Problem 5 - Calculus for Biology and Medicine

Question

The functions are defined for all \((x, y) \in \boldsymbol{R}^{2} .\) 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 critical point \((0, 3)\) is a saddle point.

Step 1: Identify Critical Points

To find critical points, we first calculate the partial derivatives of the function \(f(x, y) = -2x^2 + y^2 - 6y\). The partial derivative with respect to \(x\) is \(f_x = \frac{\partial}{\partial x}(-2x^2 + y^2 - 6y) = -4x\). The partial derivative with respect to \(y\) is \(f_y = \frac{\partial}{\partial y}(-2x^2 + y^2 - 6y) = 2y - 6\). To find critical points, set these derivatives to zero: \(f_x = 0\) gives \(x = 0\), and \(f_y = 0\) gives \(2y - 6 = 0\), which solves to \(y = 3\). Thus, the critical point is \((0, 3)\).

Step 2: Construct the Hessian Matrix

The Hessian matrix \(H\) is a square matrix of second-order partial derivatives. Calculate the second partial derivatives for the function: \(f_{xx} = \frac{\partial^2 f}{\partial x^2} = -4\), \(f_{yy} = \frac{\partial^2 f}{\partial y^2} = 2\), and \(f_{xy} = f_{yx} = 0\). The Hessian matrix thus is: \[ H = \begin{bmatrix} -4 & 0 \ 0 & 2 \end{bmatrix} \]

Step 3: Determine the Type of Critical Point

To classify the critical point using the Hessian matrix, we use the determinant \(D\) of \(H\), \(D = f_{xx}f_{yy} - (f_{xy})^2\). Substituting the Hessian values, \(D = (-4)(2) - (0)^2 = -8\). Since \(D < 0\), the critical point \((0, 3)\) is classified as a saddle point, meaning it is neither a local maximum nor a minimum.

Key Concepts

Critical Points

A critical point in a function of two variables, such as \(f(x, y) = -2x^2 + y^2 - 6y\), occurs where the function's gradient is zero. This gradient consists of the partial derivatives of the function with respect to each of its variables, expressing the rate of change of the function:

• The partial derivative with respect to \(x\) is \(-4x\).
• The partial derivative with respect to \(y\) is \(2y - 6\).

To find the critical points, you set each partial derivative to zero and solve the resulting system of equations. For our function, this results in a critical point at \((0, 3)\). Finding critical points is crucial when determining where a function may reach local maxima, minima, or saddle points. These points essentially help us find where a function behaves differently compared to its immediate surroundings.

Partial Derivatives

Partial derivatives are like the derivatives we learn in single-variable calculus, but they apply to functions with more than one variable. They show how a function changes as only one input variable is altered, keeping the others constant. In our exercise, we examined the function\(f(x, y) = -2x^2 + y^2 - 6y\). By taking partial derivatives, we computed:

• \(f_x = -4x\), indicating how the function changes with \(x\).
• \(f_y = 2y - 6\), showing how the function changes with \(y\).

These derivatives help locate critical points and give us a glimpse of the function's behavior locally. Partial derivatives are building blocks in calculus and essential for constructing tools like the Hessian matrix, which help classify the type of critical points found.

Saddle Point

A saddle point is a type of critical point that differs from local maxima or minima in its nature. Unlike maxima or minima, a saddle point is not where the function peaks or dips, but rather where it changes direction. In simpler terms, a saddle point might look like a peak in one direction and a trough in another.
In our example with critical point \((0, 3)\), the Hessian matrix was utilized to help identify it. By calculating the Hessian determinant \(D = -4 \times 2 - 0^2 = -8\), we could see that \(D < 0\), indicating a saddle point. The conclusion is that the point \((0, 3)\) doesn't represent a maximum or minimum, but rather a point of balance where the function's behavior shifts. Saddle points are important in optimization and modeling, as they can profoundly affect how a function operates over its domain. Identifying them aids in understanding the full landscape of the function's output.