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Math Help - Smaller tetrahedron possible limited by a plane

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    Smaller tetrahedron possible limited by a plane

    I need help with this:
    Find the equation of a plane that contains the point (1, 2, 1) and intersects the coordinate axes forming the smaller (least volume) tetrahedron possible.

    I really have no idea how to proceed.
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  2. #2
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    Quote Originally Posted by nvwxgn View Post
    I need help with this:
    Find the equation of a plane that contains the point (1, 2, 1) and intersects the coordinate axes forming the smaller (least volume) tetrahedron possible.

    I really have no idea how to proceed.
    Here is an outline of what needs to be done.

    Lets start with what we know
    Smaller tetrahedron possible limited by a plane-capture.jpg

    Using this we can make a generic equation of of this plane note that

    \displaystyle \frac{\partial z}{\partial x}=-\frac{c}{a}
    and
    \displaystyle \frac{\partial z}{\partial y}=-\frac{c}{b}

    This gives the equation of the plane as

    \displaystyle z=-\frac{c}{a}x-\frac{c}{b}y+c Note also since (1,2,1) must be in the plane we get

    \displaystyle 1=-\frac{c}{a}-\frac{2c}{b}+c \iff abc=ab+bc+2ac

    (we will need this constraint later)

    Note the volume of the tetrahedron is given by

    \displaystyle V(a,b,c)=\int_{0}^{a}\int_{0}^{-\frac{b}{a}x+b}\left( -\frac{c}{a}x-\frac{c}{b}y+c\right)dydx=\frac{1}{6}abc

    Now we need to minimize this using Lagrange multipliers. This gives the system

    \nabla V=\lambda\nabla (ab+bc+2ac-abc)

    Solving I get

    \lambda=\frac{1}{2},a=3,b=6,a=3 This gives a volume of 9.
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