Maximizing Sphere in Tetrahedron in FIRST Octant

[davedave]

Homework Statement

Consider the tetrahedron in the FIRST octant defined by x+y+z/2=1.
Find the maximum sphere inside the tetrahedron.

Homework Equations

I use Lagrange Multipliers. let L be lamba.

(del)f(x,y,z)=L*(del)g(x,y,z)

The Attempt at a Solution

I don't know if I can assume that the center of the sphere is (a,a,2a) where 0<a<1

reason: Since the x, y, z intercepts of the tetrahedron are 1, 1, 2 respectively, I let the z
coordinate of the sphere be twice the x and y coordinates.

f(x,y,z)=(x-a)^2+(y-a)^2+(z-2a)^2

g(x,y,z)=x+y+z/2-1=0

next, take the gradient of f and g in the equation

2(x-a)i+2(y-a)j+2(z-2a)k=L*(i+j+k/2)

solving for x, y, z gives x=L/2+a y=L/2+a z=L/4+2a

put them into the tetrahedron equation and solve for lamba L

L=8/9 * (1-3a)

now put the value of L into the x, y, z equations which gives

x=4/9*(1-3a)+a y=4/9*(1-3a)+a z=2/9*(1-3a)+2a

thus, put those equations above into f(x,y,z)= 4/9*(1-3a)^2

therefore, the radius is r=2/3*(1-3a).

Once we know the radius, we can maximize the sphere.


Is my solution correct? If not, how do you do it?

 

[Dick]

The sphere should be tangent to each of the bounding planes, right? Doesn't that mean the center should be equidistant from each of the xy, yz and xz planes? I.e. the center is at (a,a,a) for some a. Now the center should also be distance a from the plane x+y+z/2=1.