Maximum Value of the expression

Find the maximum value of (x  6)^2 (11  x)^3 for 6 < x < 11.

If you are comfortable with derivatives then you can check Part 1. Else, jump to Part 2 :)
Part 1:
f(x) = g(x)h(x)
f'(x) = g'(x)h(x) + h'(x)g(x) (f'(x) is the derivative of f(x))
So if f(x) = (ax + b)^m * (px + q)^n then
f'(x) = ma(ax + b)^(m1)(px + a)^n  np(px + q)^(n1)(ax + b)^m
Fox max of f(x), f'(x) = 0
ma(ax + b)^(m1)(px + q)^n + np(px + q)^(n1)(ax + b)^m = 0
ma(ax + b)^(m1)(px + q)^n =  np(px + q)^(n1)(ax + b)^m
ma(ax + b)^(m1)/(ax+b)^m = np(px + q)^(n1)/(px + q)^n
ma/(ax + b) = np/(px+q)Part 2:
If f(x) = (ax + b)^m * (px + q)^n then max/min occurs at values of x for which (ax + b)/ma =  (px + q)/np
So here, f(x) = (x  6)^2 (11  x)^3
max occurs at values of x such that (x  6)/(2 * 1) = (11  x)/(3 * 1)
3x  18 = 22  2x
5x = 40
x = 8max of (x6)^2 * (11x)^3 happens at x = 8 and max value = 2^2 * 3^3 = 108
So, if we want to find for what of x for maximizing f(x) = (4x  3)^5 * (18  2x)^4
We need to find x such that (4x  3)/20 = (18  2x)/8
32x  24 = 360  40x
72x = 384
x = 384/72 = 16/3
So max for f(x) occurs at x = 16/3

max occurs when there is a symmetry
so (x6)/2 = (11x)/3
so at x =8
therefore max value =2^2 * 3^3