Question

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

Short answer

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

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 \(f_y\): \(f_x = \frac{\partial f}{\partial x} = 1 + 2y - 2x\) \(f_y = \frac{\partial f}{\partial y} = 1 + 2x - 2y\)

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)

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: \(f_{xx} = \frac{\partial^2 f}{\partial x^2} = -2\) \(f_{yy} = \frac{\partial^2 f}{\partial y^2} = -2\) \(f_{xy} = f_{yx} = \frac{\partial^2 f}{\partial x \partial y} = 2\)

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.