Find the relative maximum value of $f(x, y, z)=x y z^{2}$, subject to the constraint $x+y+3 z=14$.
The relative maximum value is $f$ (Simplify your answers.)

Step 1 ：Define the function to be optimized as \(f(x, y, z) = xyz^2\) and the constraint function as \(g(x, y, z) = x + y + 3z\) with a constant value of \(c = 14\).

Step 2 ：Formulate the Lagrangian function \(L(x, y, z, \lambda) = f(x, y, z) - \lambda(g(x, y, z) - c)\).

Step 3 ：Find the critical points of the Lagrangian function by setting its partial derivatives with respect to \(x\), \(y\), \(z\), and \(\lambda\) equal to zero. This forms a system of equations.

Step 4 ：Solve the system of equations to find the values of \(x\), \(y\), \(z\), and \(\lambda\) that maximize or minimize \(f(x, y, z)\) subject to the constraint \(g(x, y, z) = c\).

Step 5 ：The solutions to the system of equations are \({l: 0, x: 14 - y, z: 0}\), \({l: 0, x: 0, y: 0, z: 14/3}\), \({l: 0, x: 7, y: 7, z: 0}\), and \({l: 343/18, x: 7/2, y: 7/2, z: 7/3}\).

Step 6 ：Calculate the value of \(f(x, y, z)\) at each of these points to find the maximum value.

Step 7 ：The relative maximum value of \(f(x, y, z)=x y z^{2}\), subject to the constraint \(x+y+3 z=14\) is \(\boxed{\frac{2401}{36}}\).