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Chapter 13: Problem 7

Find all stationary points \(\left(f_{x}=f_{y}=0\right)\) in \(1-16 .\) Separate minimum from maximum from saddle point. Test \(13 \mathrm{~K}\) applies to \(a=f_{x x}, b=f_{x y}, c=f_{y y}\) \(-x^{2}+2 x y-3 y^{2}\)

Short Answer

Expert verified
The stationary point is (0, 0) and is a local maximum.

Step by step solution

01

Find the Partial Derivatives

The function given is \( f(x, y) = -x^2 + 2xy - 3y^2 \). We find the partial derivatives with respect to \( x \) and \( y \). The partial derivative with respect to \( x \) is \( f_x = -2x + 2y \) and with respect to \( y \) is \( f_y = 2x - 6y \).
02

Set the Partial Derivatives to Zero

Stationary points occur where the first derivatives are equal to zero. Set \( f_x = 0 \) and \( f_y = 0 \): \(-2x + 2y = 0\) and \(2x - 6y = 0\).
03

Solve the System of Equations

Solve the equations \(-2x + 2y = 0\) and \(2x - 6y = 0\). From \(-2x + 2y = 0\), we get \(y = x\). Substitute \( y = x \) into \(2x - 6y = 0\) to get \(2x - 6x = 0\), which simplifies to \( -4x = 0 \). Therefore, \( x = 0 \), and subsequently, \( y = 0 \). Thus, the stationary point is \((0, 0)\).
04

Determine the Type of Stationary Point Using the Second Derivative Test

Calculate the second partial derivatives: \( f_{xx} = -2 \), \( f_{yy} = -6 \), \( f_{xy} = f_{yx} = 2 \). Now apply the second derivative test. Compute the determinant of the Hessian matrix: \( D = f_{xx} f_{yy} - (f_{xy})^2 = (-2)(-6) - (2)^2 = 12 - 4 = 8 \).
05

Classify the Stationary Point

Since \( D > 0 \), and \( f_{xx} < 0 \), the stationary point at \((0, 0)\) is a local maximum according to the second derivative test.

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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 like regular derivatives, but they're used when you have a function with more than one variable, like our function \( f(x, y) = -x^2 + 2xy - 3y^2 \).
Instead of just considering how the function changes as one variable changes, you look at how it changes **with respect to each variable** individually.
  • To find the partial derivative with respect to \( x \) (written as \( f_x \)), treat \( y \) as a constant and differentiate only with respect to \( x \). For the given function, \( f_x = -2x + 2y \).
  • Similarly, to find the partial derivative with respect to \( y \) (written as \( f_y \)), treat \( x \) as a constant and differentiate with respect to \( y \). Here, we have \( f_y = 2x - 6y \).
These partial derivatives help us find **stationary points** 鈥� points where the function doesn't increase or decrease, but the gradient is zero. This means solving \( f_x = 0 \) and \( f_y = 0 \). That's how we find the points where we need to check if there are any minimums, maximums, or saddle points.
Second Derivative Test
The second derivative test is crucial for understanding the nature of stationary points in multivariable calculus. Once we get a stationary point, in this case \((0, 0)\), we have to decide if it's a minimum, maximum, or a saddle point. This test involves computing second partial derivatives from the original function:
  • \( f_{xx} \) is the partial derivative of \( f_x \) with respect to \( x \), which is \( -2 \) for our function.
  • \( f_{yy} \) is the partial derivative of \( f_y \) with respect to \( y \), which equals \( -6 \).
  • Then, there's \( f_{xy} \) (or \( f_{yx} \)), meaning the derivative of \( f_x \) with respect to \( y \), or \( f_y \) with respect to \( x \), both results in \( 2 \).
We now use these derivatives to find the determinant \( D \) of the Hessian matrix: \[D = f_{xx}f_{yy} - (f_{xy})^2\] Using the determined values, our \(D = (-2)(-6) - (2)^2 = 12 - 4 = 8\). A positive \( D > 0 \) and a negative \( f_{xx} < 0 \) mean the stationary point is a local maximum.
Hessian Matrix
The Hessian matrix is a square matrix of second-order mixed partial derivatives of a scalar-valued function. For a function with two variables like ours, the Hessian is a 2x2 matrix that helps us assess the curvature of the function at any stationary point. The Hessian matrix for our function is \[\begin{bmatrix} f_{xx} & f_{xy} \ f_{yx} & f_{yy}\end{bmatrix}\]In this case, it translates to: \[\begin{bmatrix} -2 & 2 \ 2 & -6\end{bmatrix}\]The determinant of the Hessian, \( D = f_{xx}f_{yy} - (f_{xy})^2 \), will tell us about the nature of the stationary points.
  • If \( D > 0 \), the nature depends on \( f_{xx} \): if \( f_{xx} > 0 \), it's a local minimum; if \( f_{xx} < 0 \), it's a local maximum.
  • If \( D < 0 \), the point is a saddle point.
  • And if \( D = 0 \), the test is inconclusive.
This method gives a comprehensive insight into the behavior of multivariable functions at critical points.

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Most popular questions from this chapter

(a) Maximize \(f=a x+b y\) on the circle \(x^{2}+y^{2}=1\). (b) Minimize \(x^{2}+y^{2}\) on the linc \(a x+b y=1\).

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