jmc

By Cauchy-Schwarz, $$\begin{array}{rl}{\left(\sqrt{c(b-c)}+\sqrt{c(a-c)}\right)}^2& \le (1+1)(c(b-c)+c(a-c))\\ & =2(bc-c^2+ac-c^2)\\ & =2((a+b)c-2c^2)\\ & =2(16c-2c^2)\\ & =4(8c-c^2).\end{array}$$The maximum of $8c-c^2$ occurs at $c=4,$ for a maximum value of 16, so $$\sqrt{c(b-c)}+\sqrt{c(a-c)}\le \sqrt{4\cdot 16}=8.$$Equality occurs when $a=b=8$ and $c=4,$ so the maximum value is $\boxed{8}.$