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• Level: GCSE
• Subject: Maths
• Word count: 1817

Open Box Problem

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Introduction

In this investigation, a box without a lid must be made from a sheet of card, as shown below. Identical squares must be cut out of each corner and the dotted lines folded along to form the sides of the box. The goal of the investigation is to find out a relationship between the size of the initial piece of card, the size of the identical corner squares and the volume of the resulting open box. This will allow me to say what size corner square will produce the box of the largest volume, for any given rectangular sheet of card. To begin with, I am going to investigate the size of the corner square that must be cut out to make an open box of the largest volume, for any sized square sheet of card. Once I have found the formula that allows me to find this out easily, I will progress to using an initial piece of card that is rectangular in shape. The formula used to obtain the volume of a box is VOLUME = Length * Width * Height (where * is multiplication) To show a simple example of how this formula works with the open box, I will first of all use a initial piece of card that is 20cm by 20 cm, and a corner square (from now on called 'cut-off') ...read more.

Middle

Side Length Maximum Cut-off 5 5/6 6 1/1 7 7/6 8 4/3 9 3/2 10 5/3 11 11/6 Then I put all the fractions over a common denominator to compare them easier. Side Length Maximum Cut-off 5 5/6 6 6/6 7 7/6 8 8/6 9 9/6 10 10/6 11 11/6 Here I can see a definite pattern between the side length and the maximum cut-off. The maximum cut-off is equal to the side length divided by 6. To test this theory, I predict that with a side length of 12, the cut off that will produce the maximum volume will be 12/6, or 2. My prediction has been proved right in practice. So now I know that for a square, the cut-off that gives the greatest volume of box is equal to the length of one of the sides divided by 6. Now that I know this, I can rewrite my original formula to give the largest volume possible for a square, by substituting "Side/6" where I see "Cut-off". So: Largest Volume = (Side - 2 * Side/6)2 * Side/6 LV = (S - 2S/6)2 * S/6 To test this I will use a side length of nine, which I already know has a largest volume of 54cm3 LV = (9 - 2 * 9/6)2 * 9/6 LV = (9 - 3)2 * 9/6 LV = (6)2 * 9/6 LV = 36 * 9/6 LV = 54cm3 This proves that the formula is correct and completes the first part of my investigation. ...read more.

Conclusion

+- sqrt(352-35*20+202)] / 6 C = [55+- sqrt(1225-700+400)] / 6 C = (55 +- sqrt925) / 6 C = (55 +- 30.4138) / 6 C = 85.4138 / 6 or C = 24.5862 / 6 C = 14.2356 or C = 4.0977 Substituting these results back into my formula for the volume, I can find which gives the largest volume. (see next page) Using C = 14.2356 V = 14.2356 * (35 - 2*14.2356) * (20 - 2*14.3256) V = 14.2356 * 6.5288 * -8.4717 V = -787.3715cm3 (since the volume cannot be negative, the optimum cut-off must be 4.0977) Using C = 4.0977 V = 4.0977 * (35 - 2*4.0977) * (20 - 2*4.0977) V = 4.0977 * 26.8046 * 11.8046 V = 1296.5843cm3 So, the largest volume possible for a box of initial dimensions 35cm by 20cm, is 1296.5843cm3 (to 4 decimal places). This occurs when the cut-off is 4.0977cm (to 4 decimal places). To prove that this is correct, I will change my original spreadsheet slightly to account for the use of rectangles instead of squares. These formula are again printed at the back of the project. As you can see in the picture above, the maximum volume is attained when the cut-off is 4.0977, proving that my results were correct. This concludes my project. The formula to work out the optimum cut-off from a rectangle is C = [(L+W) +- sqrt(L2-LW+W2)] / 6 This also works for a square, but the simpler formula for a square is S/6 ...read more.

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