Question 43897
OK, so you start with a flat piece of cardboard with width 6ft and length 8ft.
If yo draw this out, you can then draw where the xft by xft squares will be cut out. Looking at the cross-shaped web you now have, we can see that the length of the box will be (6-2x)ft, the width will be (8-2x)ft, and the height of the open-topped box will be xft.

Now to work out the volume we multiply the width by the length and then multiply by the height:
V=(6-2x)*(8-2x)*(x)
V=(48-12x-16x+4x^2)*x
V=4x^3-28x^2+48x

Here's the graph:
{{{graph(300, 200, 0, 3, 0, 30, 4x^3-28x^2+48x)}}}

Note I only drew the graph uo to x=3
The reason I did this is because if x was any greater than 3, the side of the box with length (6-2x)ft would be less than zero, which is impossible!

Using the graph, we can see that thethe maximum value occurs when x is about 1.1
To be exact, V is a maximum of V=24.258 square feet when x=1.131 ft

I hope this helps.
If you have any other problems please don't hesitate to contact me at adam.chapman@student.manchester.ac.uk