Problem 15 - College Algebra and Calculus: An Applied Approach

Question

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

Step-by-Step Solution

Answer

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

Step 1: 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 \(y\): \(f_y = \frac{1}{2} x\)

Step 2: 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 \((0,0)\).

Step 3: 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 \(y\): \(f_{xy} = \frac{1}{2}\)

Step 4: 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 \((0,0)\).