For $f ( x , y ) \text {, find all values of } x \text { and } y \text { such that } f _ { x } ( x , y ) = 0 \text { and } f _ { y } ( x , y ) = 0$ simultaneously. $f ( x , y ) = 10 x ^ { 3 } - 2 x y + 10 y ^ { 3 }$
A) $( 0,0 ) , \left( \frac { 1 } { 15 } , \frac { 1 } { 15 } \right)$
B) $\left( \frac { 1 } { 15 } , \frac { 1 } { 15 } \right)$
C) $\left( - \frac { 1 } { 15 } , - \frac { 1 } { 15 } \right) , \left( \frac { 1 } { 15 } , \frac { 1 } { 15 } \right)$
D) $\left( - \frac { 1 } { 15 } , - \frac { 1 } { 15 } \right)$
E) $( 0,0 )$

Question

Quizplus · Accepted Answer

To find the critical points where $f_x(x, y) = 0$ and $f_y(x, y) = 0$, we first calculate the partial derivatives of $f(x, y)$ with respect to $x$ and $y$, respectively. For $f(x, y) = 10x^3 - 2xy + 10y^3$, $f_x(x, y) = \frac{\partial}{\partial x}(10x^3 - 2xy + 10y^3) = 30x^2 - 2y$$f_y(x, y) = \frac{\partial}{\partial y}(10x^3 - 2xy + 10y^3) = -2x + 30y^2$Setting $f_x(x, y) = 0$ and $f_y(x, y) = 0$ gives us two equations:1. $30x^2 - 2y = 0$2. $-2x + 30y^2 = 0$From equation 1, we get $y = 15x^2$. Substituting $y$ in equation 2 gives:$-2x + 30(15x^2)^2 = 0$Solving these equations simultaneously, we find that $x = 0, y = 0$ is a solution, satisfying both equations. Additionally, solving for non-zero solutions gives $x = \frac{1}{15}, y = \frac{1}{15}$ as another set of solutions. Therefore, the correct answer includes both $(0,0)$ and $\left(\frac{1}{15}, \frac{1}{15}\right)$.

For f(x,y), find all values of x and y such that fx(x,y)=0 and fy(x,y)=0f ( x , y ) \text {, find all values of } x \text { and } y \text { such that } f _ { x } ( x , y ) = 0 \text { and } f _ { y } ( x , y ) = 0f(x,y), find all values of x and y such that fx​(x,y)=0 and fy​(x,y)=0

For $f ( x , y ) \text {, find all values of } x \text { and } y \text { such that } f _ { x } ( x , y ) = 0 \text { and } f _ { y } ( x , y ) = 0$