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Chapter 16: Problem 1

Find the extrema and saddle points of \(f\). \(f(x, y)=-x^{2}-4 x-y^{2}+2 y-1\)

Short Answer

Expert verified
The function has a local maximum at \((-2, 1)\).

Step by step solution

01

Find Partial Derivatives

First, we need to find the partial derivatives of the function \( f(x, y) = -x^{2} - 4x - y^{2} + 2y - 1 \). The partial derivative of \( f \) with respect to \( x \) is: \[ f_{x}(x, y) = \frac{\partial}{\partial x}(-x^{2} - 4x - y^{2} + 2y - 1) = -2x - 4 \]The partial derivative of \( f \) with respect to \( y \) is:\[ f_{y}(x, y) = \frac{\partial}{\partial y}(-x^{2} - 4x - y^{2} + 2y - 1) = -2y + 2 \]
02

Set Partial Derivatives to Zero

Next, we set the partial derivatives obtained in Step 1 to zero to find the critical points.For \( f_{x} = 0 \):\[ -2x - 4 = 0 \]Solve for \( x \):\[ x = -2 \]For \( f_{y} = 0 \):\[ -2y + 2 = 0 \]Solve for \( y \):\[ y = 1 \]Thus, the critical point is \((x, y) = (-2, 1)\).
03

Compute Second Partial Derivatives

Calculate the second partial derivatives to analyze the critical point further.\( f_{xx} = \frac{\partial^{2}}{\partial x^{2}}(-x^{2} - 4x - y^{2} + 2y - 1) = -2 \)\( f_{yy} = \frac{\partial^{2}}{\partial y^{2}}(-x^{2} - 4x - y^{2} + 2y - 1) = -2 \)\( f_{xy} = \frac{\partial^{2}}{\partial x \partial y}(-x^{2} - 4x - y^{2} + 2y - 1) = 0 \)
04

Determine Type of Critical Point Using Second Derivative Test

Use the second derivative test for functions of two variables to determine the nature of the critical point. The test uses the following determinant:\[ D = f_{xx}f_{yy} - (f_{xy})^{2} \]Substituting the values we computed in Step 3:\[ D = (-2)(-2) - (0)^{2} = 4 \]Since \( D > 0 \) and \( f_{xx} < 0 \), the critical point \((x, y) = (-2, 1)\) 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 mathematics, partial derivatives are vital for understanding how functions change. They represent the rate of change of a multivariable function with respect to one variable at a time, while keeping others constant.
To calculate the partial derivatives of a function of two variables like \( f(x, y) = -x^{2} - 4x - y^{2} + 2y - 1 \),
  • First, treat \( y \) as a constant and differentiate with respect to \( x \), which yields \( f_{x}(x, y) = -2x - 4 \).
  • Similarly, treat \( x \) as a constant, differentiate with respect to \( y \), leading to \( f_{y}(x, y) = -2y + 2 \).
Understanding these partial derivatives is essential because they provide the slope of the tangent lines to the curve, which helps in identifying the behavior of the function at specific points.
Critical Points
Critical points are where a function's behavior changes from increasing to decreasing, or vice versa. They arise where the partial derivatives equal zero. In the example with \( f(x, y) = -x^{2} - 4x - y^{2} + 2y - 1 \):
  • Set \( f_{x} = 0 \) to find the \( x \)-component, which implies \( -2x - 4 = 0 \), solving gives \( x = -2 \).
  • Set \( f_{y} = 0 \) to find the \( y \)-component, which implies \( -2y + 2 = 0 \), solving gives \( y = 1 \).
Thus, the critical point is \((-2, 1)\). This critical point is essential for determining where a function might reach a maximum, minimum, or saddle point.
Second Derivative Test
Once critical points are identified, the second derivative test helps categorize them as minima, maxima, or saddle points using second partial derivatives and their relationships.For \( f(x, y) = -x^{2} - 4x - y^{2} + 2y - 1 \), compute the following:
  • Second partial derivative with respect to \( x \): \( f_{xx} = -2 \).
  • Second partial derivative with respect to \( y \): \( f_{yy} = -2 \).
  • Mixed partial derivative: \( f_{xy} = 0 \).
Consider the determinant \( D = f_{xx}f_{yy} - (f_{xy})^{2} \). For this function, \( D = 4 \).
Because \( D > 0 \) and \( f_{xx} < 0 \), by the second derivative test, the critical point \((-2, 1)\) is a local maximum. Understanding this test is crucial for accurate identification of the function's nature at the critical points.
Functions of Two Variables
Functions of two variables, like \( f(x, y) \), map pairs of input values into a single output. These are often used in fields such as physics, economics, and engineering to model complex scenarios.
This type of function requires tools like partial derivatives to analyze because they involve changes in multiple directions. In practice, these functions can represent surfaces in three-dimensional space, where:
  • The x-axis represents one variable.
  • The y-axis represents another variable.
  • The z-axis is often the output or the function's value.
Locating critical points and applying tests like the second derivative test are essential elements in understanding these surfaces. This allows for an analysis of regions where the functions reach their extrema or change direction abruptly, acting as saddle points.

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Most popular questions from this chapter

Use a chain rule. A function \(f\) of two variables is homogeneous of degree \(n\) if \(f(t x, t y)=t^{n} f(x, y)\) for every \(t\) such that \((t x, t y)\) is in the domain of \(f\). Show that, for such functions, \(x f_{x}(x, y)+y f_{y}(x, y)=n f(x, y)\). (Hint: Differentiate \(f(t x, t y)\) with respect to \(t .\)

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