Chapter 13: Problem 28
Find all the local maxima, local minima, and saddle points of the functions. $$f(x, y)=e^{x}\left(x^{2}-y^{2}\right)$$
Short Answer
Expert verified
The function has a local maximum at \((-2, 0)\) and a saddle point at \((0, 0)\).
Step by step solution
01
Find the Partial Derivatives
First, calculate the first-order partial derivatives of the function \( f(x, y) = e^x(x^2 - y^2) \).- For \( f_x \) (the partial derivative with respect to \( x \)): \[ f_x = e^x (2x - y^2) + e^x (x^2 - y^2) = e^x (2x + x^2 - y^2) \]- For \( f_y \) (the partial derivative with respect to \( y \)): \[ f_y = -2y e^x \]
02
Find the Critical Points
Set the first-order partial derivatives equal to zero and solve for \( x \) and \( y \):- For \( f_x = 0 \), solve: \[ e^x (2x + x^2 - y^2) = 0 \] This implies \( 2x + x^2 - y^2 = 0 \) since \( e^x eq 0 \).- For \( f_y = 0 \), solve: \[ -2y e^x = 0 \] This implies \( y = 0 \).Substitute \( y = 0 \) back into the first equation: \[ 2x + x^2 = 0 \] Factor this equation: \[ x(2 + x) = 0 \] Thus, \( x = 0 \) or \( x = -2 \).Critical points are thus \( (0, 0) \) and \( (-2, 0) \).
03
Determine the Nature of the Critical Points
Find the second-order partial derivatives:- \( f_{xx} = e^x (2 + 2x) + e^x (2x + x^2 - y^2) = e^x \, (4x + x^2 + 2) \)- \( f_{yy} = -2e^x \)- \( f_{xy} = f_{yx} = -2y e^x \)Use the second derivative test:Calculate the determinant \( D = f_{xx} f_{yy} - (f_{xy})^2 \) and evaluate it at each critical point:1. At \( (0, 0) \): - \( f_{xx} = e^0(0 + 0 + 2) = 2 \) - \( f_{yy} = -2e^0 = -2 \) - \( f_{xy} = 0 \) - \( D = (2)(-2) - 0^2 = -4 \) Since \( D < 0 \), \( (0, 0) \) is a saddle point.2. At \( (-2, 0) \): - \( f_{xx} = e^{-2}(4(-2) + (-2)^2 + 2) = e^{-2}(-8 + 4 + 2) = -2e^{-2} \) - \( f_{yy} = -2e^{-2} \) - \( f_{xy} = 0 \) - \( D = (-2e^{-2})(-2e^{-2}) - 0 = 4e^{-4} \) Since \( D > 0 \) and \( f_{xx} < 0 \), \( (-2, 0) \) is a local maximum.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
Partial derivatives are a cornerstone of multivariable calculus, providing a way to understand the rate of change of a function in relation to just one of its variables, while the other remains constant. This is similar to how derivatives function in single-variable calculus, but here, it's expanded to multiple dimensions.
For a function of two variables, like the one in our example, you will often calculate two partial derivatives:
For a function of two variables, like the one in our example, you will often calculate two partial derivatives:
- With respect to the first variable, say, \( x \), written as \( f_x \).
- With respect to the second variable, say, \( y \), written as \( f_y \).
- \( f_x = e^{x}(2x + x^2 - y^2) \), showing how \( f \) changes with \( x \).
- \( f_y = -2y e^{x} \), showing how \( f \) changes with \( y \).
Critical Points
Identifying critical points is a crucial step in analyzing a function to understand its local behavior – such as locating potential maxima, minima, or saddle points. Critical points in the context of multivariable functions are points where the first-order partial derivatives are zero.
These points are akin to the concept of turning points in single-variable calculus, marking where the function doesn’t increase or decrease immediately, indicating a potential maximum, minimum, or something in-between.
In our function example, we find that setting \( f_x = 0 \) results in \( 2x + x^2 - y^2 = 0 \). Simultaneously, setting \( f_y = 0 \) yields \( y = 0 \). Solving these equations, we find critical points at
These points are akin to the concept of turning points in single-variable calculus, marking where the function doesn’t increase or decrease immediately, indicating a potential maximum, minimum, or something in-between.
In our function example, we find that setting \( f_x = 0 \) results in \( 2x + x^2 - y^2 = 0 \). Simultaneously, setting \( f_y = 0 \) yields \( y = 0 \). Solving these equations, we find critical points at
- \( (0, 0) \)
- \( (-2, 0) \)
Second Derivative Test
The second derivative test is an essential tool to distinguish the nature of critical points found in a function. By examining the second-order partial derivatives, this test helps determine if a critical point is a local maximum, minimum, or a saddle point.
To apply the second derivative test in multivariable calculus, you need:
To apply the second derivative test in multivariable calculus, you need:
- \( f_{xx} \) – the second partial derivative with respect to \( x \)
- \( f_{yy} \) – the second partial derivative with respect to \( y \)
- \( f_{xy} \) – the mixed partial derivative with respect to \( x \) and \( y \)
- If \( D > 0 \) and \( f_{xx} < 0 \), it's a local maximum.
- If \( D > 0 \) and \( f_{xx} > 0 \), it's a local minimum.
- If \( D < 0 \), the point is a saddle point.
Saddle Point
A saddle point is a peculiar type of critical point that is neither a local maximum nor a local minimum. It is named because of its resemblance to a horse saddle, indicating regions that curve upwards and downwards.
A saddle point occurs when the test results in \( D < 0 \) in the second derivative test, reflecting a point where the function does not consistently increase or decrease in all directions. For instance, the point \( (0, 0) \) in the provided solution is a saddle point.
At this point, although it might seem the function is positioning itself at a critical position potentially leading to maximum or minimum values—by further evaluation, it's clear the directional rates of the function's change are mixed, creating a hyperbolic surface rather than a peak or valley. Understanding saddle points is crucial for optimizing complex systems, as they show direction-dependent behaviors.
A saddle point occurs when the test results in \( D < 0 \) in the second derivative test, reflecting a point where the function does not consistently increase or decrease in all directions. For instance, the point \( (0, 0) \) in the provided solution is a saddle point.
At this point, although it might seem the function is positioning itself at a critical position potentially leading to maximum or minimum values—by further evaluation, it's clear the directional rates of the function's change are mixed, creating a hyperbolic surface rather than a peak or valley. Understanding saddle points is crucial for optimizing complex systems, as they show direction-dependent behaviors.