Chapter 8: Problem 10
In Exercises 1 through 20 , find all critical points, and determine whether each point is a relative minimum, relative maximum. or a saddle point. $$ f(x, y)=2 x^{2}-x y+y^{2}-8 x+7 $$
Short Answer
Expert verified
The critical point is \( \left( \frac{16}{7}, \frac{8}{7} \right) \) and it is a relative minimum.
Step by step solution
01
Find Partial Derivatives
Calculate the first-order partial derivatives of the function with respect to both \( x \) and \( y \). We have: \[ f_x = \frac{\partial f}{\partial x} = 4x - y - 8 \]\[ f_y = \frac{\partial f}{\partial y} = -x + 2y \]
02
Set Partial Derivatives to Zero
To find the critical points, set the first-order partial derivatives equal to zero:\[ 4x - y - 8 = 0 \]\[ -x + 2y = 0 \]
03
Solve the System of Equations
Solve the system of equations from Step 2:- From \( -x + 2y = 0 \), we get \( x = 2y \).- Substitute \( x = 2y \) into \( 4x - y - 8 = 0 \): \ \( 4(2y) - y - 8 = 0 \) => \( 8y - y - 8 = 0 \) => \( 7y = 8 \) => \( y = \frac{8}{7} \).- Substitute \( y = \frac{8}{7} \) back to find \( x: x = 2 \cdot \frac{8}{7} = \frac{16}{7} \).Thus, the critical point is \( \left( \frac{16}{7}, \frac{8}{7} \right) \).
04
Calculate the Second Order Partial Derivatives
Find the second-order partial derivatives:\[ f_{xx} = \frac{\partial^2 f}{\partial x^2} = 4 \]\[ f_{yy} = \frac{\partial^2 f}{\partial y^2} = 2 \]\[ f_{xy} = f_{yx} = \frac{\partial^2 f}{\partial x \partial y} = -1 \]
05
Use the Second Derivative Test
The second derivative test for functions of two variables uses the determinant \( D \) of the Hessian matrix:\[ D = f_{xx} f_{yy} - (f_{xy})^2 \]Substitute the values:\[ D = 4 \times 2 - (-1)^2 = 8 - 1 = 7 \]Since \( D > 0 \) and \( f_{xx} > 0 \), the critical point \( \left( \frac{16}{7}, \frac{8}{7} \right) \) is a relative minimum.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
In calculus, partial derivatives are used to understand how a function behaves when one of its variables is changed, while others remain constant. This is particularly useful for functions of multiple variables. For example, when we look at a function like \[ f(x, y)=2 x^{2}-x y+y^{2}-8 x+7 \],we want to know how the function changes with respect to each variable separately.
Given the function, you can compute the partial derivative with respect to \( x \) by treating \( y \) as a constant. Similarly, compute the partial derivative with respect to \( y \) by holding \( x \) constant. This helps in simplifying the problem into easier, one-dimensional slices that are easier to analyze:
Given the function, you can compute the partial derivative with respect to \( x \) by treating \( y \) as a constant. Similarly, compute the partial derivative with respect to \( y \) by holding \( x \) constant. This helps in simplifying the problem into easier, one-dimensional slices that are easier to analyze:
- \( f_x = 4x - y - 8 \) shows how the function changes with \( x \)
- \( f_y = -x + 2y \) shows how the function changes with \( y \)
Second Derivative Test
The second derivative test is a method that utilizes second-order derivatives to determine the nature of critical points found by setting first-order partial derivatives to zero. Essentially, once you have a critical point, you use second derivatives to conclude whether it's a peak, a valley, or a saddle point.
The test checks conditions using:
The test checks conditions using:
- \( f_{xx} \) - the second derivative with respect to \( x \)
- \( f_{yy} \) - the second derivative with respect to \( y \)
- \( f_{xy} \) - the mixed partial derivative
Hessian Matrix
The Hessian matrix is a square matrix consisting of all the second-order partial derivatives of a multivariable function. It's crucial when performing the second derivative test because it encapsulates information about the curvature of the function around a critical point.
For a function of two variables \( f(x, y) \), the Hessian matrix is structured as follows:\[H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{yx} & f_{yy} \end{bmatrix}\]When calculating the determinant of this matrix, termed \( D \), we gain insights about the behavior at the critical point:
For a function of two variables \( f(x, y) \), the Hessian matrix is structured as follows:\[H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{yx} & f_{yy} \end{bmatrix}\]When calculating the determinant of this matrix, termed \( D \), we gain insights about the behavior at the critical point:
- A positive \( D \) indicates that the point could be a local min or max.
- A negative \( D \) implies it's a saddle point, characterizing different properties in different directions.