Maximum Value of the expression



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


  • Being MBAtious!


    @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


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