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Math Help - lagrange multipliers

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    lagrange multipliers

    hey, ive got this problem, which isnt that hard, but i cannot see how to get to the solution. Any help will be very much appreciated!!

    Find the maximum of f(x,y)=4xy, subject to the condition x2+4y2=1

    its a lagrange multiplier, help!!
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    Quote Originally Posted by turion645344 View Post
    hey, ive got this problem, which isnt that hard, but i cannot see how to get to the solution. Any help will be very much appreciated!!

    Find the maximum of f(x,y)=4xy, subject to the condition x2+4y2=1

    its a lagrange multiplier, help!!
    We have the condition
    x^2 + 4y^2 = 1

    So
    x^2 + 4y^2 - 1 = 0

    Thus we may consider the maximum of the function
    f(x, y) = 4xy = 4xy + \lambda(x^2 + 4y^2 - 1)
    where \lambda is an arbitrary constant.

    Now:
    \frac{\partial f}{\partial x} = 4y + \lambda \cdot 2x = 0

    \frac{\partial f}{\partial y} = 4x + \lambda \cdot 8y = 0

    \frac{\partial f}{\partial \lambda} = x^2 + 4y^2 - 1 = 0

    The first equation says
    \lambda = - \frac{2y}{x}

    So the second equation now says:
    4x + \left ( - \frac{2y}{x} \right ) \cdot 8y = 0

    4x - \frac{16y^2}{x} = 0

    x^2 - 4y^2 = 0

    So we have two equations in two unknowns:
    x^2 - 4y^2 = 0

    x^2 + 4y^2 - 1 = 0

    Adding the two equations gives
    2x^2 - 1 = 0

    x = \pm \frac{\sqrt{2}}{2}

    Thus
    \left ( \pm \frac{\sqrt{2}}{2} \right ) ^2 - 4y^2 = 0

    4y^2 = \frac{1}{2}

    y = \pm \frac{\sqrt{2}}{4}

    So there are a total of 4 solution points:
    \left ( \pm \frac{\sqrt{2}}{2}, \pm \frac{\sqrt{2}}{4} \right )
    where the \pm signs are independent.

    I leave it to you to determine which are relative maxima or relative minima.

    -Dan
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    cheers Dan that really helped. I am still a bit new to Lagrange multipliers and am not sure how to work out the relative maxima (I have worked it out as 1, but not entirely sure) and how do I check that this is the relative maxima not the minima?
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    Quote Originally Posted by turion645344 View Post
    cheers Dan that really helped. I am still a bit new to Lagrange multipliers and am not sure how to work out the relative maxima (I have worked it out as 1, but not entirely sure) and how do I check that this is the relative maxima not the minima?
    Well, there is a more complicated way using second derivatives, but two simple methods also work:
    1) Graph it
    2) Can you find any values of the function greater than 1?

    As I get the values \pm 1 when I put in the values of x and y that were derived for possibilities, I would say that
    \left ( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{4} \right ) and \left ( -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{4} \right ) generate maximum values and that \left ( -\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{4} \right ) and \left ( \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{4} \right ) generate the minimum values.

    -Dan
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    oh right, yea that wasnt too bad! also ive managed to do most of the other questions, but this one im finding hard...


    find the minimum of g(x,y) = x2+2y2+2xy+2x+3y subject to the condition x2-y=1.

    again any help would be much appreciated!
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    Quote Originally Posted by turion645344 View Post
    oh right, yea that wasnt too bad! also ive managed to do most of the other questions, but this one im finding hard...


    find the minimum of g(x,y) = x2+2y2+2xy+2x+3y subject to the condition x2-y=1.

    again any help would be much appreciated!
    We need to solve,
    \nabla g = k\nabla f where f=x^2-y.
    Thus,
    \left< 2x+2y+2, 4y+2x+3 \right> = k\left< 2x , - 1 \right>

    Thus,
    \left\{ \begin{array}{c}2x+2y+2 = 2xk \\ 4y+2x+3 = -k \\ x^2 - y = 1 \end{array} \right\}
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    yea i managed to get to these three formulas but this is where i got stuck!
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    Quote Originally Posted by The Greatest
    \left\{ \begin{array}{c}2x+2y+2 = 2xk \\ 4y+2x+3 = -k \\ x^2 - y = 1 \end{array} \right\}
    2x+2y+2 = 2x(-4y-2x-3)

