Find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in the plane x + y + 4z = 4
Volume of a box with L being the length of a diagonal in that box = L3/(3sqrt3)
Shortest distance from any point (x,y,z) to a plane with normal vector [a,b,c]
The Attempt at a Solution
L = distance from (0,0,0) to plane x +y+4z=4
Now I realized that what I did here that was wrong was I assumed that the shortest distance to the plane was that diagonal. But it doesn't have to be, or does it?
Is there only one way that the box can have a vertex on that plane? I figured that you can change the shape of the box and a vertex will hit the plane somewhere else.
Thanks for your help.