Results 1 to 3 of 3

Math Help - Body Diagonal of a Rectangular Prism

  1. #1
    Member
    Joined
    Aug 2008
    Posts
    88

    Body Diagonal of a Rectangular Prism

    The length of the body diagonal of a rectangular prism is L. What is the largest possible volume enclosed by the prism?

    So I said that the dimensions are x, y, and z. Using a Pythagorean theorem:

    L^{2}=x^{2}+y^{2}+z^{2}<br />
L=\sqrt{x^{2}+y^{2}+z^{2}}<br />

    Using this to solve for each side:
    <br />
x=\sqrt{L^{2}-y^{2}-z^{2}}<br />
y=\sqrt{L^{2}-x^{2}-z^{2}}<br />
z=\sqrt{L^{2}-x^{2}-y^{2}}<br />

    Then substituting back into the equation for volume:
    V=xyz=\sqrt{\left( L^{2}-y^{2}-z^{2}\right)+\left( L^{2}-x^{2}-z^{2}\right)+\left( L^{2}-x^{2}-y^{2}\right)}<br />

    which simplifies to:
    V=xyz=\sqrt{3L^{2}+\left(-2x^{2}-2y^{2}-2z^{2} \right)}

    But then you can say:
    -2L^{2}=-2x^{2}-2y^{2}-2z^{2}

    And then substituting into the simplified volume equation:
    V=\sqrt{3L^{2}-2L^{2}}=L

    I don't see any problems with my math, but the answer doesn't seem intuitively correct to me. Maybe it is, but it just doesn't feel right at all. Can anyone conclusively confirm or deny this answer?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,684
    Thanks
    615
    Hello, davesface!

    We usually need Calculus to maximize a quantity . . .


    The length of the body diagonal of a rectangular prism is L.

    What is the largest possible volume enclosed by the prism?

    Let the dimensions be x.\,y, z.

    Then: . x^2+y^2+z^2 \:=\:L^2 \quad\Rightarrow\quad z \:=\:(L^2 - x^2 - y^2)^{\frac{1}{2}}<br />
.[1]

    The volume is: . V \;=\;xyz .[2]


    Substitute [1] into [2]: . V \;=\;xy(L^2 - x^2 - y^2)^{\frac{1}{2}}


    Equate the partial derivatives to zero ... and solve:

    . . \frac{\partial V}{\partial x} \;=\;xy\cdot\frac{1}{2}(L^2-x^2-y^2)^{-\frac{1}{2}}(\text{-}2x) + y(L^2 - x^2 - y^2)^{\frac{1}{2}} \;=\;0 . . \Longrightarrow\qquad y\sqrt{L^2-x^2-y^2} - \frac{xy^2}{\sqrt{L^2-x^2-y^2}} \;=\;0 .[3]

    . . \frac{\partial V}{\partial y} \;=\;xy\cdot\frac{1}{2}(L^2-x^2-y^2)^{-\frac{1}{2}}(\text{-}2y) + x(L^2-x^2-y^2)^{\frac{1}{2}} \;=\;0 . . \Longrightarrow\qquad x\sqrt{L^2-x^2-y^2} - \frac{xy^2}{\sqrt{L^2-x^2-y^2}} \;=\;0 .[4]


    \begin{array}{cc}{\color{blue}[3]}\text{ simplifies to:} & L^2 - x^2 - y^2 \:=\:x^2 \\ \\[-4mm]<br /> <br />
{\color{blue}[4]}\text{ simplifies to:} & L^2 - x^2 - y^2 \:=\:y^2 \end{array}\qquad\text{Hence: }\:x^2\:=\:y^2 \quad\Rightarrow\quad x \:=\:y


    . . And we find that: . x \:=\:y\:=\:z . . . The rectangular prism is a cube.


    Hence: . x \:=\:y \:=\:z \:=\:\frac{L}{\sqrt{3}}
    . . And the maximum volume is: . V \;=\;\left(\frac{L}{\sqrt{3}}\right)^3 \;=\;\frac{L^3}{3\sqrt{3}}

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Aug 2008
    Posts
    88
    Honestly, I hadn't even thought about taking a calculus approach. Thanks a bunch.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Diagonal of a prism
    Posted in the Geometry Forum
    Replies: 1
    Last Post: December 21st 2008, 01:13 AM
  2. Rectangular Prism
    Posted in the Geometry Forum
    Replies: 1
    Last Post: August 29th 2008, 03:55 AM
  3. rectangular prism problem
    Posted in the Geometry Forum
    Replies: 1
    Last Post: July 29th 2008, 08:38 PM
  4. Rectangular Prism
    Posted in the Geometry Forum
    Replies: 1
    Last Post: June 10th 2007, 11:04 AM
  5. Rectangular Prism
    Posted in the Geometry Forum
    Replies: 3
    Last Post: January 17th 2007, 06:22 PM

Search Tags


/mathhelpforum @mathhelpforum