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    Find the maximum value of y.x^2 - x.y^2 when x and y are between 0 and 1 inclusive
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    factorising i get:

    xy(x-y)

    x=1 (for max solution)

    therefore:

    y(1-y)

    now for this to have a max value we differentiate

    y - y^2

    1 - 2y

    max occurs when this equals 0, so;

    1 - 2y = 0
    1 = 2y
    y = 1/2

    now 2nd differential gives -2, therefore the value calculated is max.

    x = 1, y = 1/2

    and max answer is 1/4
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    (Original post by IntegralNeo)
    factorising i get:

    xy(x-y)

    x=1 (for max solution) ....
    I can't get that, why for max solution we have x = 1? Can u explain?

    I did like that:
    Let y = kx, so k is in R+.
    so we have yx^2 - xy^2 = kx^3 - k^2x^3
    = x^3(k - k^2).
    So it is max if (x^3) max and (k - k^2) max
    then have x = 1, k = 1/2.
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    (Original post by BCHL85)
    I can't get that, why for max solution we have x = 1? Can u explain?

    I did like that:
    Let y = kx, so k is in R+.
    so we have yx^2 - xy^2 = kx^3 - k^2x^3
    = x^3(k - k^2).
    So it is max if (x^3) max and (k - k^2) max
    then have x = 1, k = 1/2.
    It's fairly logical, say y is fixed, x-y is maximised when x is greatest, likewise with xy...
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    (Original post by beauford)
    It's fairly logical, say y is fixed, x-y is maximised when x is greatest, likewise with xy...
    exactly
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    ok, someone please tell me which P unit this is- just to reassure myself i wasnt meant to know that!! lol
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    (Original post by faa)
    ok, someone please tell me which P unit this is- just to reassure myself i wasnt meant to know that!! lol
    its not in any P-unit...it wil probably come on a BMo paper 9unlikely coz its too simple) or on a maths challenge...so for Pure knowledge ur not supposed to know this...i would have thought :rolleyes:
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    that was a warmup

    maximise:

    yx^2 + zy^2 + xz^2 - zx^2 - xy^2 - yz^2

    if x, y and z are all between 0 and 1 inclusive
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    Im pretty sure the answer is 1/4, (x,y,z) = (0, 1, 1/2) which is quite interesting, i cant prove it tho - just wrote a little program instead.
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    Yes, I get two answers,
    f(x,y,z) = ¼, (x,y,z) = (0,1,0.5)
    f(x,y,z) = ¼, (x,y,z) = (0.5,0,1)

    But I got those results using 3-D graphs
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    To prove the above result mathematically you have to take partial derivatives of the function f(x,y,z) with respect to x, y and z seperately, and taking them equal to 0 solve the simultaneous equations, with the boundaries that they must lie between 0 and 1, and you will obtain 1/4 as the maximum value.
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    (Original post by AntiMagicMan)
    To prove the above result mathematically you have to take partial derivatives of the function f(x,y,z) with respect to x, y and z seperately, and taking them equal to 0 solve the simultaneous equations, with the boundaries that they must lie between 0 and 1, and you will obtain 1/4 as the maximum value.
    Can you quickly do it, cos thats what i tried, but I think i messed up somewhere.
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    (Original post by mik1a)
    that was a warmup

    maximise:

    yx^2 + zy^2 + xz^2 - zx^2 - xy^2 - yz^2

    if x, y and z are all between 0 and 1 inclusive
    Although the factorisation isn't obvious, if you have a good enough look (i tried for about 10-15 minutes) it factorises to (x-y)(z-x)(z-y). Using a similar argument as in the first part, set z = 1, to get (x-y)(1-x)(1-y), now set y = 0, to maximise 1-y and x-y, and so you get (x)(1-x) <= 1/4. You could argue that you could also set z = 0 to make the - signs cancel to give xy(x-y), which is the same as in q.1 so is <= 1/4.
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    (Original post by AntiMagicMan)
    To prove the above result mathematically you have to take partial derivatives of the function f(x,y,z) with respect to x, y and z seperately, and taking them equal to 0 solve the simultaneous equations, with the boundaries that they must lie between 0 and 1, and you will obtain 1/4 as the maximum value.
    You would take partial derivatives for functions of more than one variable to find the turning points, or local maxima and minima.
    This function has a limited domain. In this domain the function has no turning points, but has a maximum and a minimum value.
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    (Original post by beauford)
    Although the factorisation isn't obvious, if you have a good enough look (i tried for about 10-15 minutes) it factorises to (x-y)(z-x)(z-y). Using a similar argument as in the first part, set z = 1, to get (x-y)(1-x)(1-y), now set y = 0, to maximise 1-y and x-y, and so you get (x)(1-x) <= 1/4. You could argue that you could also set z = 0 to make the - signs cancel to give xy(x-y), which is the same as in q.1 so is <= 1/4.
    I got that factorisation also, but that was as far as I could go. I knew that z=1 gave a max value, but I just couldn't justify why I should set z=1!
    Also z=0.5 is another value giving a max for the function (value=¼), which put me further off trying to justify why I should set z=1!!
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    (Original post by Fermat)
    I got that factorisation also, but that was as far as I could go. I knew that z=1 gave a max value, but I just couldn't justify why I should set z=1!
    Also z=0.5 is another value giving a max for the function (value=¼), which put me further off trying to justify why I should set z=1!!
    Of course given the factorisation as it stands (x-y)(z-x)(z-y) then maximisation clearly occurs when you set z = 1 or 0, as you are increasing the value of the product (z-x)(z-y), clearly (1-x)(1-y) > (0.8-x)(0.8-y) or (0-x)(0-y) > (0.8-x)(0.8-y). The reason you get a max when z = 1/2, is because the factorisation is identical to (y-x)(z-x)(y-z), in this case we set y = 1 or 0 and x = 0, z = 1/2
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    OK, gottit now
    I could see that the double product would be at a maximum, but somehow or other I thought that wasn't enouogh to justify the triple product being a maximum!

    Ta.
 
 
 
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