Problem 15
Question
Examine the function for relative extrema and saddle points. $$ f(x, y)=\frac{1}{2} x y $$
Step-by-Step Solution
Verified Answer
The function \(f(x, y) = \frac{1}{2} xy\) has a saddle point at (0,0).
1Step 1: Compute the first order partial derivatives
The partial derivative of \(f\) with respect to \(x\) is denoted as \(f_x\) and the partial derivative of \(f\) with respect to \(y\) is denoted as \(f_y\). So we have \(f_x = \frac{\partial f}{\partial x} = \frac{1}{2} y\) and \(f_y = \frac{\partial f}{\partial y} = \frac{1}{2} x\). The critical points occur where both these first order partial derivatives are zero, i.e., \(f_x = f_y = 0\). This gives us the point (0,0).
2Step 2: Compute the second order partial derivatives
The second order partial derivatives are \(f_{xx} = \frac{\partial^2 f}{\partial x^2} = 0\), \(f_{yy} = \frac{\partial^2 f}{\partial y^2} = 0\) and \(f_{xy} = \frac{\partial^2 f}{\partial y \partial x} = \frac{1}{2}\).
3Step 3: Compute the determinant (D) of the Hessian Matrix
The determinant (D) of a 2x2 Hessian Matrix \([f_{xx} \quad f_{xy} \ f_{xy} \quad f_{yy}]\) is given by \(D = f_{xx} * f_{yy} - (f_{xy})^2\). From Step 2, we know \(f_{xx} = 0\), \(f_{yy} = 0\) and \(f_{xy} = \frac{1}{2}\). Therefore, \(D = 0 - (\frac{1}{2})^2 = -\frac{1}{4}\).
4Step 4: Classification of the critical point found in Step 1
The determinant (D = -1/4) is less than zero, which means that the critical point (0,0) is a saddle point.
Key Concepts
First Order Partial DerivativesSecond Order Partial DerivativesHessian Matrix DeterminantRelative Extrema
First Order Partial Derivatives
When analyzing a function of two variables like \( f(x, y) = \frac{1}{2}xy \), the first order partial derivatives act as a mathematical tool to understand how the function changes as each variable varies independently. Simply put, if you change the value of \(x\) while keeping \(y\) constant, the first order partial derivative with respect to \(x\), denoted as \(f_x\), tells you the rate of this change and vice versa for \(f_y\).
Finding these first order partial derivatives is essential; They provide critical points (when set to zero) which are potential locations of peaks, valleys, or saddle points in the multivariable landscape of the function.
Finding these first order partial derivatives is essential; They provide critical points (when set to zero) which are potential locations of peaks, valleys, or saddle points in the multivariable landscape of the function.
Second Order Partial Derivatives
Diving deeper, second order partial derivatives measure the curvature of the function's graph. They represent how the rate of change (given by the first order partial derivatives) is itself changing. If the first order partial derivatives are about velocity, think of second order partial derivatives as acceleration.
Specifically, the second order partial derivatives \(f_{xx}\), \(f_{yy}\), and \(f_{xy}\) (the mixed partial derivative) give critical insight into the concavity and interactions between the variables of the function. In the context of the given function, these values help in predicting the nature of the critical point found.
Specifically, the second order partial derivatives \(f_{xx}\), \(f_{yy}\), and \(f_{xy}\) (the mixed partial derivative) give critical insight into the concavity and interactions between the variables of the function. In the context of the given function, these values help in predicting the nature of the critical point found.
Hessian Matrix Determinant
The Hessian matrix determinant is a sophisticated concept in multivariable calculus that encapsulates the second order partial derivatives into a single value, denoted as \(D\). The Hessian matrix itself is a square matrix of these second order partial derivatives, and when we compute the determinant of this matrix, we get a scalar value that holds the key to identifying the type of critical point we are dealing with at a specific location on the function's surface.
The determinant combines the information about the function's concavity in the directions of both variables into one number. A positive determinant suggests a local minimum or maximum (depending on the signs of the second order partial derivatives), while a negative one points to the thrilling possibility of a saddle point - a point that is neither purely a peak nor a valley.
The determinant combines the information about the function's concavity in the directions of both variables into one number. A positive determinant suggests a local minimum or maximum (depending on the signs of the second order partial derivatives), while a negative one points to the thrilling possibility of a saddle point - a point that is neither purely a peak nor a valley.
Relative Extrema
Discussing relative extrema entails talking about the mountain tops and ocean floors on the graph of a function of two variables - these are the local maximums and minimums. They are 'relative' because they are not necessarily the absolute highest or lowest points on the graph, but they are the highest or lowest in their immediate vicinity.
Identifying these points involves analyzing the critical points - where the first order partial derivatives are zero - and then consulting the second order partial derivatives and the Hessian matrix determinant. If the determinant is positive, and if the second order partial derivatives are positive, we're at a local minimum; if they are negative, it's a local maximum. However, if the determinant is negative, we're on more unusual terrain - a saddle point, which is what we've encountered with the function \(f(x, y) = \frac{1}{2}xy\).
Identifying these points involves analyzing the critical points - where the first order partial derivatives are zero - and then consulting the second order partial derivatives and the Hessian matrix determinant. If the determinant is positive, and if the second order partial derivatives are positive, we're at a local minimum; if they are negative, it's a local maximum. However, if the determinant is negative, we're on more unusual terrain - a saddle point, which is what we've encountered with the function \(f(x, y) = \frac{1}{2}xy\).
Other exercises in this chapter
Problem 14
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Use the regression capabilities of a graphing utility or a spreadsheet to find the least squares regression line for the given points. $$ (0,0),(1,1),(3,4),(4,2
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