Body Diagonal of a Rectangular Prism

**davesface** · September 27th 2009, 10:06 AM

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} L=\sqrt{x^{2}+y^{2}+z^{2}} $

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

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)} $

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?

**Soroban** · September 27th 2009, 11:51 AM

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}} $ .[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]

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

. . 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: . $V \;=\;\left(\frac{L}{\sqrt{3}}\right)^3 \;=\;\frac{L^3}{3\sqrt{3}}$

**davesface** · September 27th 2009, 11:55 AM

Honestly, I hadn't even thought about taking a calculus approach. Thanks a bunch.

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