Find all points (x, y) where $f ( x , y ) = x ^ { 2 } + x y + y ^ { 2 } - 3 x + 2$ 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) (2, -1) is a relative maximum
B) (2, -1) is a relative minimum
C) (-2, 1) is a relative maximum
D) (-2, 1) is a relative minimum

Question

Quizplus · Accepted Answer

DTo find the critical points of $f(x, y)$, we first need to compute the first partial derivatives of $f$ with respect to $x$ and $y$, and then set them equal to zero.$$f_x = 2x + y - 3$$$$f_y = x + 2y$$Setting $f_x = 0$ and $f_y = 0$ gives us a system of equations:$$2x + y - 3 = 0$$$$x + 2y = 0$$Solving this system, we find $x = 2$ and $y = -1$, so one critical point is $(2, -1)$.Next, we compute the second partial derivatives:$$f_{xx} = 2$$$$f_{yy} = 2$$$$f_{xy} = f_{yx} = 1$$The second-derivative test for a function of two variables is given by $D = f_{xx}f_{yy} - (f_{xy})^2$. At $(2, -1)$, we find:$$D = (2)(2) - (1)^2 = 4 - 1 = 3$$Since $D > 0$ and $f_{xx} > 0$, the function $f$ has a relative minimum at $(2, -1)$.For the point $(-2, 1)$, it seems there was a misunderstanding in the calculation as this point does not satisfy the system of equations derived from setting the first partial derivatives to zero. Therefore, only the point $(2, -1)$ is correctly identified and analyzed according to the given function and its derivatives. The correct interpretation is that $(2, -1)$ is a relative minimum, making option B correct. The mention of $(-2, 1)$ seems to be a mistake as it does not result from solving the system of equations for the critical points of $f(x, y)$. Thus, only option B is correct based on the provided analysis.

Find All Points (X, Y) Where f(x,y)=x2+xy+y2−3x+2f ( x , y ) = x ^ { 2 } + x y + y ^ { 2 } - 3 x + 2f(x,y)=x2+xy+y2−3x+2

Find All Points (X, Y) Where $f ( x , y ) = x ^ { 2 } + x y + y ^ { 2 } - 3 x + 2$