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GCSE: Open Box Problem

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  • Marked by Teachers essays 2
  1. GCSE Maths questions

    • Develop your confidence and skills in GCSE Maths using our free interactive questions with teacher feedback to guide you at every stage.
    • Level: GCSE
    • Questions: 75
  2. Marked by a teacher

    Open Box Investigation

    5 star(s)

    The height is the same as the side length of square cut. Below are my results: 10x10 Square Side Length of Square (cm) 1 2 3 4 Length of Box 8 6 4 2 Width of Box 8 6 4 2 Height 1 2 3 4 Volume 64 72 48 16 To find the more precise length of square which I should cut, I have used the upper and lower bounds of the best length of square in the table above (in this case, 2cm). Side Length of Square 1.5 1.6 1.7 1.8 1.9 2 2.1 Length of Box 7 6.8 6.6 6.4 6.2 6 5.8 Width of Box 7 6.8 6.6

    • Word count: 1276
  3. Maths Courseowrk - Open Box

    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 do this I will generalise the formula then use a method called differentiation. After I have done this I will input the letters and numbers into the quadratic formula and work out the relationship. Throughout this piece of coursework I will be using this formula v=x(l-2x)

    • Word count: 1918
  4. 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.

    (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 size of the corner. I tried my formula for the corner length of 2cm, (18- 2 x 2)

    • Word count: 1391
  5. Kunstnik ja dekoraator.

    Detailid on need, mis ekraanimaailma elusaks muudavad ja vaataja �mbritsevat unustama panevad. Siin oli aga tegu �he suure teleteatriga. Arvan, et mina ei olnud ainuke, kellel n�iteks stseenis, kus Ahast taga aetakse, tekkis kriipiv ja kahtlustav tunne , mis �tleb:"Vaata, vot selle aia taga seisab uus mersu ja tolle kuuri taga R-Kiosk...". No ei suudetud seda ajastu h�ngu filmi sisse tuua ja k�ik. V�ib-olla on viga mise-en-scene`is. Tihti tuli ette olukordi, kus esiplaanil olevat tegevust saatis tagaplaanil olev kohatu t�hjus.

    • Word count: 1069
  6. In this coursework my task description is to see how fast I can catch a reaction time ruler. I have to do this with both my hands and see the difference between the hands.

    19 17 18 15.5 15 16.5 15 20 18 12 15 13 18 17 18 14 20 21 19 17 16 17 12 16 12 10 21.5 18.5 19 13 17.5 11 17.5 17.5 18 16 20 15 21 13 15 16 17 4 16 13 14 13 21 14 17.5 13 16 19 14 15 GIRL 1 GIRL 2 GIRL 3 RIGHT HAND LEFT HAND RIGHT HAND LEFT HAND RIGHT HAND LEFT HAND 23 20 17.5 12 15.5 11 12 21 15 13 15 23 19 23 23 14 11 19 20 17 14.5 15.5 22 9 5 8

    • Word count: 1801
  7. Open Box Problem

    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 * Width * Height, the volume of this open box is 16*16*2 = 512cm3 We can also see relationships between the cut-off and the dimensions of the net from this example. 1.

    • Word count: 1817
  8. The Open Box Problem

    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 results: I have highlighted the value of X, which allows the maximum volume in each case 12x12 square X=1 10x10x1 100 cm X=2 8x8x2 128 cm X=3 6x6x3 108 cm X=4 5x5x4 100 cm X=5 4x4x5 80

    • Word count: 1555
  9. The Open Box Problem

    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 cut became 5cm, it then decreased. There is no point investigating cuts any bigger than 5 cm because after 5 cm the volume starts to decrease.

    • Word count: 1439
  10. Area & Volume Exploration – Component proportional changes

    = 15 30 90 50 67'500 + 17.5 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 17.5. Height (Value of x cm)

    • Word count: 1145
  11. The Open Box Problem

    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 * (n - 2x)

    • Word count: 1324

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