Let f(x, y) = $x ^ { 2 }$ - xy + $y ^ { 2 }$ + 2y - 4. The point $- \frac { 2 } { 3 } , - \frac { 4 } { 3 }$ is a
A) relative maximum
B) not a relative extreme point
C) absolute maximum
D) relative minimum

Question

Quizplus · Accepted Answer

To determine if the given point is a relative maximum, relative minimum, or not a relative extreme point, we can use the second partial derivative test. First, we find the first partial derivatives of $f(x, y)$:$$f_x = 2x - y$$$$f_y = -x + 2y + 2$$Setting these equal to zero to find critical points:$$2x - y = 0$$$$-x + 2y + 2 = 0$$Solving this system of equations, we find $x = -\frac{2}{3}$ and $y = -\frac{4}{3}$, which matches the given point, indicating it's a critical point.Next, we find the second partial derivatives:$$f_{xx} = 2$$$$f_{yy} = 2$$$$f_{xy} = f_{yx} = -1$$Then, we calculate the determinant of the Hessian matrix at the critical point:$$D = f_{xx}f_{yy} - (f_{xy})^2 = (2)(2) - (-1)^2 = 4 - 1 = 3$$Since $D > 0$ and $f_{xx} > 0$, the point is a relative minimum.

Let F(x, Y) = x2x ^ { 2 }x2 - Xy

Let F(x, Y) = $x ^ { 2 }$ - Xy