# Maximum Value of the expression

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

• @deepalis727

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)^(m-1)(px + a)^n - np(px + q)^(n-1)(ax + b)^m
Fox max of f(x), f'(x) = 0
ma(ax + b)^(m-1)(px + q)^n + np(px + q)^(n-1)(ax + b)^m = 0
ma(ax + b)^(m-1)(px + q)^n = - np(px + q)^(n-1)(ax + b)^m
ma(ax + b)^(m-1)/(ax+b)^m = -np(px + q)^(n-1)/(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 = 8

max of (x-6)^2 * (11-x)^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 (x-6)/2 = (11-x)/3
so at x =8
therefore max value =2^2 * 3^3

Looks like your connection to MBAtious was lost, please wait while we try to reconnect.