15. [9] Vijay chooses three distinct integers $a, b, c$ from the set $\{1,2,3,4,5,6,7,8,9,10,11\}$. If $k$ is the minimum value taken on by the polynomial $a(x-b)(x-c)$ over all real numbers $x$, and $l$ is the minimum value taken on by the polynomial $a(x-b)(x+c)$ over all real numbers $x$, compute the maximum possible value of $k-l$.
Answer:
990
Solution: Quadratics are minimized at the average of their roots, so
$$
\begin{aligned}
k & =a\left(\frac{b+c}{2}-b\right)\left(\frac{b+c}{2}-c\right)=a\left(\frac{b-c}{2}\right)\left(\frac{c-b}{2}\right)=-\frac{a(b-c)^{2}}{4}, \text { and } \\
l & =a\left(\frac{b-c}{2}-b\right)\left(\frac{b-c}{2}+c\right)=a\left(\frac{-b-c}{2}\right)\left(\frac{b+c}{2}\right)=-\frac{a(b+c)^{2}}{4} .
\end{aligned}
$$

Therefore,
$$
k-l=-\frac{a}{4}\left((b-c)^{2}-(b+c)^{2}\right)=a b c
$$

Thus, $k-l=a b c$ is maximized when $a, b$, and $c$ are 9,10 , and 11 are some order, so the answer is $9 \cdot 10 \cdot 11=990$.