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Chapter 14: Problem 14

Find all the local maxima, local minima, and saddle points of the functions in Exercises \(1-30 .\) $$ f(x, y)=x^{3}+3 x y+y^{3} $$

Short Answer

Expert verified
The function has a local maximum at \((-1, -1)\) and a saddle point at \((0, 0)\).

Step by step solution

01

Find the First Partial Derivatives

To locate the critical points of the function \( f(x, y) = x^{3} + 3xy + y^{3} \), first find the partial derivatives with respect to \( x \) and \( y \). The partial derivative of \( f \) with respect to \( x \) is:\[ f_x = \frac{\partial}{\partial x} (x^{3} + 3xy + y^{3}) = 3x^{2} + 3y. \]The partial derivative of \( f \) with respect to \( y \) is:\[ f_y = \frac{\partial}{\partial y} (x^{3} + 3xy + y^{3}) = 3x + 3y^{2}. \]
02

Solve the System for Critical Points

Set the partial derivatives equal to zero to find the critical points:1. \( 3x^{2} + 3y = 0 \)2. \( 3x + 3y^{2} = 0 \)Solving these simultaneously:\[ x^{2} + y = 0 \] implies \( y = -x^{2} \).Substitute into the second equation:\[ x + (-x^{2})^{2} = 0 \rightarrow x + x^{4} = 0 \]\[ x(1 + x^{3}) = 0 \]Thus, \( x = 0 \) or \( x = \sqrt[3]{-1} \). Correspondingly, \( y = 0 \) for \( x = 0 \), and \( y = -(-1)^{2/3} = -1 \) for \( x = -1 \). So, we have the potential critical points \( (0, 0) \) and \( (-1, -1) \).
03

Calculate the Second Partial Derivatives

To classify the critical points, we find the second partial derivatives:\[ f_{xx} = \frac{\partial^2 f}{\partial x^2} = 6x \]\[ f_{yy} = \frac{\partial^2 f}{\partial y^2} = 6y \]\[ f_{xy} = f_{yx} = \frac{\partial^2 f}{\partial x \partial y} = 3 \]
04

Determine the Nature of Each Critical Point

Compute the Hessian determinant at each critical point:\[ H = f_{xx}f_{yy} - (f_{xy})^2 = (6x)(6y) - (3)^2 \]For \( (0, 0) \):\[ H = (6 \times 0)(6 \times 0) - 9 = -9 \]Since \( H < 0 \), \( (0, 0) \) is a saddle point.For \( (-1, -1) \):\[ H = (6 \times -1)(6 \times -1) - 9 = 36 - 9 = 27 \]Since \( H > 0 \) and \( f_{xx} = -6 < 0 \), \((-1, -1)\) 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
In calculus, partial derivatives play a crucial role when working with multivariable functions. A partial derivative represents the rate of change of a function with respect to one variable, while keeping other variables constant. For the function \[ f(x, y) = x^3 + 3xy + y^3 \] we calculate the partial derivatives with respect to \( x \) (by treating \( y \) as a constant) and with respect to \( y \) (by treating \( x \) as a constant).
  • The partial derivative with respect to \( x \) is \( f_x = 3x^2 + 3y \).
  • The partial derivative with respect to \( y \) is \( f_y = 3x + 3y^2 \).
To find critical points, we set these derivatives to zero and solve the equations. This helps us identify potential points where the function reaches local maxima, minima, or saddle points.
Hessian Determinant
The Hessian determinant helps classify the nature of critical points found from partial derivatives. It is derived from the second partial derivatives of a function. For a function \( f(x, y) \), the Hessian determinant \( H \) is given by:\[H = f_{xx}f_{yy} - (f_{xy})^2.\]The sign of \( H \) determines the nature of each critical point.
  • If \( H > 0 \) and \( f_{xx} > 0 \), the point is a local minimum.
  • If \( H > 0 \) and \( f_{xx} < 0 \), the point is a local maximum.
  • If \( H < 0 \), the point is a saddle point.
  • If \( H = 0 \), further analysis is needed, as the test is inconclusive.
In our example, we found \( H = -9 \) at \((0,0)\) indicating a saddle point, and \( H = 27 \) at \((-1, -1)\) suggesting a local maximum.
Local Maxima
Local maxima occur at points where a function reaches a peak in a small neighborhood around those points. These points are higher than any other nearby values, but not necessarily the highest point in the entire domain. To determine a local maximum, we use the Hessian determinant in conjunction with the second partial derivative test.In our exercise, \( (-1,-1) \) was identified as a local maximum:
  • The Hessian \( H = 27 \) at this point is positive.
  • We also found \( f_{xx} = -6 \) which is less than zero, confirming a peak.
This means that near \((-1, -1)\), our function does not rise above \( f(-1, -1) \), making it a local maximum point.
Saddle Points
Saddle points are interesting critical points where the function does not have a local extremum. A saddle point occurs where the graph of the function flattens before changing direction. Here, the value at a saddle point is neither a peak (maximum) nor a trough (minimum).In mathematical terms, a saddle point is where the Hessian determinant \( H \) is negative using our second derivative test.For the function we analyzed, the point \((0, 0)\) was determined to be a saddle point because:
  • The Hessian determinant \( H = -9 \) indicating that the changes in slope cancel each other out, creating this characteristic shape.
This means that the function behaves differently in multiple directions at \((0, 0)\), climbing in some directions and falling in others.

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Most popular questions from this chapter

In Exercises \(55-60,\) verify that \(w_{x y}=w_{y x}\). $$w=x \sin y+y \sin x+x y$$

Express \(v_{x}\) in terms of \(u\) and \(y\) if the equations \(x=v \ln u\) and \(y=u \ln v\) define \(u\) and \(v\) as functions of the independent variables \(x\) and \(y,\) and if \(v_{x}\) exists. (Hint: Differentiate both equations with respect to \(x\) and solve for \(v_{x}\) by eliminating \(u_{x}\).)

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