How to reduce 3 dimensional optimization to 2 dimensions?

Question

I am trying to minimize the surface area of a parallelepiped of unit volume.

Using

$$\mbox{Volume } = xyz\sqrt{1 + 2\cos(a)\cos(b)\cos(c) - \cos^2(a) - \cos^2(b) - \cos^2(c)} = 1$$

where $x$, $y$, $z$ are edge lengths and $a$, $b$, $c$ are the angles between them.

And

$$\mbox{Surface Area } = 2(xy\sin(a) + xz\sin(b) + yx\sin(c))$$

How can I eliminate one dimension (which is what the problem suggests I do)?

Answer 1

HINT:Angles a,b and c should be right angles.

Answer 2

This solution uses vectors. Take one vertex to be the origin, and define the edges originating from it as follows

$ u = x( 1 , 0, 0 ) $

$ v = y (\cos a, \sin a , 0 ) $

$ w = z ( \sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta )$

Then, the volume is given by

$ \text{Volume} = x y z \sin(a) \cos(\theta) $

And the surface area is

$ \text{Area} = 2 ( \| u \times v \| + \| u \times w \| + \| v \times w \| )$

And we have

$ u \times v = x y (0, 0, \sin a ) $

$ u \times w = x z (0, - \cos \theta, \sin \theta \sin \phi ) $

$ v \times w = y z ( \sin a \cos \theta , - \cos a \cos \theta , \sin \theta ( \cos a \sin \phi - \sin a \cos \phi ) $

So that

$ \| u \times v \| = x y \sin a \tag{1}$

$ \| u \times w \| = x z \sqrt{ \cos^2 \theta + \sin^2 \theta \sin^2 \phi } \tag{2}$

$ \| v \times w \| = y z \sqrt{ \cos^2 \theta + \sin^2 \theta \sin^2(\phi - a ) } \tag{3}$

Using the volume equation, we have

$ \cos \theta = \dfrac{\text{Volume}}{ x y z \sin a } = \dfrac{1}{x y z \sin a } \tag{4}$

Looking at $(1),(2),(3)$, we see that using $(4)$ we can elimiante $ \theta $.

And the expressions in $(2)$ and $(3)$ become

$\|u \times w \| = x z \sqrt{ \sin^2 \phi + \left( \dfrac{\cos^2 \phi}{x^2 y^2 z^2 \sin^2 a } \right) } $

$ \| v \times w \| = yz \sqrt{ \sin^2 (\phi - a) + \left( \dfrac{\cos^2(\phi - a) }{x^2 y^2 z^2 \sin^2 a } \right) } $

So now, we have completely eliminated $\theta$, and the expression for the surface area is a function of only $a$ and $\phi$.

$\text{Area} = 2x y \sin a + 2x z \sqrt{ \sin^2 \phi + \left( \dfrac{\cos^2 \phi}{x^2 y^2 z^2 \sin^2 a } \right) } +2yz \sqrt{ \sin^2 (\phi - a) + \left( \dfrac{\cos^2(\phi - a) }{x^2 y^2 z^2 \sin^2 a } \right) } $