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• Level: GCSE
• Subject: Maths
• Word count: 3105

# The Open Box Problem

Extracts from this document...

Introduction

The Open Box Problem An open box is to be made from a sheet of card. Identical squares are cut off the four corners of the card, as shown below.(fig 1) FIG 1 The card is then folded along the dotted lines to make the box. The main aim is to determine the size of the square cut which makes the volume of the box as large as possible for any given rectangular sheet of card. I am going to begin by investigating a square with a side length of 20 cm. Using this side length, the maximum whole number I can cut off each corner is 9cm, as otherwise I would not have any box left. I am going to begin by looking into whole numbers being cut out of the box corners. The formula that needs to be used to get the volume of a box is: ( * = multiply) Volume = Length * Width * Height If I am to use a square of side length 20cm, then I can calculate the side lengths minus the cut out squares using the following equation. Volume = Length -- (2 * Cut Out) * Width -- (2 * Cut Out) * Height Using a square, both the length & the width are equal. I am using a length/width of 20cm. I am going to call the cut out ""x."" Therefore the equation can be changed to: Volume = 20 -- (2x) * 20 -- (2x) * Height If I were using a cut out of length 1cm, the equation for this would be as follows: Volume = 20 -- (2 * 1) ...read more.

Middle

I shall begin with a width of 20cm, and a length of 40cm, this is a ratio of 1:2, the length being twice as long as the width. FIG 3 The square cut outs will rise in measurements of 1cm at a time, beginning with a cut out of 1 cm, and 9cm being the maximum for there to be a box left. I will then investigate various other ratios. I have again constructed a table to assist with my workings out. Firstly I am looking at whole numbers, a beginning with a width of 20cm and a length of 40cm, and cut outs from 1 to 9cm. Here are the results I got. cut out (cm�) Formula Volume (cm�) 1 Volume = 40 -- (2*1) * 20 -- (2*1) * 1 684 2 Volume = 40 -- (2*2) * 20 -- (2*2) * 2 1152 3 Volume = 40 -- (2*3) * 20 -- (2*3) * 3 1428 4 Volume = 40 -- (2*4) * 20 -- (2*4) * 4 1536 5 Volume = 40 -- (2*5) * 20 -- (2*5) * 5 1500 6 Volume = 40 -- (2*6) * 20 -- (2*6) * 6 1344 7 Volume = 40 -- (2*7) * 20 -- (2*7) * 7 1092 8 Volume = 40 -- (2*8) * 20 -- (2*8) * 8 768 9 Volume = 40 -- (2*9) * 20 -- (2*9) * 9 396 You should be able to see from this table, that the largest volume achieved is 1536cm�, and this is obtained when the amount cut off from the corners measures 4cm�. I can also see that to make my results more accurate, to more decimal places, I next need to look between 4 and 5cm� cut offs. ...read more.

Conclusion

* 10 -- (2*2.3) * 2.3 12362.868 2.4 Volume = 1000 -- (2*2.4) * 10 -- (2*2.4) * 2.4 12420.096 2.5 Volume = 1000 -- (2*2.5) * 10 -- (2*2.5) * 2.5 12437.5 2.6 Volume = 1000 -- (2*2.6) * 10 -- (2*2.6) * 2.6 12415.104 I worked out the proportion that needs to be cut off the box to give maximum volume; this was 0.2444, which is very close to 1/4. � stands for delta. Firstly we should consider a graph of y = x� as shown below. The line through X and Y has almost the correct gradient. It''s gradient is Increase in y-coordinate from X to Y Increase in x-coordinate from X to Y You have to find an expression for , which represents the gradient of the graph at the point X. So y = x� y + �y = (x + �x)(x +�x) Multiply out brackets. y+ �y = x� + x �x + x �x + (�x) � Add like terms together. y + �y = x� + 2x�x + (�x)� Now here the x� at the end is y in terms of x. �y = x� + 2x�x + (�x) � - x� Then you divide by ��x, which gives you �y 2x�x -- (�x) � �x = �x �y �x = 2x + �x Now because delta (�) is so tiny that it is insignificant, we forget all about it, which leaves us with �y/�x = 2x = 0 I can use calculators to help me complete my calculations to solve the problem, for proof through exhaustion. I have drawn a graph showing the proportions that I have worked out. I can see that they tend towards 1/4. This is the amount that should be cut off each corner to give the maximum volume possible. FIG 6 ?? ?? ?? ?? Max Box Sam Norton ...read more.

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# Related GCSE Open Box Problem essays

1. ## Investigation: The open box problem.

�x2.5 V = 90 The cut off that will leave the largest volume so far is 1.8. I will now look for the cut off to 2d.p. between 1.75 - 1.85 X = 1.75 X = 1.76 X = 1.77 V = (11-(2x1.75))

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I came to this conclusion as if the volume was going to increase then the corner square length would have to be either 3.04, 3.03, 3.02 or 3.01cm and the volume would have to increase by least 2 cm�,

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2. ## Maths Courseowrk - Open Box

The point we will look for is the first maximum which gives us the highest volume. The possible values for the cut out are between the points (0,0) and the point where the line passes downwards through the x axis.

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2. ## The Open Box Problem

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1. ## The Open Box Problem

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