The Open Box Problem
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. Using a Graph A graph can be plotted showing the cut out size and volume. With the side length of the square n, the formula Volume (V) = x *
Maths Courseowrk - Open Box
Mathematics GCSE coursework The open box problem An open box is made from a sheet of card Identical squares cut off the four corners of the card The card is then folded along lines to make an open box. The main aim of the activity 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. Part 1 I have drawn tables for my squares. I have tested four squares; 5cm square, a 10cm square, a 20cm square, a 40cm square. I have decided against trial and improvement. As this method can is time consuming so I have used gone up in 0.5cm each time for the size of x (the cut out size) then calculated the size of v (volume). In one table of values I have gone to 5 decimal places to prove my relationship is correct. The volume also depends on the size of a (the length of the square before the cut out.) which is represented by the letter l. After I have drawn the tables I will analyze the graphs. I will put the graphs under the table so on the next few pages there will be a table of values showing the different volumes depending on the cut out of the square then a graph giving us a clear picture of the maximum value. You will see this process repeated four times for each of my tested squares. After doing these stages I will try and work out a relationship between the cut out of the square and the volume. To
Investigate the volume of an open box constructed by one piece of rectangular card that has all four corners having had squares cut out of them.
Introduction The aim of this coursework is to investigate the volume of an open box constructed by one piece of rectangular card that has all four corners having had squares cut out of them. Firstly I will be studying the volume whilst changing the side of one length of the cut out square and the size of the original rectangle card. After I have investigated this relationship I will try to find out the formula for finding the cut size to get the largest volume for any specified original card size. Square card size I am going to begin by investigating a square card because this will give me a basic formula which I can elaborate on. I will start with a round number of 20cm for the length. This means that the maximum cut out square length I can cut out will be 9cm else I will have no box left. The formula for the volume of any box is as follows: Below, there is a diagram to explain where all these figures come from. I have also included in this diagram the labels c, x and y, these show the cut out size and the original length and width of the card, I will now need to show the values of the width, length and height in terms of c, x and y. Therefore I can replace these sub formulae into the first formula. However for the values of a square, y=x therefore: c V 324 2 512 3 588 4 576 5 500 6 384 7 252 8 28 9 36 As I have already said, the value of x is
Trays.The shopkeepers statement was that, When the area of the base is the same as the area of the four sides, the volume of the tray will be maximum.
Maths Coursework; Trays In this coursework candidates were given a task entitled "Trays." The task consisted of a shopkeeper's statement upon the volume of a tray which was to be made from an 18x18 piece of card. The shopkeeper's statement was that, "When the area of the base is the same as the area of the four sides, the volume of the tray will be maximum." By saying this, the shopkeeper basically meant that when the area of the base of the tray is equal to the total area of the sides the volume of the tray will be at its highest. We were told to investigate this claim. Plan. . I will investigate the different sizes of tray possible from an 18x18 piece of card. 2. After gaining my results I will then put them in a table. 3. I will try to spot any patterns from my table. 4. I will express any patterns or other formulae in mathematical notation. To investigate the different volumes given by different trays, I first decide to cut the corners in ascending order from 1-8. (The longest possible corner could only be 8 as after this there would be no base.) After this I worked out the formula needed to work out the volume for the various trays. For the corner size 1x1 the way I worked out the volume was 16x16x1 which equalled 256cm. Thus the formula to work out the volume for a tray made by an 18x18cm card is (n - 2X) x X. In this formula the letter "X" represents the
This coursework is all based on the significance of Racism.
This coursework is all based on the significance of
Boxed In.
Boxed In Coursework Paul Mitchell Mathematics Coursework Mathematics Coursework "Boxed In" For my coursework I have been given a sheet of metal measuring 20cm x 20cm. I have been asked to produce a box without a lid. By cutting one centimetre squares from the corners of the sheet, I will be able to fold the corners up and create the box. I am going to investigate further by cutting 2cm - 9cm squares off the corner of each box, and look at the results to find the biggest volume. * 2cm Squares: Volume = 162 x 2 = 512cm3 * 3cm Squares: Volume = 142 x 3 = 588cm3 * 4cm Squares: Volume = 122 x 4 = 576cm3 * 5cm Squares: Volume = 102 x 5 = 500cm3 * 6cm Squares: Volume = 82 x 6 = 384cm3 * 7cm Squares: * 8cm Squares: Volume = 42 x 8 = 126cm3 * 9cm Squares: Volume = 22 x 9 = 36cm3 From my results I can see that the largest volume is 588cm3 from the cut out of 3cm. I will now plot my results onto a graph. From my graph, I can see that the maximum volume is at its highest between the cut out of 3 and 4cm. I will now use trial and improvement method to work out the maximum volume. I will work between the cut outs 3 and 4 centimetres. To get the Length/Breath of the box; to work out the volume in my trial and improvement table; I will double the cut out number and subtract it from 20. E.g. Cut out = 3.1 20 - (3.1 + 3.1) = Length/Breath I will then take the
Applications of Differentiation
Using differentiation to solve practical problems 988 GCSE Maths Coursework Task A rectangular sheet of card is 20cm long and 12cm wide. Equal squares are cut from each corner. The flaps are then folded to make an open box in the form of a cuboid, as shown below. 2cm 20cm a) The box is to be filled with centimetre cubes. What size square should be cut from each corner so that the box will hold as many cubes as possible? b) The box is to be filled with sand. What size of square should now be cut from each corner so that the box will hold as much sand as possible? a) This can be solved using trial and error by forming a table and trying different possibilities of the squares that will be cut. Size of square Length of cuboid Width of cuboid Height of cuboid No. of cubes x 1 8 0 80 2 x 2 6 8 2 256 3 x 3 4 6 3 252 4 x 4 2 4 4 92 Looking at the 1 x 1 square; if a 1 x 1 square is removed for the corners, the length becomes 20 -1 -1 =18cm. Similarly, for the width we have 12 -1 -1 =10cm. The height is the size of the square. The no. of cubes we can fit is the volume of the resulting cuboid i.e. 18(10)(1) = 180cm3 We can see that the maximum volume occurs when a 2 x 2 square is cut from each corner and hence the answer is "a 2x2 square." METHOD 1: Without differentiation b) Similar to a) we can form a table and attempt some trial
Investigation: The open box problem.
