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Fimd the maximum value

  1. Nov 4, 2012 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    Let a,b,c be real numbers such that a+2b+c=4. Then the max value of (ab+bc+ca) is

    2. Relevant equations

    3. The attempt at a solution
    I am trying to substitute for b into the second expression but that won't give me the answer. Any help would be appreciated.
     
  2. jcsd
  3. Nov 4, 2012 #2

    mfb

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    Why?
    The approach is fine, what do you get as result?
     
  4. Nov 4, 2012 #3
    Note that a + c = 4-2b
    and ab + bc + ac is the same as b(a + c) + ac

    So you can write X = b(4-2b) + ac.
    That's a quadratic that you can differentiate wrt b to find a maximum for X at a particular b. You can also find what ac must be from the same result.

    After that, a few substitutions gives you the answer.
     
    Last edited: Nov 4, 2012
  5. Nov 4, 2012 #4

    utkarshakash

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    [itex]b=\dfrac{4-a-c}{2}[/itex]
    Now substituting this into the second expression and simplifying I get
    [itex]\dfrac{4(a+c)-(a^2+c^2)}{2}[/itex]

    Whats next?
     
  6. Nov 4, 2012 #5

    utkarshakash

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    Here's what I did

    [itex]-2b^2+4b+ac[/itex]
    Differentiating wrt b and setting it to 0
    -4b+4=0
    b=1
    a+c=2
    Now , ac=c(2-c)
    Diff again wrt c and setting it to 0
    c=1
    a=1
    Substituting a,b and c I get maximum value = 3

    But the answer is 4. I can't find out my mistake. Maybe there is something wrong with differentiation.
     
  7. Nov 4, 2012 #6

    HallsofIvy

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    The derivative with respect to a is 2- a and the derivative wth respect to c is 2- c. Set those equal to 0 and you get a= c= 2 which then gives b= 0.
     
  8. Nov 6, 2012 #7

    utkarshakash

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    Gold Member

    Thanks
     
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