Question

Examine the function for relative extrema and saddle points. $$ f(x, y)=\frac{1}{2} x y $$

Short answer

The function \(f(x, y) = \frac{1}{2}xy\) has a saddle point at the origin \((0,0)\).

Step-by-step solution

Find the first partial derivatives

Calculate the first partial derivatives of the function \(f(x, y)\) with respect to \(x\) and \(y\).\n\n Derivative with respect to \(x\): \(f_x = \frac{1}{2} y\)\n\n Derivative with respect to \(y\): \(f_y = \frac{1}{2} x\)

Find the critical points

Critical points occur where the first partial derivatives are both zero or undefined. However, neither \(f_x\) nor \(f_y\) are ever undefined in this case. Thus set both \(f_x\) and \(f_y\) to zero and solve for \(x\) and \(y\). This gives the point \((0,0)\).

Find the second partial derivatives

We calculate the second partial derivatives of the function \(f(x, y)\). We also need to calculate the mixed second-order derivative.\n\n Second derivative with respect to \(x\): \(f_{xx} = 0\)\n\n Second derivative with respect to \(y\): \(f_{yy} = 0\)\n\n Second derivative with respect to \(x\) and then \(y\): \(f_{xy} = \frac{1}{2}\)

Applied the second derivative test

To find out whether our critical point \((0,0)\) is a local maximum, minimum or saddle point, we apply the second derivative test. We need to evaluate the determinant of the Hessian matrix, \(D = f_{xx} f_{yy} - (f_{xy})^2\), at the critical point. However, due to the fact that both \(f_{xx}\) and \(f_{yy}\) are zero, the determinant \(D\) will always be negative, indicating that we have a saddle point at \((0,0)\).