Problem 19 - College Algebra and Calculus: An Applied Approach

Question

Examine the function for relative extrema and saddle points. $$ f(x, y)=4 e^{x y} $$

Step-by-Step Solution

Answer

The function \(f(x, y) = 4e^{xy}\) has a saddle point at (0,0).

Step 1: Evaluate the first order partial derivatives

First, calculate the partial derivatives of \( f(x, y) \) with respect to \( x \) and \( y \) respectively. For \(f_x\) we get: \(f_x = 4ye^{xy}\) and for \(f_y\): \(f_y = 4xe^{xy}\).

Step 2: Find critical points

Now set the first order derivatives equal to zero and solve for \( x \) and \( y \) to find the critical points. We get: \(0 = 4ye^{xy}\) and \(0 = 4xe^{xy}\). Here, e^{xy} is never zero, so we get \(y = 0\) from the first equation and \(x = 0\) from the second. Thus, the only critical point is (0,0).

Step 3: Evaluate the second order partial derivatives

Calculate the second order partial derivatives of \( f(x, y) \) with respect to \( x \), \( y \) and mixed \( xy \). We get: \(f_{xx} = 4y^2e^{xy}\), \(f_{yy} = 4x^2e^{xy}\), and \(f_{xy} = f_{yx} = 4xe^{xy} + 4ye^{xy}\). Now, evaluate these second order partial derivatives at the critical point (0,0). We get \(f_{xx}(0,0) = 0\), \(f_{yy}(0,0) = 0\), and \(f_{xy}(0,0) = 0\).

Step 4: Apply the second partial derivative test

The second partial derivative test classifies critical points by examining the sign of \(D = f_{xx} f_{yy} - (f_{xy})^2\). Calculate the determinant \(D\) using the second order partial derivatives at (0,0). We get \(D = (0 * 0) - (0)^2 = 0\). When \(D < 0\), the function has a saddle point at the critical point.