11. Let $a, b, c, d$ be four distinct real numbers such that $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=4$, and $a c=$ $b d$. Find the maximum value of $\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}$.
11. Let $x=\frac{a}{b}, y=\frac{b}{c}$, then from $a c=b d$ we get $\frac{c}{d}=\frac{b}{a}=\frac{1}{x}, \frac{d}{a}=\frac{c}{b}=\frac{1}{y}$, so the problem becomes finding the maximum value of $x y+\frac{y}{x}+\frac{1}{x y}+\frac{x}{y}$ under the constraint $x \neq 1, y \neq 1, x+y+\frac{1}{x}+\frac{1}{y}=4$. Let $x+\frac{1}{x}=e, y+\frac{1}{y}=f$, then $x y+\frac{y}{x}+\frac{1}{x y}+\frac{x}{y}=e f$. When $t>0$, $t+\frac{1}{t} \geqslant 2$; when $t<0$, $t+\frac{1}{t} \leqslant -2$. If $x>0, y>0$, then $e \geqslant 2, f \geqslant 2, e f \geqslant 4$, which does not satisfy the condition. If $x<0, y>0$, then $e \leqslant -2, f \geqslant 2, e f \leqslant -4$, which also does not satisfy the condition. If $x>0, y<0$, then $e \geqslant 2, f \leqslant -2, e f \leqslant -4$, which does not satisfy the condition. If $x<0, y<0$, then $f \leqslant -2, e=4-f \geqslant 6, e f \leqslant -12$, with equality holding if and only if $y=-1, x=3 \pm 2 \sqrt{2}$. Specifically, when $a=3+2 \sqrt{2}, b=1, c=-1, d=-3-2 \sqrt{2}$, equality holds. Thus, $(e f)_{\max }=-12$.