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Math Help - Larange Multipliers Problem

  1. #1
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    Larange Multipliers Problem

    I am having a bit of trouble setting up a problem:

    (1) Find the volume of the largest regtangular box with edges parallel to the axes that can be inscribed in the ellipsoid: 9x^2 + 36y^2 + 4z^2 = 36

    I thought it would be simply:

    f(x,y,z) = xyz && g(x,y,z) = 9x^2 + 36y^2 + 4z^2 = 36

    However, this seems to give me the wrong solution.

    Any help would be appreciated.
    Last edited by machi4velli; June 4th 2007 at 11:33 PM.
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  2. #2
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    Quote Originally Posted by machi4velli View Post
    I am having a bit of trouble setting this problem up:

    Find the volume of the largest regtangular box with edges parallel to the axes that can be inscribed in the ellipsoid: 9x^2 + 36y^2 + 4z^2 = 36

    Any help would be appreciated.
    Let the box have width w, depth d, and height h.

    Then:

    <br />
(h/2)^2 \le 36/4<br />

    Also:

    <br />
(w/2)^2 \le \frac{36-4(h/2)^2}{9}<br />

    and:

    <br />
(d/2)^2 = \frac{36-4(h/2)^2-9(w/2)^2}{36}<br />
.

    So our problem becomes:

    Maximise f(h,w,d) = h\times w \times d subject to:

    <br />
(h/2)^2 \le 9<br />

    <br />
(w/2)^2 \le \frac{36-4(h/2)^2}{9}<br />

    <br />
(d/2)^2 = \frac{36-4(h/2)^2-9(w/2)^2}{36}<br />
.

    RonL
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  3. #3
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    Quote Originally Posted by machi4velli View Post
    I am having a bit of trouble setting up a problem:

    (1) Find the volume of the largest regtangular box with edges parallel to the axes that can be inscribed in the ellipsoid: 9x^2 + 36y^2 + 4z^2 = 36

    I thought it would be simply:

    f(x,y,z) = xyz && g(x,y,z) = 9x^2 + 36y^2 + 4z^2 = 36

    However, this seems to give me the wrong solution.

    Any help would be appreciated.
    I agree with your setup except that you need to multiply xyz by a constant to get the volume.

    In two dimensions with an ellipse,

    max xy subject to 9x^2 + 36y^2 = 36

    would produce a point (x,y) on the ellipse in the upper right quadrant. The inscribed rectangle would have corners (x,y), (x,-y), (-x,y) and (-x,-y). The area of the rectangle would be 4xy.



    In three dimensions, the constant to multiply xyz by is 8.
    Last edited by JakeD; June 5th 2007 at 12:39 AM.
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  4. #4
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    Hello, machi4velli!

    (1) Find the volume of the largest regtangular box with edges parallel to the axes
    that can be inscribed in the ellipsoid: . 9x^2+36y^2+4z^2\:=\:36
    This is the way I was taught to use Lagrange Multipliers.


    The length of the box is 2x, the width 2y, the height 2z.
    . . The volume of the box is: . (2x)(2y)(2z) \,=\,8xyz.

    We have: . V(x,y,z)\:=\:8xyz .and . g(x) \:=\:9x^2+36y^2+4z^2 -36

    . . Then: . f(x,y,z,\lambda) \;=\;8xyz + \lambda(9x^2 + 36y^2 + 4z^2 - 36)


    Set the four partial derivatives equal to zero, and solve.

    [1] . f_x \:=\:8yz + 18x\lambda\:=\:0

    [2] . f_y \:=\:8xz + 72y\lambda\:=\:0

    [3] . f_z \:=\:8xy + 8z\lambda \:=\:0

    [4] . f_{\lambda} \:=\:9x^2+36y^2 + 4z^2 - 36 \:=\:0


    From [1], we have: . \lambda \:=\:-\frac{8yz}{18x}\:=\:-\frac{4y^2}{9x} .[5]

    From [2], we have: . \lambda \:=\:-\frac{8xz}{72y}\:=\:-\frac{xz}{9y} .[6]

    From [3], we have: . \lambda\:=\:-\frac{8xy}{8z}\:=\:-\frac{xy}{z} .[7]


    Equate [5] and [6]: . -\frac{4yz}{9x}\:=\:-\frac{xz}{9y}\quad\Rightarrow\quad x^2\,=\,4y^2 .[8]

    Equate [6] and [7]: . -\frac{xz}{9y} \:=\:-\frac{xy}{9z}\quad\Rightarrow\quad z^2 \,=\,9y^2 .[9]


    Substitute [8] and [9] into [4]: . 9(4y^2) + 36y^2 + 4(9y^2) - 36 \:=\:0

    . . and we get: . 108y^2\:=\:36\quad\Rightarrow\quad y^2\:=\:\frac{1}{3}

    Substitute into [8]: . x^2\:=\:4\left(\frac{1}{3}\right) \:=\:\frac{4}{3}
    Substitute into [9]: . z^2 \:=\:9\left(\frac{1}{3}\right)\:=\:3

    . . Hence: . x\:=\:\frac{2\sqrt{3}}{3},\;y \:=\:\frac{\sqrt{3}}{3},\;z\:=\:\sqrt{3}


    Therefore, the maximum volume is: . 8\left(\frac{2\sqrt{3}}{3}\right)\left(\frac{\sqrt  {3}}{3}\right)\left(\sqrt{3}\right) \:=\:\frac{16\sqrt{3}}{3} unitsł

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  5. #5
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    heheh, the 8 threw me off, i was getting 2sqrt3/3, thanks a lot guys
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