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Chapter 9: Problem 22

Give a similar discussion for $$ f(x, y)=2 x^{3}+6 x y^{2}-3 x^{2}+3 y^{2} $$

Short Answer

Expert verified
Question: Determine whether the critical points found for the function $$f(x, y) = 2x^3 + 6xy^2 - 3x^2 + 3y^2$$ are local maxima, local minima, or saddle points. Solution: 1. Compute the first partial derivatives of the function, $$f_x(x, y) = 6x^2 + 6y^2 - 6x$$ and $$f_y(x, y) = 12xy + 6y$$. 2. Find the critical points by solving the system of equations: $$6x^2 + 6y^2 - 6x = 0$$ and $$12xy + 6y = 0$$. 3. Compute the second partial derivatives: $$f_{xx}(x, y) = 12x - 6, f_{xy}(x, y) = 12y, f_{yx}(x, y) = 12y, f_{yy}(x, y) = 12x + 6$$. 4. Construct the Hessian matrix: $$ H = \begin{bmatrix} 12x-6 & 12y \\ 12y & 12x+6 \end{bmatrix} $$ 5. Calculate the eigenvalues of the Hessian matrix using the equation: $$ \begin{vmatrix} 12x-6-\lambda & 12y \\ 12y & 12x+6-\lambda \end{vmatrix} = (12x-6-\lambda)(12x+6-\lambda)- 144y^2 = 0 $$ 6. Study the signs of the eigenvalues for each critical point: both positive eigenvalues correspond to a local minimum, both negative eigenvalues correspond to a local maximum, and opposite-signed eigenvalues correspond to a saddle point.

Step by step solution

01

Compute the first partial derivatives

The first step is to compute the first partial derivatives of the function with respect to x and y, which are denoted as \(f_{x}(x, y)\) and \(f_{y}(x, y)\): $$ f_{x}(x, y) = \frac{\partial f}{\partial x} = 6x^2 + 6y^2 - 6x $$ $$ f_{y}(x, y) = \frac{\partial f}{\partial y} = 12xy + 6y $$
02

Find the critical points

Now, we need to find the critical points by setting both first partial derivatives equal to zero and solving the system of equations: $$ 6x^2 + 6y^2 - 6x = 0 $$ $$ 12xy + 6y = 0 $$ Solving this system of equations will give us the critical points (x, y) for which the function has local extrema or saddle points.
03

Compute the second partial derivatives

Next, we need to compute the second partial derivatives \(f_{xx}(x, y)\), \(f_{xy}(x, y)\), \(f_{yx}(x, y)\), and \(f_{yy}(x, y)\): $$ f_{xx}(x, y) = \frac{\partial^2 f}{\partial x^2} = 12x - 6 $$ $$ f_{xy}(x, y) = \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial f_y}{\partial x} = 12y $$ $$ f_{yx}(x, y) = \frac{\partial^2 f}{\partial y \partial x} = \frac{\partial f_x}{\partial y} = 12y $$ $$ f_{yy}(x, y) = \frac{\partial^2 f}{\partial y^2} = 12x + 6 $$
04

Construct the Hessian matrix

We will now construct the Hessian matrix using the second partial derivatives we found in step 3, which is a 2x2 matrix given by: $$ H = \begin{bmatrix} f_{xx}(x, y) & f_{xy}(x, y) \\ f_{yx}(x, y) & f_{yy}(x, y) \end{bmatrix} = \begin{bmatrix} 12x-6 & 12y \\ 12y & 12x+6 \end{bmatrix} $$
05

Analyze the eigenvalues of the Hessian matrix

To determine whether the critical points found in step 2 are local maxima, minima, or saddle points, we need to calculate the eigenvalues of the Hessian matrix and study their signs. If both eigenvalues are positive, the critical point is a local minimum; if both are negative, it is a local maximum; if they have opposite signs, it is a saddle point. Calculate the eigenvalues by finding the determinant of the matrix \(H - \lambda I\), and set it to zero: $$ \begin{vmatrix} 12x-6-\lambda & 12y \\ 12y & 12x+6-\lambda \end{vmatrix} = (12x-6-\lambda)(12x+6-\lambda)- 144y^2 = 0 $$ Once the eigenvalues are found, we can now analyze their signs for each critical point.

