Examine the function $f ( x , y ) = 5 x ^ { 2 } - 8 y ^ { 2 } - 3 x + 8 y + 4$ for relative extrema.
A)saddle point at $\left( \frac { 3 } { 10 } , \frac { 1 } { 2 } \right)$
B)relative minimum at $\left( \frac { 3 } { 10 } , \frac { 1 } { 2 } \right)$
C)saddle point at $\left( - \frac { 3 } { 10 } , \frac { 1 } { 2 } \right)$
D)relative maximum at $\left( \frac { 3 } { 10 } , - \frac { 1 } { 2 } \right)$
E)saddle point at $\left( - \frac { 3 } { 10 } , - \frac { 1 } { 2 } \right)$

Question

Quizplus · Accepted Answer

To determine the nature of the critical point for the function $f(x, y) = 5x^2 - 8y^2 - 3x + 8y + 4$, we first find its partial derivatives and then use the second derivative test. The first partial derivatives are $f_x = 10x - 3$ and $f_y = -16y + 8$. Setting these equal to zero gives the critical point $\left(\frac{3}{10}, \frac{1}{2}\right)$. The second partial derivatives are $f_{xx} = 10$, $f_{yy} = -16$, and $f_{xy} = f_{yx} = 0$. The determinant of the Hessian matrix at the critical point is $D = f_{xx}f_{yy} - (f_{xy})^2 = (10)(-16) - (0)^2 = -160$. Since $D < 0$, the critical point is a saddle point.

Examine the Function f(x,y)=5x2−8y2−3x+8y+4f ( x , y ) = 5 x ^ { 2 } - 8 y ^ { 2 } - 3 x + 8 y + 4f(x,y)=5x2−8y2−3x+8y+4

Examine the Function $f ( x , y ) = 5 x ^ { 2 } - 8 y ^ { 2 } - 3 x + 8 y + 4$