Question 63266

 Imagine the 6 by 8 cardboard with squares cut out of it.

|-------6---------|
...._______ 
__|.............|__ ....__ 
|.x............... x | .....|
|......................| .....|
|......................| .....| 8
|__..............__| .....|
..x..|______| x ..._|_

When you fold it up, you have a cube of have the height is x, the width is 6-2x, and the length is 8 - 2x.

The volume is, (6-2x)(8-2x)x so ...
V = (6-2x)(8-2x)x = 4x^3 - 28x^2 +48x

Let us first find the domain of x. x has to be bigger than 0, because it is a length, and less than 3, because otherwise the side 6-2x would be negative. So the domain of x is 0<x<3, so only graph over this section.

We find that the maximum volume occurs at x = about 1.1315, using a graph. This creates a volume of about 24.2584

We can calculate this value exactly using algebra as well. We realize that the max value must be a double root of the equation 4x^3 - 28x^2 +48x + D for some value of D. 

We do the algebra, and find out that this means that
x = 7/3 - &#8730;(13)/3, and at this value, the volume is (104&#8730;13)/27 + 280/27. These values do approximate to the approximations above.