jmc

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, $$\begin{array}{rl}1+a& =1+\frac{a}{3}+\frac{a}{3}+\frac{a}{3}\ge 4\sqrt[4]{\frac{a^3}{27}},\\ a+b& =a+\frac{b}{3}+\frac{b}{3}+\frac{b}{3}\ge 4\sqrt[4]{\frac{a{b}^3}{27}},\\ b+c& =b+\frac{c}{3}+\frac{c}{3}+\frac{c}{3}\ge 4\sqrt[4]{\frac{b{c}^3}{27}},\\ c+81& =c+27+27+27\ge 4\sqrt[4]{c\cdot 27^3}.\end{array}$$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 $$\frac{\frac{1}{20}abc}{(1+a)(a+b)(b+c)(c+81)}\le \frac{\frac{1}{20}abc}{256abc}\le \frac{1}{5120}.$$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 $\boxed{\frac{1}{5120}}.$