Find the relative maximum and minimum values.
\[
f(x, y)=x^{3}+y^{3}-12 x y
\]
Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.
A. The function has a relative maximum value of $f(x, y)=\square$ at $(x, y)=\square$ (Simplify your answers. Type exact answers. Type an ordered pair in the second answer box.)
B. The function has no relative maximum value.
Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.
A. The function has a relative minimum value of $f(x, y)=\square$ at $(x, y)=\square$. (Simplify your answers. Type exact answers. Type an ordered pair in the second answer box.)
B. The function has no relative minimum value.

Step 1 ：First, we need to find the critical points of the function. The critical points are where the first derivatives of the function are equal to zero. So we calculate the first derivatives of the function with respect to \(x\) and \(y\).

Step 2 ：The first derivative of \(f(x, y)\) with respect to \(x\) is \(f_x = 3x^2 - 12y\).

Step 3 ：The first derivative of \(f(x, y)\) with respect to \(y\) is \(f_y = 3y^2 - 12x\).

Step 4 ：We set these two derivatives equal to zero and solve for \(x\) and \(y\).

Step 5 ：Setting \(f_x = 0\) gives \(3x^2 - 12y = 0\), or \(x^2 = 4y\).

Step 6 ：Setting \(f_y = 0\) gives \(3y^2 - 12x = 0\), or \(y^2 = 4x\).

Step 7 ：Solving these two equations simultaneously, we get \(x = 0, y = 0\) and \(x = 4, y = 4\).

Step 8 ：These are the critical points of the function. Next, we need to determine whether these points are relative maximum, minimum, or saddle points. We do this by calculating the second derivatives of the function and using the second derivative test.

Step 9 ：The second derivative of \(f(x, y)\) with respect to \(x\) is \(f_{xx} = 6x\).

Step 10 ：The second derivative of \(f(x, y)\) with respect to \(y\) is \(f_{yy} = 6y\).

Step 11 ：The mixed second derivative of \(f(x, y)\) with respect to \(x\) and \(y\) is \(f_{xy} = -12\).

Step 12 ：We calculate the determinant of the Hessian matrix, \(D = f_{xx}f_{yy} - f_{xy}^2\).

Step 13 ：For the point \((0, 0)\), we have \(D = 6*0*6*0 - (-12)^2 = -144 < 0\), so \((0, 0)\) is a saddle point.

Step 14 ：For the point \((4, 4)\), we have \(D = 6*4*6*4 - (-12)^2 = 576 - 144 = 432 > 0\), and since \(f_{xx} = 6*4 = 24 > 0\), \((4, 4)\) is a relative minimum.

Step 15 ：We substitute \((4, 4)\) into the function to get the minimum value, \(f(4, 4) = 4^3 + 4^3 - 12*4*4 = 64 - 192 = -128\).

Step 16 ：So, the function has a relative minimum value of \(f(x, y) = \boxed{-128}\) at \((x, y) = \boxed{(4, 4)}\).

Step 17 ：The function has no relative maximum value.