The Open Box Problem
The Open Box Problem An open box is to be made from a sheet of card as shown below. The corner squares are to be cut-off. The card is then folded along the dotted lines to make the box. ---------------------------------------------------------------------------------------------------------------------------- Investigation 1 - Square shaped pieces of card Aim - To find the length of the cut-out corners that gives the maximum volume for the open box formed for any sized piece of square card. The length of the square cut will be to 3 significant figures of accuracy. Method - I will investigate what length of cut-out corners will give the largest volume ofr square pieces of card with dimensions 12 x 12, 18 x 18, 24 x 24 and 30 x 30. NOTE - when 'small side' is mentioned, it refers to the size of the cut-out corners. When 'Length', 'Width' and 'Height' are mentioned, they refer to the dimensions of the open box. When 'Volume' is mentioned, it refers to the volume of the open box. Rows in Italics are those which contain the correct cut-out corner size for the maximum volume of the open box. Square piece of card with dimensions 12 x 12 Small Side Volume Length Width Height 00 0 0 2 28 8 8 2 3 08 6 6 3 2.1 27.764 7.8 7.8 2.1 2.2 27.072 7.6 7.6 2.2 2.3 25.948 7.4 7.4 2.3 2.4 24.416 7.2 7.2 2.4 2.5 22.5 7 7 2.5 .9
THE OPEN BOX PROBLEM
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. 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, but first I am going to experiment with a square to make it easier for me to investigate rectangles. I am going to begin by investigating a square with a side length of 10 cm. Using this side length, the maximum whole number I can cut off each corner is 4.9cm, as otherwise I would not have any box left. I am going to begin by looking into going up in 0.1cm from 0cm being the cut out of the box corners. The formula that needs to be used to get the volume of a box is: Volume = Length * Width * Height If I am to use a square of side length 10cm, 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 10cm. I am going to call the cut out "x." Therefore the equation can be changed to: Volume = 10 - (2x) * 10 - (2x) * x If I were using a cut out of length 1cm, the equation for this would be as follows: Volume = 10 - (2 * 1) * 10 -
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. The card is then folded along the dotted lines to make the box. My 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 square sheet of card. I am going to begin by investigating a square with a side length of 24 cm. Using this side length, the maximum whole number I can cut off each corner is 11cm, as otherwise I would not be able to make an open cube. I am going to begin by looking into whole numbers being cut out of the box corners. The formula needed to get the volume of a box is: Volume = Length x Width x Height If I am to use a square of side length 24cm, then I can calculate the side lengths minus the cut out squares using the following equation. Volume = Length - (2 x Cut Out) x Width - (2 x Cut Out) x Height (Cut Out) Using a square, both the length & the width are equal. I am using a length and width of 24cm. I am going to call the cut out "x." Therefore the equation can be changed to: When x = 1, Volume = (Width - X) x (Length - X) x (Width -> also known as X) My formula allows me to construct a spreadsheet, which would allow me to quickly and accurately calculate the volume of the box. Below are the results I got through this spreadsheet. Here I have tabulated my
The Open Box Problem
Mathematics GCSE The Open Box Problem Tiers F, I and H Introduction 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 in figure 1. Figure 1: The card is then folded along the dotted lines to make the box. The main aim of this activity is to determine the size of the square cut out which makes the volume of the box as large as possible for any given rectangular sheet of card. . For any sized square sheet of card, investigate the size of the cut out square which makes an open box of the largest volume. 2. For any sized rectangular sheet of card, investigate the size of the cut out square which makes an open box of the largest volume. Question 1 I began work on question 1, which was to investigate the correct cut out size to satisfy the largest possible volume of the open box on any sized square piece of card. I started by using a square piece of card measuring 10 x 10. To work out the volume for the open box with the variable being 'x' I made a formula to help: V=x(l-2x)(l-2x) Here are my results: size of cut out 'x' (cm) volume (cm³) 64 2 72 3 48 4 6 The highest volume is 72cm³ with a cut out size of 2cm. I found that the highest volume must have a cut out size of between 1 and 2cm so I tried the formula for cut out sizes between 1 and 2cm: size of cut out 'x'
The open box problem
The open box problem Introduction The aim of my algebraic investigation into the open box problem is to determine the size of square cut which makes the volume of the box as large as possible for any given rectangular sheet of card. The problem itself is simple, an open box is made from a sheet of card, identical squares are then cut off each of the four corners, the sheet is then folded to make box. It is my aim to find out the maximum square cut which gives me the maximum volume box. Strategy . Try to find the size of cut-out that will give me the maximum volume of a piece of card 6cm x 6cm, progressing onto algebra. 2. Look for limitations in the results. 3. Devise a plan for overcoming these limitations. 4. Extend the work to find a general formula that will help me to work with all squares and rectangles. Technique In this situation there is no need for us to go even close to a pair of scissors or piece of card, for this investigation I am to use Excel. Excel is a computer programme in which I can input information; it will then calculate this information and give me results for what different size cut outs for different sized card. The set-up of this is quite simple; Size of card is out into cell c3 in this case Volume is calculated With a real box by Size of cut-out in Width x Depth x Height Cells a6, a7 etc... The same happens
The Open Box Problem.
