Example 13 Given that $x, y, z$ are positive numbers, and $4x + 5y + 8z = 30$. Find the minimum value of $W = 8x^2 + 15y^2 + 48z^2$.
Solution: Let $x=3+t_{1}, y=2+t_{2}, z=1+t_{3}$, substituting into the given equation yields $4 t_{1}+5 t_{2}+8 t_{3}=0$.
$$
\begin{aligned}
W= & 8 x^{2}+15 y^{2}+48 z^{2} \\
= & 180+12\left(4 t_{1}+5 t_{2}+8 t_{3}\right)+8 t_{1}^{2} \\
& +15 t_{2}^{2}+48 t_{3}^{2} \\
\geqslant & 180,
\end{aligned}
$$
(when and only when $t_{1}=t_{2}=t_{3}=0$)
That is, when $x=3, y=2, z=1$, the minimum value of $W$ is 180.