jmc

We have that $q=-p^2+6p-5,$ so by the Shoelace Theorem, the area of triangle $ABC$ is $$\begin{array}{rl}& \frac{1}{2}|(1)(3)+(4)(-p^2+6p-5)+(p)(0)-(0)(4)-(3)(p)-(-p^2+6p-5)(1)|\\ & =\frac{1}{2}|-3p^2+15p-12|\\ & =\frac{3}{2}|p^2-5p+4|\\ & =\frac{3}{2}|(p-1)(p-4)|.\end{array}$$Since $1\le p\le 4,$ $|(p-1)(p-4)|=(p-1)(4-p),$ so we want to maximize $$\frac{3}{2}(p-1)(4-p).$$The maximum value occurs at $p=\frac{5}{2},$ so the maximum area is $$\frac{3}{2}\left(\frac{5}{2}-1\right)\left(4-\frac{5}{2}\right)=\boxed{\frac{27}{8}}.$$