11. [8] Let $a, b, c, d$ be real numbers such that
$$
\min (20 x+19,19 x+20)=(a x+b)-|c x+d|
$$
for all real numbers $x$. Find $a b+c d$.
Answer:
380

In general, $\min (p, q)=\frac{p+q}{2}-\left|\frac{p-q}{2}\right|$. Letting $p=20 x+19$ and $q=19 x+20$ gives $a=b=19.5$ and $c=d= \pm 0.5$. Then the answer is $19.5^{2}-0.5^{2}=19 \cdot 20=380$.