The function $f ( x , y ) = x ^ { 2 } - 2 y ^ { 2 }$ has one critical point. Determine its location and type.
A) $( 2,1 )$ , saddle point
B) $( 2,1 )$ , minimum point
C) $( 2,1 )$ , maximum point
D) $( \sqrt { 2 } , 1 )$ , saddle point
E) $( \sqrt { 2 } , 1 )$ , minimum point
F) $( \sqrt { 2 } , 1 )$ , maximum point
G) $( 0,0 )$ , saddle point
H) $( 0,0 )$ , minimum point

Question

Quizplus · Accepted Answer

To find the critical points of $f(x, y) = x^2 - 2y^2$, we calculate the partial derivatives and set them equal to zero. The partial derivatives are $f_x = 2x$ and $f_y = -4y$. Setting these equal to zero gives $x = 0$ and $y = 0$, so the critical point is at $(0,0)$. To determine the type of critical point, we use the second derivative test. The second partial derivatives are $f_{xx} = 2$, $f_{yy} = -4$, and $f_{xy} = f_{yx} = 0$. The determinant of the Hessian matrix is $D = f_{xx}f_{yy} - (f_{xy})^2 = (2)(-4) - (0)^2 = -8$. Since $D < 0$, the critical point at $(0,0)$ is a saddle point.

The Function f(x,y)=x2−2y2f ( x , y ) = x ^ { 2 } - 2 y ^ { 2 }f(x,y)=x2−2y2

The Function $f ( x , y ) = x ^ { 2 } - 2 y ^ { 2 }$