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Find the maximum of 'arbitrary power' function

  1. Oct 21, 2014 #1
    1. If a and b are positive numbers, find the maximum value of f(x)=x^a(2-x)^b D={0<=x<=2}

    attempt of solution
    I did this question more intuitively.
    So I first differentiated and found it to be axa-1(2-x)b + xa (2-x)bb-1
    and I figured it will only be <0 when x>=2 and b is odd, so it will be >0 on the left which means maxima is at x=2 .

    Is this right? and how do i prove it rigorously?
     
  2. jcsd
  3. Oct 22, 2014 #2

    haruspex

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    Your differentiation is not quite right (sign error).
     
  4. Oct 22, 2014 #3

    RUber

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    To find the zeros of the derivative, (once you have the right derivative), you will be able to factor out ##x^{a-1}## and ##(2-x)^{b-1}## terms from the derivative. These have zeros when a, b are not 1.
    The remaining terms should give you a zero in terms of a and b.
     
  5. Oct 22, 2014 #4

    RUber

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    Also, I think your logic is wrong for x=2 being a maximum. Plug 2 and 0 into the original function. Then over the remainder of the domain, you can see that f(x) > 0 for x in (0,2).
     
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