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Math Help - Lagrange Multiplier issues

  1. #1
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    Lagrange Multiplier issues

    hey everyone, i'm currently stumped on a problem that seems impossible to solve.
    the formula is ((x^2)/9)+((y^2)/4)=1, and it asks to find the maximum area of a rectangle within that limit (equation)

    so far ive gotten:
    y=(lambda)((2x)/9)
    x=(lambda)(y/2)
    ((x^2)/9)+((y^2)/4)-1=0 as my equations.

    i am having major problems finding the individual values of x,y and lambda. every other problem ive seen on the web has only 2 variables, which is extremely simple.

    Can anyone solve this? thanks!
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  2. #2
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    Re: Lagrange Multiplier issues

    Hello, robman!

    Find the maximum area of the rectangle inscribed in the ellipse \frac{x^2}{9} + \frac{y^2}{4} \:=\:1

    Code:
                     |
                  *  *  *
              *      |      *
            * - - -  +  - - - *
           *|        |        |*
            |        |       y|
          * |        |        | *
          * | - - -  +  - - - | *
          * |        |    x   | *
            |        |        |
           *|        |        |*
            * - - -  +  - - - *
              *      |      *
                  *  *  *
    \text{The area of the rectangle is: }\:A \:=\:(2x)(2y) \:=\:4xy

    \text{The constraint is: }\:\frac{x^2}{9}+\frac{y^2}{4} \:=\:1


    \text{The function is: }\:f(x,y,\lambda) \;=\;4xy + \lambda\left(\tfrac{x^2}{9} + \tfrac{y^2}{4} - 1\right)


    Can you finish it now?

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  3. #3
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    Re: Lagrange Multiplier issues

    still stuck...i don't think its possible to solve for 3 variables with only 3 equations
    now i have 4y=(lambda)(2x)/9
    4x=(lambda) (y/2)
    however i try to twist around the equations still end up with 2 unknown variables...
    how did you know to use 4xy as the area? isn't the area just xy?
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  4. #4
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    Re: Lagrange Multiplier issues

    Hi robman.

    You have 4y=\frac{2\lambda x }{9} and 4x=\frac{\lambda y }{2}

    You can use these to find x in terms of y and then use the equation of the ellipse.
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  5. #5
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    Re: Lagrange Multiplier issues

    i got up to there easily but im having problems. i can only get 1 variable to cancel out, no matter what i do i still get 2 unknowns

    thanks!
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  6. #6
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    Re: Lagrange Multiplier issues

    Hello, robman!


    \text{The function is: }\:f(x,y,\lambda) \;=\;4xy + \lambda\left(\tfrac{x^2}{9} + \tfrac{y^2}{4} - 1\right)

    \begin{Bmatrix}f_x &=& 4y + \frac{2\lambda}{9}x &=& 0 & [1] \\f_y &=& 4x + \frac{\lambda}{2}y &=& 0 & [2] \\ f_{\lambda} &=& \frac{x^2}{9} + \frac{y^2}{4} - 1 &=& 0 & [3] \end{Bmatrix}

    From [1], we have: . \lambda \,=\,-\frac{18y}{x}\;\;[4]

    From [2], we have: . \lambda \,=\,-\frac{8x}{y}\;\;[5]

    Equate [4] and [5]: . -\frac{18y}{x} \:=\:-\frac{8x}{y} \quad\Rightarrow\quad 18y^2 \:=\:8x^2

    . . . . . . . . . . . . . . . . . . y^2 \:=\:\frac{4x^2}{9} \quad\Rightarrow\quad y \:=\:\pm\frac{2}{3}x\;\;[6]

    Substitute into [3]: . \frac{x^2}{9} + \frac{(\frac{2}{3}x)^2}{4} \:=\:1 \quad\Rightarrow\quad \frac{2}{9}x^2 \:=\:1

    . . . . . . . . . . . . . . . . . . . . . x^2 \:=\:\frac{9}{2} \quad\Rightarrow\quad x \:=\:\pm\frac{3}{\sqrt{2}}

    Substitute into [6]: . y \;=\;\pm\frac{2}{3}\left(\pm\frac{3}{\sqrt{2}} \right) \quad\Rightarrow\quad y \;=\;\pm\sqrt{2}


    The area of the rectangle is: . A \;=\;4xy \;=\;4\left(\frac{3}{\sqrt{2}}\right)\left(\sqrt{2  }\right) \;=\;12
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  7. #7
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    Re: Lagrange Multiplier issues

    I find it often simplest to start by eliminating the \lambda by dividing one equation by another.

    Here, you have 4y= \frac{2\lambda x}{9} and 4x= \frac{\lambda y}{2}.

    Dividing the first by the second gives \frac{y}{x}= \frac{4}{9}\frac{x}{y} which is the same as y^2= \frac{4}{9}x^2. That gives y= \pm\frac{2}{3}x.

    And, of course, we must have \frac{x^2}{9}+ \frac{y^2}{4}= \frac{x^2}{9}+ \frac{x^2}{9}= \frac{2}{9}x^2= 1 so that x= \pm\frac{3}{\sqrt{2}}. Since the " \pm" in y= \pm\frac{2}{3}x goes with both positive and values of x, there are four solutions.
    Last edited by HallsofIvy; June 27th 2012 at 09:40 AM.
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  8. #8
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    Re: Lagrange Multiplier issues

    wow thanks...what program do you use to solve the equations? I sure could use something like that on my tests also, why did you use 4xy if the area of a rectangle is xy?
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  9. #9
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    Re: Lagrange Multiplier issues

    I don't think anyone is using any software to solve the equations. Perhaps you're referring to the LaTeX which is used to produce easy to read mathematical notation. You can read about it here:
    LaTex Tutorial

    The area of the rectangle is 2x\times 2y = 4xy.
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