First, we make the terms in the denominator identical. For example, we can multiply the factor $ 4x + 3y $ by $ \frac{5}{4} $ (and we also multiply the numerator by $ \frac{5}{4} $), which gives us \[\frac{\frac{5}{4} xyz}{(1 + 5x)(5x + \frac{15}{4} y)(5y + 6z)(z + 18)}.\]We then multiply the factor $ 5y + 6z $ by $ \frac{3}{4} $ (and the numerator), which gives us \[\frac{\frac{15}{16} xyz}{(1 + 5x)(5x + \frac{15}{4} y)(\frac{15}{4} y + \frac{9}{2} z)(z + 18)}.\]Finally, we multiply the factor $ z + 18 $ by $ \frac{9}{2} $ (and the numerator), which gives us \[\frac{\frac{135}{32} xyz}{(1 + 5x)(5x + \frac{15}{4} y)(\frac{15}{4} y + \frac{9}{2} z)(\frac{9}{2} z + 81)}.\]Let $ a = 5x, $ $ b = \frac{15}{4} y, $ and $ c = \frac{9}{2} z $. Then $ x = \frac{1}{5} a, $ $ y = \frac{4}{15} b, $ and $ z = \frac{2}{9} c, $ so the expression becomes \[\frac{\frac{1}{20} abc}{(1 + a)(a + b)(b + c)(c + 81)}.\]By AM-GM, \begin{align*} 1 + a &= 1 + \frac{a}{3} + \frac{a}{3} + \frac{a}{3} \ge 4 \sqrt[4]{\frac{a^3}{27}}, \\ a + b &= a + \frac{b}{3} + \frac{b}{3} + \frac{b}{3} \ge 4 \sqrt[4]{\frac{a b^3}{27}}, \\ b + c &= b + \frac{c}{3} + \frac{c}{3} + \frac{c}{3} \ge 4 \sqrt[4]{\frac{b c^3}{27}}, \\ c + 81 &= c + 27 + 27 + 27 \ge 4 \sqrt[4]{c \cdot 27^3}.\end{align*}Then \[(1 + a)(a + b)(b + c)(c + 81) \ge 4 \sqrt[4]{\frac{a^3}{27}} \cdot 4 \sqrt[4]{\frac{a b^3}{27}} \cdot 4 \sqrt[4]{\frac{b c^3}{27}} \cdot 4 \sqrt[4]{c \cdot 27^3} = 256abc,\]so \[\frac{\frac{1}{20} abc}{(1 + a)(a + b)(b + c)(c + 81)} \le \frac{\frac{1}{20} abc}{256 abc} \le \frac{1}{5120}.\]Equality occurs when $ a = 3, $ $ b = 9, $ and $ c = 27, $ or $ x = \frac{3}{5}, $ $ y = \frac{12}{5}, $ and $ z = 6, $ so the maximum value is $ \boxed{\frac{1}{5120}} $.