jmc

With word problems, the first step is to translate the words into equations. Let the two numbers be $a$ and $b$. Then their sum is $a+b$ and their product is $ab$. The sum of their product and their sum is $a+b+ab$. So we know $$\begin{array}{rl}ab+a+b& =454\Rightarrow \\ a(b+1)+(b+1)& =454+1\Rightarrow \\ (a+1)(b+1)& =455.\end{array}$$The prime factorization of $455$ is $5\cdot 7\cdot 13$. Since the equation is symmetric with $a$ and $b$, we may (without loss of generality) suppose that $a<b$. Thus $a+1<b+1$, so in each factor pair the smaller factor is equal to $a+1$. We list all possibilities: \begin{array}{c|c|c|c} $a+1%%DISP_1%%amp;$b+1%%DISP_1%%amp;$a%%DISP_1%%amp;$b$\\ \hline $1%%DISP_1%%amp;$455%%DISP_1%%amp;$0%%DISP_1%%amp;$454$\\ $5%%DISP_1%%amp;$91%%DISP_1%%amp;$4%%DISP_1%%amp;$90$\\ $7%%DISP_1%%amp;$65%%DISP_1%%amp;$6%%DISP_1%%amp;$64$\\ $13%%DISP_1%%amp;$35%%DISP_1%%amp;$12%%DISP_1%%amp;$34$ \end{array}We must find the largest possible value of "the product of their sum and their product", or $ab\cdot (a+b)$. We know the first possibility above gives a value of zero, while all the others will be greater than zero. We check: $$\begin{array}{rl}4\cdot 90\cdot (4+90)& =4\cdot 90\cdot 94=33840\\ 6\cdot 64\cdot (6+64)& =6\cdot 64\cdot 70=26880\\ 12\cdot 34\cdot (12+34)& =12\cdot 34\cdot 46=18768.\end{array}$$Thus the largest possible desired value is $\boxed{33840}$, achieved when $(a,b)=(4,90)$.