Find all points (x, y) where $f ( x , y ) = e ^ { \left( x ^ { 2 } + y ^ { 2 } \right) }$ has a possible relative maximum or minimum. Use the second-derivative test to determine, if possible, the nature of f(x, y) at each of these points.
A) (0, 1) gives a relative minimum point
B) (0, 0) gives a relative minimum point
C) (0, 1) gives a relative maximum point
D) (0, 0) gives a relative maximum point

Question

Quizplus · Accepted Answer

To determine the points where $f(x, y) = e^{(x^2 + y^2)}$ has a possible relative maximum or minimum, we first find the critical points by taking the partial derivatives of $f$ with respect to $x$ and $y$, setting them equal to zero, and solving for $x$ and $y$.The first partial derivative with respect to $x$ is $f_x = 2xe^{(x^2 + y^2)}$, and with respect to $y$ is $f_y = 2ye^{(x^2 + y^2)}$. Setting $f_x = 0$ and $f_y = 0$ gives $x = 0$ and $y = 0$, so the only critical point is $(0, 0)$.To determine the nature of this critical point, we use the second derivative test for functions of two variables, which involves computing the second partial derivatives $f_{xx}$, $f_{yy}$, and the mixed partial derivative $f_{xy}$, and then evaluating the determinant of the Hessian matrix $D = f_{xx}f_{yy} - (f_{xy})^2$ at the critical point.The second partial derivatives are $f_{xx} = (2 + 4x^2)e^{(x^2 + y^2)}$, $f_{yy} = (2 + 4y^2)e^{(x^2 + y^2)}$, and $f_{xy} = 4xye^{(x^2 + y^2)}$. At the point $(0, 0)$, these derivatives simplify to $f_{xx} = 2$, $f_{yy} = 2$, and $f_{xy} = 0$. Thus, $D = (2)(2) - (0)^2 = 4$, which is positive.Since $D > 0$ and $f_{xx} > 0$ at $(0, 0)$, the function $f(x, y)$ has a relative minimum at this point.

Find All Points (X, Y) Where f(x,y)=e(x2+y2)f ( x , y ) = e ^ { \left( x ^ { 2 } + y ^ { 2 } \right) }f(x,y)=e(x2+y2)

Find All Points (X, Y) Where $f ( x , y ) = e ^ { \left( x ^ { 2 } + y ^ { 2 } \right) }$