jmc

Let $$f(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_{1}x+a_0.$$Then from the given information, $$\begin{array}{rl}{a}_n\cdot 6^n+a_{n-1}\cdot 6^{n-1}+\cdots +a_1\cdot 6+a_0& =24,\\ a_n\cdot 24^n+a_{n-1}\cdot 24^{n-1}+\cdots +a_1\cdot 24+a_0& =1536.\end{array}$$Then by Cauchy-Schwarz, $$\begin{array}{rl}& (a_n\cdot 6^n+a_{n-1}\cdot 6^{n-1}+\cdots +a_1\cdot 6+a_0)(a_n\cdot 24^n+a_{n-1}\cdot 24^{n-1}+\cdots +a_1\cdot 24+a_0)\\ & \ge (a_n\cdot 12^n+a_{n-1}\cdot 12^{n-1}+\cdots +a_1\cdot 12+a_0{)}^2.\end{array}$$In other words, $[f(12){]}^2\le 24\cdot 1536=36864,$ so $f(12)\le 192.$

Equality occurs for $f(x)=\frac{x^3}{9},$ so the maximum value is $\boxed{192}.$