Its easier than you think.!

$\dfrac{a}{2} + \dfrac{a}{2} + \dfrac{b}{3} + \dfrac{b}{3} + \dfrac{b}{3} + \dfrac{c}{4} + \dfrac{c}{4} + \dfrac{c}{4} + \dfrac{c}{4} = 18$

Now applying A.M - G.M ;

$\dfrac{\frac{a}{2} + \frac{a}{2} + \frac{b}{3} + \frac{b}{3} + \frac{b}{3} + \frac{c}{4} + \frac{c}{4} + \frac{c}{4} + \frac{c}{4}}{9} = \dfrac{18}{9} \geq \sqrt[9]{\dfrac{a^2 \times b^3 \times c^4}{4 \times 27 \times 256}}$

$2^9 \geq \dfrac{a^2b^3c^4}{2^{10} 3^3} \Rightarrow 2^{19} 3^3 \geq a^2b^3c^4$

Hence the maximum value is: $\boxed{2^{19}3^3}$

$w = 2 , x = 19 , y = 3 , z = 3 \Rightarrow w + x + y + z = 27$

Let $F = a^2 b^3 (18-a-b)^4$

Then

$\dfrac{d}{da}F = 2ab^3(a+b-18)^3(3a+b-18) = 0$
$\dfrac{d}{db}F = 2a^2b^2(a+b-18)^3(3a+7b-54) = 0$

We solve the simultaneous equations to find $a, b, c$

$3a+b-18 = 0$
$3a+7b-54 = 0$
$a+b+c=18$

$a=4$
$b=6$
$c=8$

and then work out $w, x, y, z$ to be $2, 19, 3, 3$

To maximize $a^2b^3c^4$ is the same as to maximize $\ln(a^2b^3c^4)=2ln(a)+3ln(b)+4ln(c)=S$ .

If we make slight changes in $a$ , $b$ and $c$ such that $da+db+dc=0$ , we see that

$dS=\frac{2}{a}da+\frac{3}{b}db+\frac{4}{c}dc$

But when we are at a maxima, $dS$ has to be $0$ for changes in every "direction". We can easly see that when $c=8$ , $b=6$ and $a=4$

$a+b+c=18$ and $dS=\frac{1}{2}(da+db+dc)=0$

So the maximum value is $4^26^38^4=2^{19}3^3$ and $w+x+y+z=27$

Hint:Split a+b+c as a/2+a/2+b/3+b/3+b/3+c/4+c/4+c/4+c/4 & Apply AM>=GM.The way of splitting is apparent from the indices of a,b,c in the expression which is to be maximized.