Denote the side of the removed squares as . Then the length and width of the base of the open box will each be as will be removed from each side.
The height of the open box will be .
The volume of the box, , is the product of its length, width and height, and therefore
To find the value of that maximises the volume, first differentiate with respect to .
Stationary points occur when , so set the derivative equal to 0 and solve the resulting quadratic equation for :
The value of is the value that would give the minimum volume, as this would gives us
Therefore, the maximum volume is achieved when the side of the square removed is .