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GCSE: Open Box Problem

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  1. The Open Box Problem

    * 20 -- (2 * 1) * 1 So we can work out through this method that the volume of a box with corners of 1cm� cut out would be: (20 -- 2) * (20 -- 2) * 1 18 * 18 * 1 = 324cm� I used these formulae to construct a table, which would allow me to quickly and accurately calculate the volume of the box. Cut out (cm�) formula volume (cm�) 1 Volume = 20 -- (2*1)

    • Word count: 3105
  2. 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
  3. The Open Box Problem

    * Width - (2 * Cut Size) * Cut Size I am now going to substitute the cut out size with the sign "?" Therefore the equation can be changed to: Volume = 20 - (2?) * 18* ? 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 1cm� cut out would be: (20 - 2)

    • Word count: 5227
  4. 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
  5. 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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