Create the Hessian matrix using the second partial derivatives:\[ H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{yx} & f_{yy} \end{bmatrix} = \begin{bmatrix} -4 & 0 \ 0 & 2 \end{bmatrix} \]

Step 5: Determine Type of Extrema Using the Hessian

Calculate the determinant of the Hessian to determine the nature of the stationary point:\[ \text{det}(H) = (-4)(2) - (0)(0) = -8 \]Since \( \text{det}(H) < 0 \), the stationary point \((0, 3)\) is a saddle point.

Key Concepts

Partial Derivatives

In calculus, partial derivatives are a fundamental concept, especially when dealing with functions of multiple variables like our example function, \( f(x, y) = -2x^2 + y^2 - 6y \).
Partial derivatives allow us to understand how a function changes as we vary one of the variables, while holding the others constant.

• For a function \( f(x, y) \), the partial derivative with respect to \( x \), denoted \( f_x \), tells us the rate of change of the function as \( x \) changes, with \( y \) held constant. In our example, \( f_x = -4x \).


• Similarly, \( f_y = 2y - 6 \) provides the rate of change of the function with respect to \( y \), with \( x \) held constant.

Finding the partial derivatives is the first step in locating stationary points, which are crucial for identifying potential local extrema.
To find these points, we set the partial derivatives equal to zero.

Local Extrema

Local extrema of a function are points where the function reaches a local minimum or maximum value. At a local minimum, the function value is lower than at any nearby point, while at a local maximum, it is higher.
Finding these points involves calculating the first and second partial derivatives and requires setting the first partial derivatives to zero to locate the stationary points.

• For our function, by setting \( f_x = -4x \) and \( f_y = 2y - 6 \) to zero, we find the stationary point \((0, 3)\).

Understanding the nature of these points, whether they are maxima, minima, or something else, is determined using the second derivatives and, specifically, the Hessian matrix.
This informs us about the concavity and helps in distinguishing between different types of extrema.

Saddle Point

A saddle point is a type of stationary point that is not a local maximum or minimum. Instead, it resembles a saddle found on a horse; the function increases in one direction and decreases in another.
In mathematical optimization and calculus, saddle points are significant because they indicate a change in the behavior of the function.

• For our function, \( f(x, y) \), after finding the stationary point \((0, 3)\), we compute the Hessian matrix using the second partial derivatives.


• The Hessian matrix for our function is \( H = \begin{bmatrix} -4 & 0 \ 0 & 2 \end{bmatrix} \).

The determinant of this matrix, \( \text{det}(H) = -8 \), is negative.
This tells us that the stationary point \((0, 3)\) is a saddle point, confirming mixed behavior in different directions.
Saddle points like these are vital in understanding the geometry and topology of a function's graph across its domain.