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Math Help - Absolute Max and Min

  1. #1
    Junior Member
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    Absolute Max and Min

    Hey i'm having difficulty solving this problem:

    Find the absolute maximum and minimum values of f on the set D.
    f(x, y) = xy2 + 7D = {(x, y) | x ≥ 0, y ≥ 0, x2 + y2 ≤ 3}

    I tried doing this:
    fx = y^2
    fy = 2xy
    critical point: (0,0)

    then i got y^2 = 3-x^2 and plugging that in to the original to get g(x) = f(x,y) = -x^3+3x+7. setting that equal to zero i get x=1,-1. but i'm totally lost on what to do from here (or even if i've started correctly). Any and all help is greatly appreciated.
    Thanks.
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  2. #2
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    Allahabad,Uttarpradesh
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    Smile please expalin your process ..i don't understand

    I tried doing this:
    fx = y^2
    fy = 2xy
    critical point: (0,0).....even i don't understand what u have written.
    what u have meant by fx,fy???
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  3. #3
    MHF Contributor matheagle's Avatar
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    I'm guessing that he/she means the partial derivatives by fx and fy.
    BUT they are missing powers here.
    I'm guessing f= xy^2 and the constraint is the interior of the circle x^2+y^2 =3.
    Otherwise I have no idea what x2 and y2 are.
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  4. #4
    MHF Contributor chisigma's Avatar
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    If the function under study is...

    f(x,y)= x \cdot y^{2} (1)

    ... [we are omitting the constant term + 7...] the partial derivatives are effectively...

    f_{x}(*,*)= y^{2}

    f_{y}(*,*)= 2\cdot x\cdot y (2)

    Both the f(*,*) and the derivatives are non negative in the quarter of circle [x\ge 0, y\ge 0, x^{2} + y^{2} \le 3], so that the maximum is on the circular contour, i.e. a point in which is...

    y^{2}= 3 - x^{2} (3)

    Substituting (3) in (1) we obtain the function of the only x...

    f(x,3-x^{2}) = 3\cdot x - x^{3} (4)

    ... the derivative of which vanishes in x=0, x=-1, and x=1. But in x=0 is f(*,*)=0 and x=-1 is outside the quarter of circle, so that x=1 is the only solution. The maximum of f(*,*) in D is then in the point...

    [x,y]= [1,\sqrt{2}]

    ... and its value is f(1,\sqrt{2})= 2 [+ 7]...

    Since on the x and y axes is f(*,*)=0, the (1) has no minimum points...

    Kind regards

    \chi \sigma
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