For those who don't know the max box problem here is a quick recap... A lidless box is to be constructed from a square sheet of paper measuring 15cm by 15cm, by cutting out squares from the corners and then folding up the sides. Let x cm be the size of one of the sides of the small corner squares. Question: prove the following "the maximum volume will always be of a height which is one sixth of the length of one side of the square piece of paper used". If S is the length of one side of the large square and x as before, write down the volume in terms of S and x and use calculus to prove that indeed this is right? Here is my work so far.... Volume is x (S-2x)^2 = x (S-2x)(S-2x) = x (S^2 - 4Sx + 4x^2) = 4x^3 - 4Sx^2 + xS^2 Can anyone help with the rest please?