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Chapter 7: Problem 35

Find all points \((x, y)\) where \(f(x, y)\) has a possible relative maximum or minimum. Then, use the second-derivative test to determine, if possible, the nature of \(f(x, y)\) at each of these points. If the second-derivative test is inconclusive, so state. $$f(x, y)=x^{4}-12 x^{2}-4 x y-y^{2}+16$$

Short Answer

Expert verified
The points are (0, 0) 鈥� relative maximum, (2, -4) and (-2, 4) 鈥� saddle points.

Step by step solution

01

- Find the first partial derivatives

Compute the first partial derivatives of the function with respect to both x and y. These derivatives will help find the critical points. \[ f_x = \frac{\text{d}}{\text{dx}}(x^4 - 12x^2 - 4xy - y^2 + 16) = 4x^3 - 24x - 4y \] \[ f_y = \frac{\text{d}}{\text{dy}}(x^4 - 12x^2 - 4xy - y^2 + 16) = -4x - 2y \]
02

- Find the critical points

Set the first partial derivatives equal to zero to find the critical points. \[ 4x^3 - 24x - 4y = 0 \] \[ -4x - 2y = 0 \] From the second equation: \[ y = -2x \] Substitute this back into the first equation: \[ 4x^3 - 24x - 4(-2x) = 0 \] Simplify: \[ 4x^3 - 16x = 0 \] Factor: \[ 4x(x^2 - 4) = 0 \] Thus, \[ x = 0 \] or \[ x = \text{卤} 2 \]. Hence, the solutions are: \[ (0, 0) \] \[ (2, -4) \] \[ (-2, 4) \]
03

- Compute second partial derivatives

Compute the second partial derivatives needed for the second-derivative test. \[ f_{xx} = \frac{\text{d}}{\text{dx}}(4x^3 - 24x - 4y) = 12x^2 - 24 \] \[ f_{yy} = \frac{\text{d}}{\text{dy}}(-4x - 2y) = -2 \] \[ f_{xy} = \frac{\text{d}}{\text{dx}}(-4x - 2y) = -4 \]
04

- Apply the second-derivative test

Use the second-derivative test at each critical point. The determinant of the Hessian matrix will help determine the nature of each critical point. The Hessian determinant is given by: \[ D = f_{xx}f_{yy} - (f_{xy})^2 \] At (0, 0): \[ f_{xx} = 12(0)^2 - 24 = -24 \] \[ D = (-24)(-2) - (-4)^2 = 48 - 16 = 32 \] \[ D > 0 \] and \[ f_{xx} < 0 \], so (0,0) is a relative maximum. At (2, -4): \[ f_{xx} = 12(2)^2 - 24 = 24 \] \[ D = (24)(-2) - (-4)^2 = -48 - 16 = -64 \] \[ D < 0 \], so (2, -4) is a saddle point. At (-2, 4): \[ f_{xx} = 12(-2)^2 - 24 = 24 \] \[ D = (24)(-2) - (-4)^2 = -48 - 16 = -64 \] \[ D < 0 \], so (-2, 4) is a saddle point.

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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 a way to extend the concept of a derivative to functions of multiple variables. They measure how a function changes as each individual variable changes, while keeping the other variables constant. For a function of two variables, say \( f(x, y) \), the partial derivatives with respect to \( x \) and \( y \) are noted as \( f_x \) and \( f_y \).

To compute the partial derivatives of the given function \( f(x, y) \), we treat \( y \) as a constant when differentiating with respect to \( x \), and treat \( x \) as a constant when differentiating with respect to \( y \). In our example, these are:
  • First partial derivative with respect to \( x \): \( f_x = 4x^3 - 24x - 4y \)
  • First partial derivative with respect to \( y \): \( f_y = -4x - 2y \)
Hessian determinant
To classify the nature of the critical points found from the first partial derivatives, we use the second-derivative test, which involves the Hessian determinant. The Hessian matrix \( H \) for a function of two variables \( f(x, y) \) is a 2x2 matrix composed of the second partial derivatives:

\begin{matrix} H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{yx} & f_{yy} \ \ \ \ \ \ \ \ \ \ \ \ H is \text{forms } _(

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Most popular questions from this chapter

Four hundred eighty dollars are available to fence in a rectangular garden. The fencing for the north and south sides of the garden costs \(\$ 10\) per foot, and the fencing for the east and west sides costs \(\$ 15\) per foot. Find the dimensions of the largest possible garden.

Find all points \((x, y)\) where \(f(x, y)\) has a possible relative maximum or minimum. $$f(x, y)=-8 y^{3}+4 x y+4 x^{2}+9 y^{2}$$

Find all points \((x, y)\) where \(f(x, y)\) has a possible relative maximum or minimum. $$f(x, y)=-8 y^{3}+4 x y+9 y^{2}-2 y$$

Find the values of \(x\) and \(y\) that minimize \(18 x^{2}+12 x y+4 y^{2}+6 x-4 y+5\) subject to the constraint \(3 x+2 y-1=0.\)

Find the values of \(x\) and \(y\) that minimize \(f(x, y)=x-x y+2 y^{2}\) subject to the constraint \(x-y+1=0.\)
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