2. Let real numbers $x, y, z, w$ satisfy $x \geqslant y \geqslant z \geqslant w \geqslant 0$, and $5 x+4 y+3 z+6 w=100$. Denote the maximum value of $x+y+z+w$ as $a$, and the minimum value as $b$. Then $a+b=$ $\qquad$
2. 45 .

From $x \geqslant y \geqslant z \geqslant w \geqslant 0$, we know
$$
\begin{array}{l}
100=5 x+4 y+3 z+6 w \geqslant 4(x+y+z+w) \\
\Rightarrow x+y+z+w \leqslant 25 .
\end{array}
$$

When $x=y=z=\frac{25}{3}, w=0$, the equality holds.
$$
\begin{array}{l}
\text { Also } 100=5 x+4 y+3 z+6 w \leqslant 5(x+y+z+w) \\
\Rightarrow x+y+z+w \geqslant 20 .
\end{array}
$$

When $x=20, y=z=w=0$, the equality holds. Therefore, $a=25, b=20$.
Thus, $a+b=45$.