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Chapter 8: Problem 15

In Exercises 1 through 20 , find all critical points, and determine whether each point is a relative minimum, relative maximum. or a saddle point. $$ f(x, y)=x y-x^{3}-y^{2} $$

Short Answer

Expert verified
Critical points: (0,0) is a saddle point; \( \left( \frac{1}{6}, \frac{1}{12} \right) \) is a relative maximum.

Step by step solution

01

Find the Gradient

The gradient \( abla f(x, y) \) of the function \( f(x, y) = xy - x^3 - y^2 \) is given by the partial derivatives. Calculate the partial derivative with respect to \( x \): \( f_x = \frac{\partial}{\partial x}(xy - x^3 - y^2) = y - 3x^2 \). Now, calculate the partial derivative with respect to \( y \): \( f_y = \frac{\partial}{\partial y}(xy - x^3 - y^2) = x - 2y \). Thus, the gradient is \( abla f(x, y) = (y - 3x^2, x - 2y) \).
02

Find Critical Points

Set the gradient equal to zero to find the critical points. Solve the system of equations:1. \( y - 3x^2 = 0 \)2. \( x - 2y = 0 \)From (2), we express \( y \) in terms of \( x \): \( y = \frac{x}{2} \). Substituting into (1), we get \( \frac{x}{2} - 3x^2 = 0 \). Simplifying gives \( x(1 - 6x) = 0 \), so \( x = 0 \) or \( x = \frac{1}{6} \). For \( x = 0 \), \( y = 0 \); for \( x = \frac{1}{6} \), \( y = \frac{1}{12} \). The critical points are \( (0, 0) \) and \( \left(\frac{1}{6}, \frac{1}{12}\right) \).
03

Find Second Derivatives and the Hessian

Compute the second partial derivatives: \( f_{xx} = \frac{\partial^2 f}{\partial x^2} = -6x \), \( f_{yy} = \frac{\partial^2 f}{\partial y^2} = -2 \), and \( f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 1 \). The Hessian matrix \( H \) is:\[ H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{xy} & f_{yy} \end{bmatrix} = \begin{bmatrix} -6x & 1 \ 1 & -2 \end{bmatrix}. \]
04

Classify Critical Point (0,0)

At \( (0, 0) \): The Hessian is \( H = \begin{bmatrix} 0 & 1 \ 1 & -2 \end{bmatrix} \) with determinant \( \text{det}(H) = (0)(-2) - (1)(1) = -1 \). Since the determinant is negative, \( (0, 0) \) is a saddle point.
05

Classify Critical Point (1/6, 1/12)

At \( \left( \frac{1}{6}, \frac{1}{12} \right) \): The Hessian is \( H = \begin{bmatrix} -1 & 1 \ 1 & -2 \end{bmatrix} \) with determinant \( \text{det}(H) = (-1)(-2) - (1)(1) = 1 \). Since \( \text{det}(H) > 0 \) and \( f_{xx} = -1 < 0 \), the point \( \left( \frac{1}{6}, \frac{1}{12} \right) \) is a relative maximum.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Critical Points
Critical points of a function are the locations where the gradient (or derivative in one-dimensional cases) is zero or undefined. Identifying these points is crucial because they indicate potential maximums, minimums, or points of inflection.
  • To find critical points, set the gradient equal to zero. This provides a system of equations to solve for the variables of the function.
  • For example, with the function \( f(x, y) = xy - x^3 - y^2 \), calculate its gradient and solve the equations \( y - 3x^2 = 0 \) and \( x - 2y = 0 \).
  • From these equations, the solutions determine the critical points, which are \( (0, 0) \) and \( \left(\frac{1}{6}, \frac{1}{12}\right) \).
After determining the critical points, further analysis is needed to understand the nature of each point (e.g., local maxima, minima, or saddle points).
Gradient Calculation
The gradient of a multivariable function is a vector consisting of all its partial derivatives. It points in the direction of the greatest rate of increase of the function.
  • For the function \( f(x, y) = xy - x^3 - y^2 \), the gradient \( abla f(x, y) \) is calculated using the partial derivatives with respect to \( x \) and \( y \):\[ abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = (y - 3x^2, x - 2y). \]
This tells us how the function changes locally at any given point.
By setting the gradient to zero, you derive the conditions that determine the critical points of the function.
This process is foundational to understanding calculus optimization problems, as it highlights the necessary steps to thoroughly explore the behavior of multivariate functions.
Hessian Matrix Analysis
The Hessian matrix is a square matrix of second-order partial derivatives of a function, essential for determining the nature of critical points in multivariable calculus.
  • For the given function, calculate the second derivatives: \( f_{xx} = -6x \), \( f_{yy} = -2 \), and \( f_{xy} = 1 \).
  • The Hessian matrix \( H \) is constructed as follows:\[ H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{xy} & f_{yy} \end{bmatrix} = \begin{bmatrix} -6x & 1 \ 1 & -2 \end{bmatrix}. \]
Use the Hessian determinant to classify the critical points:
- If \( \text{det}(H) > 0 \): Check \( f_{xx} \). If negative, it's a local maximum. If positive, a local minimum.
- If \( \text{det}(H) < 0 \): It's a saddle point.For points like \( (0, 0) \) with a negative determinant, it's a saddle point.
But if \( \text{det}(H) > 0 \) and one part is less than zero, as at \( \left( \frac{1}{6}, \frac{1}{12} \right) \), it indicates a local maximum. This thorough analysis is invaluable for classifying points in optimization problems.

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