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. 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: 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) * 18 * x If I were using a cut out of length 1cm, the equation for this would be as follows: Volume = 20 - (2 * 1) * 20 - *(2 * 1) * 1 So we can work out through this method that the volume of a box with corners of 1cm² cut out would be: (20 - 2)
The Open Box Problem
Part 1: Square Ryan Simmons For any sized square sheet of card investigate the size of the cut out square that makes an open box of the largest volume. The first square to be tested has measurements of 30 x 30 cm. For cut sizes I will start from the smallest (whole) number possible (1cm) I will then work my way up to find which size cut gives the box the largest volume. So far as the size of the cuts increase the volume increases. I predict that a cut size of 3 x 3 cm will give an even bigger volume for the box. A cut of 3 x 3 cm gives a volume of 1728 cm3, hence my prediction was right. 3cm x 24cm x 24cm = 1728 cm³ The prediction was right and so far there is no obvious pattern between the cut sizes and the volume of the box. To save me from drawing a diagram for every cut size I will record my results in a table Cut size (cm) W L V (cm3) 4 22 22 936 5 20 20 2000 6 8 8 944 7 6 6 792 8 4 4 568 This is a spreadsheet, where the value of the volume is a product of the cut size, the width and the length. The formula used in the spreadsheet is: V = Cut size x W x L (on the spreadsheet the formula I used was: F6 = C6 * D6 * E6. The volume (V) being F6, the length being E6, the width being D6 and the height being C6). The volume of the box increased with increasing cut size until the
Area & Volume Exploration – Component proportional changes
Area & Volume Exploration - Component proportional changes Question 2: Suppose that you are required to design an open steel tray from a sheet of steel that measures 120cm by 80cm by cutting squares from each corner, and folding to form a tray. What size should the squares be cut from each corner of the sheet so that the maximum volume is obtained for the tray. Height = x cm Width = 80cm - 2x cm Length = 120cm - 2 x cm The task is to find the optimal size of the squares that need to be cut out of the corners in order to find the maximum obtainable Volume inside the tray once it has been folded. To work this out there are a number of different methods for doing this. One method is Trial and Improvement. Trial and Improvement Height (Value of x cm) 2 x cm Length (cm) (120cm - 2 x cm) Width (cm) (80cm - 2x cm) Volume (cm3) (120x - 2 x 2)?(80 - 2x) +/- 5 0 10 70 38'500 7.5 5 05 65 51'187.5 + 0 20 00 60 60'000 + 2.5 25 95 55 65'312.5 + = 5 30 90 50 67'500 + 7.5 35 85 45 66'937.5 - There is no point in increasing x any more than 17.5 as the volume is now clearly decreasing. This means the Maximum volume lies below 17.5. As the volume rises between 12.5 and 15 we know that the maximum volume must lie after 12.5. We can now continue with the trial and improvement table to fin the maximum volume, working only between 12.5 and
Open Box Problem.
Open Box Problem Aim During this project I will be 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. What is an Open Box 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. The card is then folded along the dotted lines to make the box. [image001.gif] [image002.gif] Names of things needed for investigation I will write-up my investigation in Microsoft Word and all formulae shall be calculated on Microsoft Excel and all table and graph will be produce in spreadsheets again in Microsoft Excel. Structure of investigation 1. Evidence: · Table · Graphs · Formulas 2. Evaluation 1. Evidence To obtain evidence I will be used a series of methods: · Table · Graphs · Formulae Part 1, Square I am going investigate 3 different sizes for the square open box. Once I have obtained all information on the 3 sizes I will look for patterns and try to formulate a rule to work out the largest volume for an open box square. The sizes that I will be using are: 1. 20 x 20 2. 40 x 40 3. 25 x 25 Because it is a square the length = width so, we can write this as L=1W therefore there is a ratio 1:1. I am going to begin by investigating a square with a side length of 20cm. Using
Open Box Problem
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') of 2cm by 2cm. Once the cut-offs are taken away, the net will look like this. From this we can see that when the dotted lines are folded along, there is a height of 2cm, a length of 16cm and a width of 16cm. Since Volume = Length *