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## Homework Statement

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

## Homework Equations

Volume of a box with L being the length of a diagonal in that box = L

^{3}/(3sqrt3)

Shortest distance from any point (x,y,z) to a plane with normal vector [a,b,c]

abs(ax+by+cz+d)/(sqrt(a

^{2}+b

^{2}+c

^{2}))

## 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.