Problem 8 - 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). $$ f(x, y)=y x e^{-(x+y)} $$

Step-by-Step Solution

Answer

The critical points are \((0, 1)\), \((1, 0)\), and \((1, 1)\). \((0, 1)\) and \((1, 0)\) are saddle points, while the test is inconclusive for \((1, 1)\).

Step 1: Finding the Critical Points

To find the critical points of the function \( f(x, y) = y x e^{-(x+y)} \), we need to calculate the first partial derivatives and set them equal to zero:1. Compute \(\frac{\partial f}{\partial x}\):\[\frac{\partial f}{\partial x} = y (1 - x) e^{-(x+y)}\]Setting \(\frac{\partial f}{\partial x} = 0\) gives us \(y (1 - x) e^{-(x+y)} = 0\), leading to two conditions: \(y = 0\) or \(x = 1\).2. Compute \(\frac{\partial f}{\partial y}\):\[\frac{\partial f}{\partial y} = x (1 - y) e^{-(x+y)}\]Setting \(\frac{\partial f}{\partial y} = 0\) gives us \(x (1 - y) e^{-(x+y)} = 0\), leading to two conditions: \(x = 0\) or \(y = 1\).Thus, solving the system formed by these conditions, the critical points are \((0, 1)\), \((1, 0)\), and \((1, 1)\).

Step 2: Constructing the Hessian Matrix

To determine the nature of the critical points, we need to use the Hessian matrix. First, calculate the second partial derivatives:- \(\frac{\partial^2 f}{\partial x^2} = -y (1 - x) e^{-(x+y)} + y e^{-(x+y)} = y e^{-(x+y)} (x - 2)\).- \(\frac{\partial^2 f}{\partial y^2} = -x (1 - y) e^{-(x+y)} + x e^{-(x+y)} = x e^{-(x+y)} (y - 2)\).- \(\frac{\partial^2 f}{\partial x \partial y} = (1 - x - y) e^{-(x+y)}\).The Hessian matrix \(H\) is given by:\[H = \begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \\frac{\partial^2 f}{\partial x \partial y} & \frac{\partial^2 f}{\partial y^2} \end{bmatrix}\]

Step 3: Evaluate the Hessian Determinant at Critical Points

Evaluate the Hessian determinant \(D\) at each critical point:1. **At \((0, 1)\):** \(H = \begin{bmatrix} 0 & -e^{-1} \ -e^{-1} & 0 \end{bmatrix}\) \(D = (0)(0) - (-e^{-1})(-e^{-1}) = -e^{-2} < 0\). Since \(D < 0\), \((0, 1)\) is a saddle point.2. **At \((1, 0)\):** \(H = \begin{bmatrix} 0 & e^{-1} \ e^{-1} & 0 \end{bmatrix}\) \(D = (0)(0) - (e^{-1})(e^{-1}) = -e^{-2} < 0\). Since \(D < 0\), \((1, 0)\) is a saddle point.3. **At \((1, 1)\):** \(H = \begin{bmatrix} -e^{-2} & -e^{-2} \ -e^{-2} & -e^{-2} \end{bmatrix}\) \(D = (-e^{-2})(-e^{-2}) - (-e^{-2})(-e^{-2}) = 0\). Since \(D = 0\), the test is inconclusive for \((1, 1)\).

Key Concepts

Understanding Critical Points

A critical point in the context of functions of several variables is a point where the gradient of the function equals zero. This means that both first partial derivatives of the function with respect to each variable are zero at this point. In simpler terms, at a critical point, the slope of the tangent line (or plane) to the function's graph is flat. For example, when you have a mountain peak or a valley in a graph, these are typically around where critical points occur.
To find the critical points of a function like

• Compute the first partial derivative with respect to each variable.
• Set each of these derivatives to zero.
• Solve the resulting system of equations to find combinations of variables that satisfy these conditions.

In the provided problem, this method is used to determine that the critical points of the function are \((0, 1)\), \((1, 0)\), and \((1, 1)\). These are the points where further investigation is necessary to determine if they lead to maxima, minima, or saddle points.

The Role of Partial Derivatives

Partial derivatives are essential in calculus for understanding how a function changes when you alter one of its input variables, keeping all other variables constant. For a function of two variables, the first partial derivative with respect to one variable gives you the rate of change of the function concerning that specific variable.
In the exercise given, the partial derivatives

• \(\frac{\partial f}{\partial x}\) tells us how the function changes with small changes in \(x\) when \(y\) is held constant.
• \(\frac{\partial f}{\partial y}\) represents the change in the function as \(y\) changes, with \(x\) held constant.

Setting these partial derivatives to zero provides conditions for finding critical points, which helps us explore where the function might have maxima, minima, or other interesting features. The understanding of partial derivatives is key for forming the Hessian matrix, a fundamental step in classifying critical points.

Local Extrema and their Classification

Local extrema refer to the maximum or minimum points of a function within a particular region. In the context of multi-variable calculus, determining whether a critical point is a local extremum involves further analysis using the second derivative test, which involves the Hessian matrix.
The Hessian matrix provides a structured way to classify critical points based on the second derivatives of the function. Analyzing the determinant of the Hessian matrix at a critical point can indicate the nature of the critical point.

• If the determinant is positive and the second partial derivative with respect to \(x\) is positive, it's a local minimum.
• If the determinant is positive and the second partial derivative is negative, it's a local maximum.
• If the determinant is negative, the critical point is a saddle point, indicating a change in concavity.
• If the determinant is zero, the test is inconclusive.

In the problem solution, at points like \((0,1)\) and \((1,0)\), the Hessian determinant is negative, confirming these points as saddle points. At \((1,1)\), the determinant is zero, providing no decisive conclusion about the nature of the critical point. Understanding these distinctions is vital for identifying local maxima, minima, or other behaviors of a function.