For $f ( x , y )$ , find all values of x and y such that $f _ { x } ( x , y ) = 0$ and $f _ { y } ( x , y ) = 0$ simultaneously. $f ( x , y ) = 16 x ^ { 3 } - 4 x y + 16 y ^ { 3 }$
A) $\left( - \frac { 1 } { 12 } , - \frac { 1 } { 12 } \right)$
B) $\left( \frac { 1 } { 12 } , \frac { 1 } { 12 } \right)$
C)(0,0)
D)Both B and C
E)Both A and B

Question

Quizplus · Accepted Answer

To find the values of $x$ and $y$ for which $f_x(x, y) = 0$ and $f_y(x, y) = 0$, we first need to compute the partial derivatives of $f(x, y)$ with respect to $x$ and $y$.Given $f(x, y) = 16x^3 - 4xy + 16y^3$,1. Compute $f_x(x, y)$:$$f_x(x, y) = \frac{\partial}{\partial x}(16x^3 - 4xy + 16y^3) = 48x^2 - 4y$$2. Compute $f_y(x, y)$:$$f_y(x, y) = \frac{\partial}{\partial y}(16x^3 - 4xy + 16y^3) = -4x + 48y^2$$Setting $f_x(x, y) = 0$ and $f_y(x, y) = 0$ gives us two equations:$$48x^2 - 4y = 0$$$$-4x + 48y^2 = 0$$Solving these equations simultaneously, we find that $x = y = 0$ is a solution, which corresponds to choice C.Additionally, by manipulating the equations, we can find that $x = y = \frac{1}{12}$ is also a solution, which corresponds to choice B.Therefore, both B and C are correct, making D the correct choice.

For f(x,y)f ( x , y )f(x,y) , Find All Values of X and Y Such That

For $f ( x , y )$ , Find All Values of X and Y Such That