19. Let real numbers $x, y$ satisfy $x^{2}+y^{2} \leqslant 5$, find the extremum of $f(x, y)=3|x+y|+|4 y+9|+|7 y-3 x-18|$.
19. Let $x=r \cos \theta, y=r \sin \theta(r \geqslant 0,0 \leqslant \theta < 2\pi)$, then $7 y-3 x-18=7 r \sin \theta-3 r \cos \theta-18=r \cdot \sqrt{58} \sin (\theta-\varphi)-18 \leqslant$ $\sqrt{290}-18<0$.
Therefore, $f(x, y)=3|x+y|+4 y+9-7 y+3 x+18$
$$
=3 \sqrt{2} r\left|\sin \left(\theta+\frac{\pi}{4}\right)\right|+3 \sqrt{2} r \cos \left(\theta+\frac{\pi}{4}\right)+27.
$$
(i) When $\sin \left(\theta+\frac{\pi}{4}\right) \geqslant 0$, i.e., $0 \leqslant \theta \leqslant \frac{3}{4} \pi$ or $\frac{7}{4} \pi \leqslant \theta < 2 \pi$, $f(x, y)=3 \sqrt{2} r \sin \left(\theta+\frac{\pi}{4}\right)+$ $3 \sqrt{2} r \cos \left(\theta+\frac{\pi}{4}\right)+27=6 r \cos \theta+27$
Thus, when $\cos \theta=1$, $f(x, y)_{\text {max }}=6 \sqrt{5}+27$, and when $\cos \theta=-\frac{\sqrt{2}}{2}$, $f(x, y)_{\min }=27-3 \sqrt{10}$.
(ii) When $\sin \left(\theta+\frac{\pi}{4}\right) \leqslant 0$, i.e., $\frac{3}{4} \pi \leqslant \theta \leqslant \frac{7 \pi}{4}$,
$$
f(x, y)=-3 \sqrt{2} r \sin \left(\theta+\frac{\pi}{4}\right)+3 \sqrt{2} r \cos \left(\theta+\frac{\pi}{4}\right)+27=-6 r \sin \theta+27,
$$

Thus, when $\sin \theta=-1$, $f(x, y)_{\max }=27+6 \sqrt{5}$,
and when $\sin \theta=\frac{\sqrt{2}}{2}$, $f(x, y)_{\text {min }}=27-3 \sqrt{10}$,
In summary, $f(x, y)_{\operatorname{max}}=27+6 \sqrt{5}, f(x, y)_{\min }=27-3 \sqrt{10}$.