Determine the nature of the critical points of the function $f ( x , y ) = 3 x ^ { 5 } y + x y ^ { 5 } + x y$
A)f has a saddle point at (1, 0).
B)f has a local maximum at (0, 0).
C)f has a saddle point at (0, 1).
D)f has a saddle point at (0, 0).
E)f has a local minimum at (0, 0).

Question

Quizplus · Accepted Answer

To determine the nature of the critical points of the function $f(x, y) = 3x^5y + xy^5 + xy$, we first find the first partial derivatives of $f$ with respect to $x$ and $y$, and then set them equal to zero to find the critical points.$$f_x = \frac{\partial f}{\partial x} = 15x^4y + y^5 + y$$$$f_y = \frac{\partial f}{\partial y} = 3x^5 + 5xy^4 + x$$Setting $f_x = 0$ and $f_y = 0$ gives us the system of equations:$$15x^4y + y^5 + y = 0$$$$3x^5 + 5xy^4 + x = 0$$From these equations, it's clear that $x = 0$ and $y = 0$ is a solution, indicating a critical point at $(0, 0)$.Next, we find the second partial derivatives to use in the second derivative test:$$f_{xx} = \frac{\partial^2 f}{\partial x^2} = 60x^3y$$$$f_{yy} = \frac{\partial^2 f}{\partial y^2} = 20xy^3$$$$f_{xy} = f_{yx} = \frac{\partial^2 f}{\partial x \partial y} = 15x^4 + 5y^4 + 1$$At the critical point $(0, 0)$, these derivatives are:$$f_{xx}(0, 0) = 0$$$$f_{yy}(0, 0) = 0$$$$f_{xy}(0, 0) = 1$$The determinant of the Hessian matrix $D$ at $(0, 0)$ is given by:$$D = f_{xx}(0, 0)f_{yy}(0, 0) - (f_{xy}(0, 0))^2 = (0)(0) - (1)^2 = -1$$Since $D < 0$, the second derivative test tells us that the function has a saddle point at $(0, 0)$.

Determine the Nature of the Critical Points of the Function f(x,y)=3x5y+xy5+xyf ( x , y ) = 3 x ^ { 5 } y + x y ^ { 5 } + x yf(x,y)=3x5y+xy5+xy

Determine the Nature of the Critical Points of the Function $f ( x , y ) = 3 x ^ { 5 } y + x y ^ { 5 } + x y$