# Thread: Smaller tetrahedron possible limited by a plane

1. ## 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.

2. 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.

Using this we can make a generic equation of of this plane note that

$\displaystyle \displaystyle \frac{\partial z}{\partial x}=-\frac{c}{a}$
and
$\displaystyle \displaystyle \frac{\partial z}{\partial y}=-\frac{c}{b}$

This gives the equation of the plane as

$\displaystyle \displaystyle z=-\frac{c}{a}x-\frac{c}{b}y+c$ Note also since $\displaystyle (1,2,1)$ must be in the plane we get

$\displaystyle \displaystyle 1=-\frac{c}{a}-\frac{2c}{b}+c \iff abc=ab+bc+2ac$

(we will need this constraint later)

Note the volume of the tetrahedron is given by

$\displaystyle \displaystyle V(a,b,c)=\int_{0}^{a}\int_{0}^{-\frac{b}{a}x+b}\left( -\frac{c}{a}x-\frac{c}{b}y+c\right)dydx=\frac{1}{6}abc$

Now we need to minimize this using Lagrange multipliers. This gives the system

$\displaystyle \nabla V=\lambda\nabla (ab+bc+2ac-abc)$

Solving I get

$\displaystyle \lambda=\frac{1}{2},a=3,b=6,a=3$ This gives a volume of $\displaystyle 9$.