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Chapter 13: Problem 19

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

Short Answer

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

Step by step solution

01

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 \(f_y\): \(f_y = 4xe^{xy}\).
02

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).
03

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 \(f_{xy}(0,0) = 0\).
04

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.

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