Problem 7
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). $$ f(x, y)=e^{-x^{2}-y^{2}} $$
Step-by-Step Solution
Verified Answer
The critical point \((0, 0)\) is a local maximum for the function \(f(x, y)=e^{-x^2-y^2}\).
1Step 1: Find Critical Points
To find critical points, we need the first-order partial derivatives of \(f(x, y)\). The partial derivative with respect to \(x\) is \(f_x(x, y) = \frac{\partial}{\partial x} e^{-x^2-y^2} = -2xe^{-x^2-y^2}\) and with respect to \(y\) is \(f_y(x, y) = \frac{\partial}{\partial y} e^{-x^2-y^2} = -2ye^{-x^2-y^2}\). Set both partial derivatives to zero: \(-2xe^{-x^2-y^2} = 0\) and \(-2ye^{-x^2-y^2} = 0\). Solving these equations, we get \(x = 0\) and \(y = 0\). Thus, the critical point is \((0,0)\).
2Step 2: Compute the Hessian Matrix
The Hessian matrix is used to determine the type of critical point. The second partial derivatives are \(f_{xx} = \frac{\partial^2}{\partial x^2} e^{-x^2-y^2} = (4x^2 - 2)e^{-x^2-y^2}\), \(f_{yy} = \frac{\partial^2}{\partial y^2} e^{-x^2-y^2} = (4y^2 - 2)e^{-x^2-y^2}\), and \(f_{xy} = f_{yx} = \frac{\partial^2}{\partial x \partial y} e^{-x^2-y^2} = 4xye^{-x^2-y^2}\). At the critical point \((0, 0)\), the Hessian matrix becomes: \[ H(0, 0) = \begin{bmatrix} -2 & 0 \ 0 & -2 \end{bmatrix} \].
3Step 3: Determine the Type of Critical Point
To determine the type of critical point, check the determinant of the Hessian matrix \(H\) at \((0, 0)\): \(\det(H) = (-2)(-2) - (0)(0) = 4\), which is positive. Since \(f_{xx} = -2 < 0\), this indicates that the critical point \((0, 0)\) is a local maximum.
Key Concepts
Local ExtremaPartial DerivativesCritical Points
Local Extrema
When dealing with a function of several variables, such as the one we have here, local extrema refer to points where the function reaches either a local maximum or local minimum. These extremas are crucial because they provide insight about regions where the function's output is either unusually high or low compared to nearby points.
To find these local extrema in two-dimensional functions, we use methods from calculus, similar to the techniques for single-variable functions but with some differences. Specifically, we seek out critical points first, which are potential candidates for local extrema.
Once the critical points are established, mathematical tools, like the Hessian matrix, are used to determine the nature of these points. It verifies whether these candidates are indeed local maxima, minima, or just saddle points. Overall, identifying local extrema is important for optimization problems and understanding the behavior of the function around specific points.
To find these local extrema in two-dimensional functions, we use methods from calculus, similar to the techniques for single-variable functions but with some differences. Specifically, we seek out critical points first, which are potential candidates for local extrema.
Once the critical points are established, mathematical tools, like the Hessian matrix, are used to determine the nature of these points. It verifies whether these candidates are indeed local maxima, minima, or just saddle points. Overall, identifying local extrema is important for optimization problems and understanding the behavior of the function around specific points.
Partial Derivatives
Partial derivatives are an extension of the derivative concepts from single to multi-variable calculus, which allow us to measure how a function changes when only one of its variables is varied while others are held constant.
For the function given by \(f(x, y) = e^{-x^2 - y^2}\), the partial derivative with respect to \(x\) is \(f_x(x, y) = -2xe^{-x^2-y^2}\), and the partial derivative with respect to \(y\) is \(f_y(x, y) = -2ye^{-x^2-y^2}\). Here, they tell us how the function value changes as \(x\) or \(y\) changes, holding the other constant.
Setting these partial derivatives to zero helps us find critical points, essential for identifying potential local extrema. The method mimics finding stationary points in single-variable calculus, where the derivative is zero. Understanding partial derivatives is vital in multi-variable calculus, as they form the foundation for more advanced concepts like gradients and directional derivatives.
For the function given by \(f(x, y) = e^{-x^2 - y^2}\), the partial derivative with respect to \(x\) is \(f_x(x, y) = -2xe^{-x^2-y^2}\), and the partial derivative with respect to \(y\) is \(f_y(x, y) = -2ye^{-x^2-y^2}\). Here, they tell us how the function value changes as \(x\) or \(y\) changes, holding the other constant.
Setting these partial derivatives to zero helps us find critical points, essential for identifying potential local extrema. The method mimics finding stationary points in single-variable calculus, where the derivative is zero. Understanding partial derivatives is vital in multi-variable calculus, as they form the foundation for more advanced concepts like gradients and directional derivatives.
Critical Points
Critical points in multi-variable functions are similar to those in single-variable functions: they are points where the first partial derivatives are zero or undefined, indicating potential local maxima, minima, or saddle points.
In our exercise, finding critical points involved calculating the first-order partial derivatives of the function \(f(x, y) = e^{-x^2 - y^2}\), and then equating them to zero. Solving the system of equations \(-2xe^{-x^2-y^2} = 0\) and \(-2ye^{-x^2-y^2} = 0\), we found that both partial derivatives equal zero when \(x = 0\) and \(y = 0\).
Thus, the point \((0, 0)\) emerged as a critical point. To classify this critical point, further tools like the Hessian matrix are needed to decide if this point is a local minimum, maximum, or neither. Recognizing and analyzing critical points are pivotal in understanding the landscape of a multi-variable function.
In our exercise, finding critical points involved calculating the first-order partial derivatives of the function \(f(x, y) = e^{-x^2 - y^2}\), and then equating them to zero. Solving the system of equations \(-2xe^{-x^2-y^2} = 0\) and \(-2ye^{-x^2-y^2} = 0\), we found that both partial derivatives equal zero when \(x = 0\) and \(y = 0\).
Thus, the point \((0, 0)\) emerged as a critical point. To classify this critical point, further tools like the Hessian matrix are needed to decide if this point is a local minimum, maximum, or neither. Recognizing and analyzing critical points are pivotal in understanding the landscape of a multi-variable function.
Other exercises in this chapter
Problem 7
Locate the following points in a three-dimensional Cartesian coordinate system: (a) \((1,3,2)\) (b) \((-1,-2,1)\) (c) \((0,1,-1)\) (d) \((2,0,2)\)
View solution Problem 7
The tangent plane at the indicated poini \(\left(x_{0}, y_{0}, z_{0}\right)\) exists. Find its equation. \(f(x, y)=e^{2 x^{2}+y^{2}} ;\left(1,0, e^{2}\right)\)
View solution Problem 7
In Problems \(1-8\), find the gradient of each function. $$ f(x, y)=\ln \left(\frac{x}{y}+\frac{y}{x}\right) $$
View solution Problem 7
7\. Write down an expression for \(\frac{d z}{d t}\) where \(z=f(x, y)\) with \(x=u(t)\) and \(y=v(t) .\)
View solution