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Chapter 13: Problem 7

Examine the function for relative extrema and saddle points. $$ f(x, y)=2 x^{2}+2 x y+y^{2}+2 x-3 $$

Short Answer

Expert verified
The function \(f(x, y)=2 x^{2}+2 x y+y^{2}+2 x-3\) has a relative minimum at point (1, -1).

Step by step solution

01

- Find the first partial derivatives

They are, \(f_x = 4x + 2y + 2\) and \(f_y = 2x + 2y\).
02

- Solve \(f_x = 0\) and \(f_y = 0\)

From \(4x + 2y + 2 = 0\), we get \(2x + y = -1\) and from \(2x + 2y = 0\), we get \(x + y = 0\). From solving the system of equations \(2x + y = -1\) and \(x + y = 0\), we get \(x=1, y=-1\).
03

- Calculate the second order partial derivatives

The second order partial derivatives are \(f_{xx} = 4\), \(f_{xy}=f_{yx}=2\), and \(f_{yy}=2\).
04

- Calculate the determinant of the Hessian matrix and analyze critical points

The Hessian matrix, H, is given by \(H = [[f_{xx}, f_{xy}], [f_{yx}, f_{yy}]] = [[4, 2], [2, 2]]\). The determinant of H, denoted as D, is calculated by \(D = f_{xx}*f_{yy} - (f_{xy})^2 = 4*2 - (2)^2 = 4\). Since \(D>0\) and \(f_{xx}>0\), the critical point (1, -1) is a relative minimum.

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