Introduction
The Open Box Problem
A square piece of card has four equal sized squares cut out from each of it's four corners and is folded to make an open top box. The problem is to find out which value of x (the side length of the cut out) will maximise the volume in the resulting box, which is equivalent to the height (x) * the width (the side length of the square - 2x) * the depth (the same value as the width). This problem can be expanded to finding the maximised volume for a box that is made from a rectangular piece of card.
Methods of Finding the Solution to the Problem - Trial and Error Approach
There are two main ways of finding the solution to the problem. The first is a trial and error approach. For square pieces of card, a table is constructed showing the cut out side length and the resulting volume.
In the example, the square being investigated has a side length of 20 cm. The length of the cut out is increased until the resulting volume goes down. This step is repeated to one and then two decimal places, giving the optimum side length to 3.33 cm as the maximum volume. The results for different length squares can be worked out, collected and then any relationships between square side and cut out can then be worked out.
The Open Box Problem
A square piece of card has four equal sized squares cut out from each of it's four corners and is folded to make an open top box. The problem is to find out which value of x (the side length of the cut out) will maximise the volume in the resulting box, which is equivalent to the height (x) * the width (the side length of the square - 2x) * the depth (the same value as the width). This problem can be expanded to finding the maximised volume for a box that is made from a rectangular piece of card.
Methods of Finding the Solution to the Problem - Trial and Error Approach
There are two main ways of finding the solution to the problem. The first is a trial and error approach. For square pieces of card, a table is constructed showing the cut out side length and the resulting volume.
In the example, the square being investigated has a side length of 20 cm. The length of the cut out is increased until the resulting volume goes down. This step is repeated to one and then two decimal places, giving the optimum side length to 3.33 cm as the maximum volume. The results for different length squares can be worked out, collected and then any relationships between square side and cut out can then be worked out.
