The maximum value of $a^2{b}^3{c}^4$  subject to $a+b+c=18$  is

Keeping in view that $a+b+c=18$

$a^2{b}^3{c}^4$  is max. when ${\left(\frac{a}{2}\right)}^2{\left(\frac{b}{3}\right)}^3{\left(\frac{c}{4}\right)}^4$  is max.

$\therefore a=4,b=6,c=8$

$\therefore$  Max, value of $a^2{b}^3{c}^4$ is $4^2{6}^3{8}^4$ .