# Smaller tetrahedron possible limited by a plane

• Dec 7th 2010, 07:42 AM
nvwxgn
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.
• Dec 7th 2010, 09:00 AM
TheEmptySet
Quote:

Originally Posted by nvwxgn
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.

Attachment 20010

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 $9$.