1. The problem statement, all variables and given/known data 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 2. Relevant equations 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] abs(ax+by+cz+d)/(sqrt(a2+b2+c2)) 3. The attempt at a solution L = distance from (0,0,0) to plane x +y+4z=4 = abs(-4)/sqrt(18) = 4/(3sqrt2) V= (4/(3sqrt2))3/(3sqrt3) = 32/81sqrt6 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.