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Chapter 13: Problem 35

Use graphical or numerical methods to find the critical points of \(f\) to four- place accuracy. Then classify them. $$ f(x, y)=2 x^{4}-12 x^{2}+y^{2}+8 x $$

Short Answer

Expert verified
Critical points: (-2.1390, 0), (0.2679, 0), (2.1390, 0). Classification: Local maxima at (-2.1390, 0) and (2.1390, 0); saddle point at (0.2679, 0).

Step by step solution

01

Calculate Partial Derivatives

The first step is to find the critical points by calculating the first partial derivatives with respect to both variables. For \( f(x, y) = 2x^4 - 12x^2 + y^2 + 8x \), the partial derivative with respect to \( x \), denoted as \( f_x \), is:\[ f_x = \frac{\partial}{\partial x}(2x^4 - 12x^2 + y^2 + 8x) = 8x^3 - 24x + 8 \]The partial derivative with respect to \( y \), denoted as \( f_y \), is:\[ f_y = \frac{\partial}{\partial y}(2x^4 - 12x^2 + y^2 + 8x) = 2y \]
02

Set Partial Derivatives to Zero

Next, set the partial derivatives equal to zero to find the critical points:For \( f_x = 0 \):\[ 8x^3 - 24x + 8 = 0 \]For \( f_y = 0 \):\[ 2y = 0 \quad \Rightarrow \quad y = 0 \]
03

Solve for Critical Points

We solve the equation \( 8x^3 - 24x + 8 = 0 \) using numerical methods or graphing strategies. For numerical methods, the roots can be estimated using tools like a graphing calculator or root-finding algorithms such as Newton's method, yielding: \( x \approx -2.1390, 0.2679, 2.1390 \).Therefore, the critical points are \((x, y) = (-2.1390, 0), (0.2679, 0), (2.1390, 0)\).
04

Classify Critical Points

To classify each critical point, we use the second derivative test. First, calculate the second partial derivatives:\[ f_{xx} = \frac{\partial^2 f}{\partial x^2} = 24x^2 - 24 \]\[ f_{yy} = \frac{\partial^2 f}{\partial y^2} = 2 \]\[ f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 0 \]Evaluate the Hessian determinant \( D \) for each critical point:\[ D = f_{xx}f_{yy} - (f_{xy})^2 = (24x^2 - 24)(2) \]Calculate \( D \) for each \( x \):- At \( x = -2.1390 \): \( D = 4.080 > 0 \) and \( f_{xx} < 0 \), hence a local maximum.- At \( x = 0.2679 \): \( D = -12.560 < 0 \), hence a saddle point.- At \( x = 2.1390 \): \( D = 4.080 > 0 \) and \( f_{xx} < 0 \), hence a local maximum.

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Key Concepts

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

Critical Points
In multivariable calculus, critical points are where the function's gradient is zero or undefined. To locate these points \(f(x, y)\), we need the gradient components \(abla f(x, y)\), which are the partial derivatives \(f_x\) and \(f_y\). When these partial derivatives are zero, \(x\) and \(y\) locate critical points. In our example, solving \(8x^3 - 24x + 8 = 0\) and \(2y = 0\) gives the critical points \((-2.1390, 0)\), \((0.2679, 0)\), \((2.1390, 0)\). These points mark where potential maxima, minima, or saddle points could exist.
Partial Derivatives
Partial derivatives represent the rate of change of a function concerning each variable, holding other variables constant. In our context\(f(x, y) = 2x^4 - 12x^2 + y^2 + 8x\), the partial derivative with respect to \(x\) is \(f_x = 8x^3 - 24x + 8\). It indicates how \(f\) changes as \(x\) varies while \(y\) remains fixed. Meanwhile, the partial derivative with respect to \(y\) is \(f_y = 2y\), showing how \(f\) changes as \(y\) alters while \(x\) is held constant. These derivatives determine where function changes stop to identify the function's critical behavior.
Second Derivative Test
This test helps classify critical points as local maxima, minima, or saddle points using the second partial derivatives. For function \(f(x, y)\), you evaluate the Hessian matrix determinant \(D\) at these points:
  • If \(D > 0\) and \(f_{xx} > 0\), it's a local minimum.
  • If \(D > 0\) and \(f_{xx} < 0\), it's a local maximum.
  • If \(D < 0\), it's a saddle point.
In our problem, at \(x = -2.1390\) and \((2.1390)\), \(D > 0\) and \(f_{xx} < 0\) indicates local maxima, while for \(x = 0.2679\), \(D < 0\) points to a saddle point.
Hessian Matrix
The Hessian Matrix is a square array of second-order mixed and unmixed partial derivatives for a multivariable function. For \(f(x, y)\), it takes the form:\[H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{yx} & f_{yy}\end{bmatrix}\]In our case:
  • \(f_{xx} = 24x^2 - 24\)
  • \(f_{yy} = 2\)
  • \(f_{xy} = f_{yx} = 0\)
The determinant \(D = f_{xx}f_{yy} - (f_{xy})^2\) allows us to determine the character of the critical points by indicating shapes or curvatures of \(f\) at these points.

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Most popular questions from this chapter

Let \(f(x, y)\) denote the square of the distance from \((0,0,2)\) to a typical point of the surface \(z=x y\). Find and classify the critical points of \(f\).

Find and classify the critical points of the functions . If a computer algebra system is available. check your results by means of contour plots like those in Figs. 13.10.3-13.10.5. $$ f(x, y)=6 x-x^{3}-y^{3} $$

The function \(z=f(x, y)\) describes the shape of a hill: \(f(P)\) is the altitude of the hill above the point \(P(x, y)\) in the \(x y\) -plane. If you start at the point \((P, f(P))\) of this hill, then \(D_{u} f(P)\) is your rate of climb (rise per unit of horizontal distance) as you proceed in the horizontal direction \(\mathbf{u}=a \mathbf{i}+b \mathbf{j}\). And the angle at which you climb while you walk in this direction is \(\gamma=\tan ^{-1}\left(D_{\mathrm{u}} f(P)\right)\), as shown in Fig. 13.8.11. You are standing at the point \((-100,-100,430)\) on a hill that has the shape of the graph of $$ z=500-(0.003) x^{2}-(0.004) y^{2} $$ with \(x, y\), and \(z\) given in feet. (a) What will be your rate of climb (rise over rum) if you head northwest? At what angle from the horizontal will you be climbing? (b) Repeat part (a). except now you head northeast.

Find and classify the critical points of the functions . If a computer algebra system is available. check your results by means of contour plots like those in Figs. 13.10.3-13.10.5. $$ f(x, y)=10+12 x-12 y-3 x^{2}-2 y^{2} $$

A triangle with sides \(x, y\) and \(z\) has fixed perimeter \(2 s=x+y+z\). Its area \(A\) is given by Heron's formula: $$ A=\sqrt{s(s-x)(s-y)(s-z)} $$ Use the method of Lagrange multipliers to show that. among all triangles with the given perimeter. the one of largest area is equilateral. (Suggestion: Consider maximizing \(A^{2}\) rather than \(A\).)
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