Question 52505
To find the volume of such a box, multiply the width by the length by the height.
The width, after cutting out the corner squares of x by x feet, will be (6-2x) feet and the length will be (8-2x) feet, and, of course, the height of the box will be x feet. So, the volume is expressed by:
{{{V = (8-2x)(6-2x)x}}} Expand this to get:
{{{V = (48-16x-12x+4x^2)x}}} Simplifying this:
{{{V = 4x^3-28x^2+48x}}} This represents the volume of the box expressed in terms of x.

The graph of this cubic function looks like:
{{{graph(300,200,-5,6,-5,25,4x^3-28x^2+48x)}}}
The valid range of x is x = 0 to x = 3

Using the graph to find the value of x that will produce the maximum volume is a little more than x = 1.
The actual value can be found with a little elementary differential calculus, because it's a matter of finding the "relative" maximum of the cubic curve.
Take the first derivative of the function and set it equal to zero, then solve for x.
{{{dV/dx = 12x^2-56x+48}}} Set this equal to zero.
{{{12x^2-56x+48 = 0}}}
The roots are:
x = 3.535... Ignore this solution as x is too large.
x = 1.131... This is the approximate value of x that will produce the largest volume.