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Magazine Chapter 14: Problem 14
Find all the local maxima, local minima, and saddle points of the functions. $$f(x, y)=x^{3}+3 x y+y^{3}$$
Short Answer
Expert verified
The function has a saddle point at (0,0), a saddle point at (1,-1), and a local maximum at (-1,-1).
Step by step solution
01
Find the First-Order Partial Derivatives
To find the critical points, first find the partial derivatives of function \( f(x, y) = x^3 + 3xy + y^3 \) with respect to both variables: - The partial derivative with respect to \( x \) is \( f_x = \frac{\partial f}{\partial x} = 3x^2 + 3y \).- The partial derivative with respect to \( y \) is \( f_y = \frac{\partial f}{\partial y} = 3x + 3y^2 \).
02
Solve for Critical Points
Set the partial derivatives equal to zero to find the critical points:- \( f_x = 3x^2 + 3y = 0 \) gives \( y = -x^2 \).- \( f_y = 3x + 3y^2 = 0 \) gives \( x = -y^2 \).Substitute \( y = -x^2 \) into \( x = -y^2 \), which results in \( x = -(x^2)^2 = -x^4 \). Solve for \( x \), we find that \( x = 0, -1, 1 \). Using \( y = -x^2 \), we get the critical points: \( (0,0), (1,-1), (-1,-1) \).
03
Find the Second-Order Partial Derivatives
The second-order partial derivatives are needed for the Hessian determinant:- \( f_{xx} = \frac{\partial^2 f}{\partial x^2} = 6x \)- \( f_{yy} = \frac{\partial^2 f}{\partial y^2} = 6y \)- \( f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 3 \).
04
Evaluate the Hessian and Classification
Calculate the Hessian determinant \( H = f_{xx}f_{yy} - (f_{xy})^2 \) at each of the critical points.- At \( (0,0) \), \( f_{xx} = 0 \), \( f_{yy} = 0 \), and \( f_{xy} = 3 \), thus \( H = 0 \cdot 0 - 3^2 = -9 \). Since \( H < 0 \), \( (0,0) \) is a saddle point.- At \( (1,-1) \), \( f_{xx} = 6 \), \( f_{yy} = -6 \), and \( f_{xy} = 3 \), thus \( H = 6 \cdot -6 - 3^2 = -45 \). Since \( H < 0 \), \( (1,-1) \) is a saddle point.- At \( (-1,-1) \), \( f_{xx} = -6 \), \( f_{yy} = -6 \), and \( f_{xy} = 3 \), thus \( H = -6 \cdot -6 - 3^2 = 27 \). Since \( H > 0 \) and \( f_{xx} < 0 \), \( (-1,-1) \) 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 essential in finding critical points of a multivariable function like \( f(x, y) = x^3 + 3xy + y^3 \). Imagine them as tools that help us understand how a function changes as we slightly tweak one of its variables at a time. In this exercise, we compute the partial derivatives with respect to both \( x \) and \( y \).
- First, the partial derivative with respect to \( x \) is \( f_x = 3x^2 + 3y \), capturing how \( f \) changes with \( x \) while keeping \( y \) constant.
- Second, the partial derivative with respect to \( y \) is \( f_y = 3x + 3y^2 \), which shows how \( f \) changes with \( y \) while fixing \( x \).
Local Maxima and Minima
Local maxima and minima are the peaks and valleys of a multivariable function. These are the points where the function \( f(x, y) \) does not appear to increase or decrease, making them stationary or flat.
To find these points, we:
To find these points, we:
- Set the first-order partial derivatives \( f_x \) and \( f_y \) equal to zero. Doing so helps identify critical points where the function "turns."
- In the given solution, critical points \( (0,0), (1,-1), (-1,-1) \) were derived by solving \( y = -x^2 \) and \( x = -y^2 \).
Saddle Points
Saddle points are more intriguing compared to standard peaks or valleys. Think of a saddle point as a surface that's part-way between an upward and downward bend, like a horse saddle. In the function \( f(x, y)\), when we locate a point where the Hessian determinant \( H \) is less than zero, it indicates a saddle point.
- In our exercise, evaluating at points \((0,0)\) and \((1,-1)\), we observed that the Hessian determinant was negative, revealing these points as saddle points.
- A saddle point is a location where the function surface is neither a maximum nor a minimum but a combination exhibiting both characteristics at the same point, depending on the direction.
Hessian Determinant
The Hessian determinant is a tool that gives us deeper insights into the behavior of critical points, crucial for distinguishing the nature of these points. It stems from the Hessian matrix, which contains second-order partial derivatives of the function \( f(x, y) \).
Computing the Hessian determinant involves:
In this exercise, we identified the critical point \( (-1,-1) \) as a local maximum and the points \( (0,0) \) and \( (1,-1) \) as saddle points, thanks to the Hessian determinant calculations which placed them critically between these classifications.
- Determining \( f_{xx} \), \( f_{yy} \), and \( f_{xy} \).
- For our function, \( f_{xx} = 6x \), \( f_{yy} = 6y \), and \( f_{xy} = 3 \).
- The Hessian determinant is \( H = f_{xx}f_{yy} - (f_{xy})^2 \).
In this exercise, we identified the critical point \( (-1,-1) \) as a local maximum and the points \( (0,0) \) and \( (1,-1) \) as saddle points, thanks to the Hessian determinant calculations which placed them critically between these classifications.
