Problem 18 - College Algebra and Calculus: An Applied Approach

Question

Examine the function for relative extrema and saddle points. $$ f(x, y)=3 e^{-\left(x^{2}+y^{2}\right)} $$

Step-by-Step Solution

Answer

The function \(f(x, y)=3 e^{-\left(x^{2}+y^{2}\right)}\) has a relative maximum at (0, 0). There are no saddle points.

Step 1: Calculate the 1st Partial Derivatives

Calculate the first partial derivatives, \(f_x\) and \(f_y\), of the function \(f(x, y)\): \[ f_x = \frac{∂f}{∂x}= -6xe^{-(x^2+y^2)} \] \[ f_y = \frac{∂f}{∂y}= -6ye^{-(x^2+y^2)} \]

Step 2: Locate Critical Points

Find the critical points by setting the first partial derivatives to zero and solving for x and y respectively: For \(f_x = 0\) and \(f_y = 0\), it yields \(x = 0\) and \(y = 0\) respectively.

Step 3: Calculate the 2nd Partial Derivatives

Calculate the second partial derivatives \(f_{xx}\), \(f_{yy}\), and \(f_{xy}\): \[ f_{xx} = \frac{∂^2f}{∂x^2} = 6e^{-(x^2+y^2)}(2x^2-1) \] \[ f_{yy} = \frac{∂^2f}{∂y^2} = 6e^{-(x^2+y^2)}(2y^2-1) \] \[ f_{xy} = \frac{∂^2f}{∂x∂y} = 12xye^{-(x^2+y^2)} \]

Step 4: Conduct the Second Derivative Test

Use the second derivative test to classify the critical points: At (x=0,y=0), the Hessian Matrix is given by: \[ D = f_{xx}(0,0) \cdot f_{yy}(0,0) - f_{xy}(0,0)^2 = -6^2 - 0 = 36\] Since D>0 and \(f_{xx}(0,0)\)<0, the function f has a relative maximum at (0, 0).