Investigation: The open box problem Problem: An open box is to be made from a piece of card. Identical squares are to be cut off the four corners of the card to make the box. (As shown below) Cut off Fold lines Aim: Determine the size or the square cut which makes the volume of the box as large as possible for any given rectangular sheet of card. Plan: To start of with I will be using the trial and improvement method to experiment with different sizes of a square boxes. By doing this I will find out the size of cut off that will leave me with the largest volume inside the box. To find out the volume I will need to know the size of the cut off side and the base length. x = length off the square cut off L = original length off the square card The formula that I will use to work out the volume is: Volume = (L-2X) ²X. The different sizes of cards that I will be using are 10cm, 11cm, 12cm, 13cm and 14cm. I will determine the size of x that will give the highest volume to 2d.p. After finding the highest value of X I will prove that my answer if right by using differentiation. Finally I will try and find a rule that allows me to find the highest value of X for a piece of square card and check that it works with any size of square card. Trail and improvement Size of card - 10cm by 10cm X must be 0<X<5: This is because if X is 0 there would not be a side to fold and if
Maximum box investigation
maximum box project Ellie Roy Mrs. Satguru MP2b Maths Introduction: This project is about finding the maximum volume of different boxes by investigating what the length of the corner square is. For part 1 we had to find the maximum volume of a box made from a 20 x 20 cm piece of paper. You had to make a box (without a lid) by cutting squares from the corners. The diagram to the left shows where the corner squares are (they are shaded in grey.) It is the size of these corner squares which impact the volume of the finished box. To solve part 1's problem I tried out different corner square lengths and recorder my results in a table. For part 2 we had to do the same thing, however we had to find the volume of a box made from a 24 x 24 cm, 15 x 15 cm, 10 x 10 cm and 36 x 36 cm piece of paper instead of a 20 x 20 cm paper. We also had to try out different corner square lengths and draw tables to show our results for this as well. Part 3 was a bit harder. We had to try and find a connection between the size of the corners cut out and the size of the original piece of paper. But I did manage to find something. Aim: my aim is to find out the largest possible volume of a box made from a 20x20 cm piece of paper. I also want to find the largest possible volume for a box that is made from a 15x15 cm piece of paper, a 24x24 cm piece of paper, a 10x10 piece of paper and a 36x36
Open box Problem.
Open box Problem Introduction Now I am going to investigate the size of the cut out square which makes an open box of the largest volume. Let the length of the square be 24cm, the card is shown as below: Method Now, I am going to set the length of the square to be 24cm and the other one will be 12cm. Then I will find out the relation between the two experiment and find out a formula to calculate the max. volume for any size of square. Then I will explain why the formula works for all squares. I have to find out the value of X to give the maximum of volume. First, I have to find out the formula to calculate the volume of the open box. Base on the formula, the volume= height x length x width. Let the height be X. Then the length is 24cm - X- X . Therefore we get X(24cm-2X) With a cut out of 1cm,X=1cm, 11cm is the maximum for there to be a box left. Here are the results: X(cm) X(24cm-2X)(cm) 484 2 800 3 972 4 024 5 980 6 864 7 700 8 512 9 324 0 60 1 44 We found out that when X=4, we can get the maximum volume in integer. Now I am going to find out the volume when X is between 3.5 and 4.5. Here are the results: X(cm) X(24cm-2X)(cm) 3.5 011.5 3.6 016.064 3.7 019.572 3.8 022.048 3.9 023.516 4.0 024 4.1 023.524 4.2 022.112 4.3 019.788.4.4 4.4 016.516 4.5 012.5 We found out that 4cm is still the biggest value whereas we