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Key Concepts

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

Critical Points
A critical point is where the derivative of a function is zero or undefined. In the context of multivariable calculus, it means looking for points where both partial derivatives are zero. This is crucial because at these points, the function might change its behavior, like reaching a peak or a valley. To find the critical points for the given function, we set the first partial derivatives with respect to both variables (\(x\) and \(y\)) to zero. This results in a system of equations:
  • \(6x^2 + 6y^2 - 6x = 0\)
  • \(12xy + 6y = 0\)
Solving these equations helps us find the \((x, y)\) pairs where critical points might occur, hinting at important aspects of the function curve such as local extrema or saddle points.
Understanding this process involves recognizing that partial derivatives represent the slope of the function in the direction of \(x\) or \(y\). Where these slopes are zero, we have points worthy of further investigation for potential extrema.
Hessian Matrix
The Hessian matrix is a square matrix of second-order partial derivatives. It provides valuable information about the curvature of a function at a given point. Constructing the Hessian matrix involves arranging these second partial derivatives in a systematic way:
  • \(f_{xx}(x, y) = \frac{\partial^2 f}{\partial x^2} = 12x - 6\)
  • \(f_{xy}(x, y) = \frac{\partial^2 f}{\partial x \partial y} = 12y\)
  • \(f_{yx}(x, y) = \frac{\partial^2 f}{\partial y \partial x} = 12y\)
  • \(f_{yy}(x, y) = \frac{\partial^2 f}{\partial y^2} = 12x + 6\)
The matrix itself looks like this:\[H = \begin{bmatrix}12x-6 & 12y \12y & 12x+6\end{bmatrix}\]This matrix helps decide the nature of the critical points found previously by telling us about the function's local behavior at those points. If it's positive definite, it's likely a local minimum; if negative definite, a local maximum. If neither, it could be a saddle point.
Eigenvalues
Eigenvalues of the Hessian matrix serve as indicators of the type of extremum present at a critical point. They are calculated by solving the characteristic polynomial derived from the determinant of the matrix \( H - \lambda I \):\[\begin{vmatrix}12x-6-\lambda & 12y \12y & 12x+6-\lambda\end{vmatrix} = 0\]The solutions, \(\lambda\), indicate how the function behaves at the critical points:
  • If both eigenvalues are positive, the point is a local minimum.
  • If both eigenvalues are negative, the point is a local maximum.
  • If eigenvalues are of opposite signs, the point is a saddle point.
Eigenvalues give a clear, algebraic method of determining such characteristics, making them a powerful tool when analyzing multivariable functions.
Local Maxima and Minima
Local maxima and minima are points in which the function reaches its highest or lowest value, compared to points immediately around them. The determination between the function being at a maximum or a minimum at a critical point can often be made by examining the eigenvalues of the Hessian matrix.
  • If both eigenvalues are positive, the function curves upwards, indicating a trough or valley at the critical point鈥攖his is a local minimum.
  • If both are negative, the function curves downwards, suggesting a peak or crest鈥攖his is a local maximum.
Contrast this with saddle points, where the directionality of the curve changes. At such points, one eigenvalue is positive and the other negative, indicating the function behaves differently in different directions.
In problem-solving, identifying these points helps to predict the behavior of a system modeled by the function, making the understanding of maxima and minima integral in fields like optimization and economics.

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Most popular questions from this chapter

Define \(f(0,0)=0\) and $$ f(x, y)=\frac{x^{3}}{x^{2}+y^{2}} \quad \text { if }(x, y) \neq(0,0) $$ (a) Prove that \(D_{1} f\) and \(D_{2} f\) are bounded functions in \(R^{2}\). (Hence \(f\) is continuous.) (b) Let u be any unit vector in \(R^{2}\). Show that the directional derivative \(\left(D_{w} f\right)(0,0)\) exists, and that its absolute value is at most 1 . (c) Let \(\gamma\) be a differentiable mapping of \(R^{1}\) into \(R^{2}\) (in other words, \(\gamma\) is a differentiable curve in \(\left.R^{2}\right)\), with \(\gamma(0)=(0,0)\) and \(\left|\gamma^{\prime}(0)\right|>0 .\) Put \(g(t)=f(\gamma(t))\) and prove that \(g\) is differentiable for every \(t \in R^{1}\). If \(\gamma \in \mathscr{C}^{\prime}\), prove that \(g \in \mathscr{8}\) '. (d) In spite of this, prove that \(f\) is not differentiable at \((0,0)\). Hint: Formula (40) fails.

Define \(f\) in \(R^{2}\) by $$ f(x, y)=2 x^{3}-3 x^{2}+2 y^{3}+3 y^{2} $$ (a) Find the four points in \(R^{2}\) at which the gradient of \(f\) is zero. Show that \(f\) has exactly one local maximum and one local minimum in \(R^{2}\). (b) Let \(S\) be the set of all \((x, y) \in R^{2}\) at which \(f(x, y)=0\). Find those points of \(S\) that have no neighborhoods in which the equation \(f(x, y)=0\) can be solved for \(y\) in terms of \(x\) (or for \(x\) in terms of \(y\) ). Describe \(S\) as precisely as you can.

If \(f(0,0)=0\) and $$ f(x, y)=\frac{x y}{x^{2}+y^{2}} \quad \text { if }(x, y) \neq(0,0) $$ prove that \(\left(D_{1} f\right)(x, y)\) and \(\left(D_{2} f\right)(x, y)\) exist at every point of \(R^{2}\), although \(f\) is not continuous at \((0.0)\).

Prove (as asserted in Sec. 9.6) that \(B A\) is linear if \(A\) and \(B\) are linear transformations. Prove also that \(A^{-1}\) is linear and invertible.

Fix two real numbers \(a\) and \(b, 0
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