    2x+2y+2=-8xy-4x^2-6x

    8xy+4x^2+8x+2y+2=0

    4xy+2x^2+4x+y+1=0

    4x(x^2-1)+2x^2+4x+(x^2-1)+1=0

    4x^3 - 4x + 2x^2 + 4x + x^2 =0

    4x^3 + 3x^2 = 0

    x^2(4x+3)=0

    You are big boy you can do it from here.
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    cheers ive got x=-0.75 and then y= -7/16 therefore ive worked out the minimum as -1.2109 (-1,27/128)

    is this right? and how do i prove its the minimum value?
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by turion645344 View Post
    cheers ive got x=-0.75 and then y= -7/16 therefore ive worked out the minimum as -1.2109 (-1,27/128)

    is this right? and how do i prove its the minimum value?
    It looks to me like you have another possibility:
    (x, y) = (0, -1)

    So of these possibilities, which would be the minimum?

    -Dan
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    Quote Originally Posted by topsquark View Post
    It looks to me like you have another possibility:
    (x, y) = (0, -1)

    So of these possibilities, which would be the minimum?

    -Dan
    g(0, -1) = -1,

    so assuming that the sums are right then the other is certainly smaller than this.

    To show it is a minimum we substitute y=x^2-1 into the objective, and show that the resulting function has a minimum at x=-3/5 (and it does).

    RonL
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    To show it is a minimum we substitute y=x^2-1 into the objective, and show that the resulting function has a minimum at x=-3/5 (and it does).

    ive tried to this, but keep on getting the maths wrong! help!
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by turion645344 View Post
    To show it is a minimum we substitute y=x^2-1 into the objective, and show that the resulting function has a minimum at x=-3/5 (and it does).

    ive tried to this, but keep on getting the maths wrong! help!
    Plug each (x, y) point into your g(x, y) function. (The thing, after all, you are trying to minimize.) The two possibilities you have are the only ones where the function is either at a relative maximum or a relative minimum. I see that
    g(0, -1) = 0^2 + 2(-1)^2 + 2 \cdot 0 \cdot (-1) +2 \cdot 0 + 3(-1) = 2 - 3 = -1

    g \left ( -\frac{3}{5}, -\frac{16}{25} \right ) =  ~ ... ~ = -\frac{733}{625}

    Since -1 > -\frac{733}{625} the <br />
\left ( -\frac{3}{5}, -\frac{16}{25} \right ) point is your minimum.

    -Dan
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    Quote Originally Posted by topsquark View Post
    Plug each (x, y) point into your g(x, y) function. (The thing, after all, you are trying to minimize.) The two possibilities you have are the only ones where the function is either at a relative maximum or a relative minimum. I see that
    g(0, -1) = 0^2 + 2(-1)^2 + 2 \cdot 0 \cdot (-1) +2 \cdot 0 + 3(-1) = 2 - 3 = -1

    g \left ( -\frac{3}{5}, -\frac{16}{25} \right ) = ~ ... ~ = -\frac{733}{625}

    Since -1 > -\frac{733}{625} the <br />
\left ( -\frac{3}{5}, -\frac{16}{25} \right ) point is your minimum.

    -Dan
    What we have shown is that the curves defined by the objective and the
    constraint kiss at the points we have found. A kissing point may be the
    analog of a local maxima, minima or point of inflection . We still need to
    classify the point, or at least justify our conclusion that one is a minimum.

    In this case we can conclude that the function does have a minima as when
    x goes of to infinity the objective becomes arbitrily large, also that any
    minima for this objective and constraint must be smooth, so we then know
    that one of the kissing points must be a minima, and we have identified which
    it is.

    RonL

    RonL
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    i was just wondering where this came from, as i thought that the other values were (-0.75,-7/16) which gives a value of -1,27/128. this is smaller than -1 so is the minimum value right?

    also to test the minimum using the idea that as
    x goes of to infinity the objective becomes arbitrily large. can you actually use this concept to test for a minimum value?!
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