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Chapter 10: Problem 4

The functions are defined for all \((x, y) \in \boldsymbol{R}^{2} .\) Find all candidates for local extrema, and use the Hessian matrix to determine the type (maximum, minimum, or saddle point). \(f(x, y)=x y-2 y^{2}\)

Short Answer

Expert verified
The critical point is (0, 0) and it is a saddle point.

Step by step solution

01

Find the First Partial Derivatives

Determine the first partial derivatives of the function. For \(f(x, y) = x y - 2 y^2\), we find:- \( f_x = \frac{\partial}{\partial x}(x y - 2y^2) = y \)- \( f_y = \frac{\partial}{\partial y}(x y - 2y^2) = x - 4y \)
02

Solve System of Equations

Find the critical points by setting the first partial derivatives to zero:- \( f_x = y = 0 \)- \( f_y = x - 4y = 0 \)Solving these gives \( y = 0 \) and substituting into \( x - 4y = 0 \) gives \( x = 0 \). Hence, the critical point is (0, 0).
03

Find the Second Partial Derivatives

Calculate the second partial derivatives needed for the Hessian matrix:- \( f_{xx} = \frac{\partial}{\partial x}(f_x) = 0 \)- \( f_{yy} = \frac{\partial}{\partial y}(f_y) = -4 \)- \( f_{xy} = \frac{\partial}{\partial y}(f_x) = 1 \)- \( f_{yx} = \frac{\partial}{\partial x}(f_y) = 1 \)
04

Construct the Hessian Matrix

The Hessian matrix \( H \) is as follows:\[H = \begin{bmatrix}0 & 1 \1 & -4\end{bmatrix}\]
05

Calculate the Determinant of the Hessian

Compute the determinant of the Hessian matrix to determine the nature of the critical point:\( \text{det}(H) = (0)(-4) - (1)(1) = -1 \).
06

Determine the Nature of the Critical Point

Since \( \text{det}(H) = -1 < 0 \), the critical point (0, 0) is a saddle point.

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Key Concepts

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

Hessian Matrix
The Hessian matrix holds a crucial role in determining the nature of critical points in multivariable functions. It is a square matrix comprised of second-order partial derivatives. This means, for a function like ours, \(f(x, y) = x y - 2 y^2\), the Hessian matrix captures how the function curves in space.
To construct it, perform the following steps:
  • Calculate all relevant second-order partial derivatives: \(f_{xx}, f_{yy}, f_{xy}, \text{and} \ f_{yx}\).
  • For our function, these derivatives are: \(f_{xx} = 0\), \(f_{yy} = -4\), \(f_{xy} = 1\), and \(f_{yx} = 1\).
  • Assemble them into the matrix \[ H = \begin{bmatrix} f_{xx} & f_{xy} \ {f_{yx} & f_{yy} \end{bmatrix} = \begin{bmatrix} 0 & 1 \ 1 & -4 \end{bmatrix}\].
The determinant of this Hessian matrix, \( \text{det}(H) \), helps identify the type of critical point. If it is positive, the point might be a local minimum or maximum, and if negative, the point is generally a saddle point. In this case, \(\text{det}(H) = -1\), thus confirming a saddle point.
Partial Derivatives
Partial derivatives are vital to understanding how a function changes with respect to each variable individually. They are the building blocks for determining the behavior of multivariable functions.
For the function \(f(x, y) = x y - 2 y^2\), the first-order partial derivatives are calculated by differentiating with respect to one variable while holding the other constant:
  • The partial derivative with respect to \(x\) is \(f_x = y\).
  • The partial derivative with respect to \(y\) is \(f_y = x - 4y\).
To find critical points, set these first-order derivatives to zero. The equations become:
  • \(f_x = y = 0\)
  • \(f_y = x - 4y = 0\)
Solving these simultaneously gives the critical point (0, 0). Partial derivatives thus guide us in exploring where the function could potentially have maxima, minima, or saddle points.
Critical Points
Critical points occur where the first partial derivatives of a function equal zero. These points are potential locations for local maxima, minima, or saddle points in the function landscape.
To find them for \(f(x, y) = x y - 2 y^2\), follow these steps:
  • Find the partial derivatives: \(f_x = y\) and \(f_y = x - 4y\).
  • Set these derivatives equal to zero to identify potential critical points.
  • This results in the system of equations: \(y = 0\) and \(x - 4y = 0\).
Solving the above, we find the critical point at (0, 0).
At this point, the Hessian matrix and its determinant reveal the point's nature. Since \(\text{det}(H) = -1\), which is less than zero, we classify the point (0, 0) as a saddle point, indicating a region of mixed riding high and low directions in the function's graph.

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Most popular questions from this chapter

Find \(\frac{d y}{d x}\) if \((\sin x+\cos y) x^{2}=0\)

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Refer to the negative binomial host-parasitoid model. Problems \(7,8,11\), and 12 are best done with the help of a spreadsheet, but can also be done with a calculator. The negative binomial model is a discrete-generation host- parasitoid model of the form $$ \begin{array}{l} N_{t+1}=b N_{t}\left(1+\frac{a P_{t}}{k}\right)^{-k} \\ P_{t+1}=c N_{t}\left[1-\left(1+\frac{a P_{t}}{k}\right)^{ k}\right] \end{array} $$ for \(t=0,1,2, \ldots\) Show that when the initial parasitoid density is \(P_{0}=0\), the negative binomial model reduces to $$ N_{t+1}=b N_{t} $$ With \(N_{0}\) denoting the initial host density, find an expression for \(N_{t}\) in terms of \(N_{0}\) and the parameter \(b\).
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