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Chapter 13: Problem 16

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

Short Answer

Expert verified
The critical point of the given function is at (1, 1). However, Hessian determinant equals zero at this point, providing no definite conclusion about whether this point is a relative extremum or a saddle point. Further analysis or graphing the function might provide a definitive answer.

Step by step solution

01

Find the first-order partial derivatives

These are found by differentiating \(f(x, y)\) partially with respect to \(x\) and \(y\) respectively. Let's denote the first-order partial derivatives as \(f_x\) and \(f_y\): \(f_x = \frac{\partial f}{\partial x} = 1 + 2y - 2x\) \(f_y = \frac{\partial f}{\partial y} = 1 + 2x - 2y\)
02

Find the critical points

Set the first-order partial derivatives equal to zero. This will give the critical points: \(1 + 2y - 2x = 0\) \(1 + 2x - 2y = 0\)Solving these equations simultaneously gives \(x = 1\) and \(y = 1\) So, the critical point is (1, 1)
03

Find the second-order partial derivatives

Differentiate the first-order partial derivatives \(f_x\) and \(f_y\) with respect to \(x\) and \(y\) respectively. Then, differentiate \(f_x\) with respect to \(y\) and \(f_y\) with respect to \(x\). These are \(f_{xx}, f_{yy}, f_{xy}, f_{yx}\) respectively: \(f_{xx} = \frac{\partial^2 f}{\partial x^2} = -2\) \(f_{yy} = \frac{\partial^2 f}{\partial y^2} = -2\) \(f_{xy} = f_{yx} = \frac{\partial^2 f}{\partial x \partial y} = 2\)
04

Analyze the critical point

Evaluate the second-order derivatives at the critical point (1, 1). \(f_{xx}(1,1) = -2, f_{yy}(1,1) = -2, f_{xy}(1,1) = 2\)The Hessian determinant, given by \(D = f_{xx} * f_{yy} - f_{xy}^2 = -2 * -2 - 2^2 = 4 - 4 = 0\)The Hessian determinant should be \(D > 0\) for a point to be local extremum and \(D < 0\) for saddle point. However, here \(D = 0\) which does not give definite conclusion. We have to plot the function or analyze it furthermore to derive proper conclusion.

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