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Homework Help: Volume in first octant cut off by a plane

  1. Dec 2, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the equation of the plane through the point [1,2,2] that cuts off the smallest possible volume in the first octant.

    2. Relevant equations

    Volume of a pyramid = 1/3Ah

    3. The attempt at a solution

    The plane is going to cut out a pyramid with the x-, y-, and z-intercepts, so let x, y, and z be the intercepts. Then V = 1/6xyz. But since the plane must go through [1,2,2] and three points define a plane, we can write one of x, y, and z in terms of the other two. Any plane passing through intercepts x, y, and z has a general point [a,b,c] so that:
    a/x + b/y + c/z = 1
    Since [1,2,2] is on the plane, plug that in for [a,b,c]:
    1/x + 2/y + 2/z = 1
    Solve for x (just because I'm guessing that it would be the easiest):
    x = -1/(2y + 2z - 1)
    Now plug that into the volume formula:
    V = -yz/(12y + 12z - 6)

    Is this right so far? If not, what did I do wrong? If so, how can I continue?
     
    Last edited: Dec 2, 2008
  2. jcsd
  3. Dec 2, 2008 #2

    Dick

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    I think it's ok so far. Now take the partial derivatives dV/dy and dV/dz and set them both to zero. That's two equations in two unknowns.
     
  4. Dec 2, 2008 #3
    All right, so from there I take the partials and I get:

    dV/dy = (-12z^2 + 6z)/(12y + 12z - 6)^2
    dV/dz = (-12y^2 + 6y)/(12y + 12z - 6)^2

    Setting these equal to zero, I get:

    y = z = 0 and y = z = 1/2

    However, I know that y and z can't be 0, and if y and z are 1/2, then by my earlier result:

    x = -1/(1 + 1 - 1) = -1

    I can't have x = -1, because the plane has to intersect the positive x-axis or else the volume will be infinite. What did I do wrong here?
     
  5. Dec 3, 2008 #4

    Dick

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    You have a mistake solving for x in 1/x + 2/y + 2/z = 1. 1/x=1-(2/y+2/z). That's
    x=1/(1-(2/y+2/z)). Not the same as your expression. I missed it. Sorry.
     
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