# Maths Courseowrk - Open Box

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Introduction

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. ...read more.

Middle

For my graphs below the volume is shown along the y-axis and the cut out size along the x-axis hence the equation of the line is y=x(l-2x)2. The value of a depended on the square I was using so for the first table the value for l was 5. So to get a graph I only needed to enter the l value. I had to use y= and not v= as stated in the original formula as in autograph you could only use x= or y= and x was already in my formula so I had to use y= On the next page is the table of values for when the l=5 X L v=x(5-2x) 2 0.5 5 8 1 5 9 1.5 5 6 2 5 2 0.75 5 9.1875 0.8 5 9.248 0.85 5 9.2565 0.9 5 9.216 0.95 5 9.1295 0.831 5 9.259204764 0.832 5 9.259241472 0.833 5 9.259258148 0.834 5 9.259254816 0.835 5 9.2592315 0.836 5 9.259188224 0.837 5 9.259125012 0.838 5 9.259041888 0.839 5 9.258938876 0.8331 5 9.259258715 0.8332 5 9.259259081 0.8333 5 9.259259248 0.8334 5 9.259259215 0.8335 5 9.259258982 0.8336 5 9.259258548 0.8337 5 9.259257915 0.8338 5 9.259257082 0.8339 5 9.259256049 0.83331 5 9.259259254 0.83332 5 9.259259257 0.83333 5 9.259259259 0.83334 5 9.259259259 0.83335 5 9.259259256 0.83336 5 9.259259252 0.83337 5 9.259259246 0.83338 5 9.259259237 0.83339 5 9.259259227 This table of value is for a square with sides of 5cm in length. ...read more.

Conclusion

So when X=0.83333 the length of the box is 5. As you can see there is an obvious relationship between the L and X values as when L is divided by X the answer comes to roughly 6. I have discovered that the relationship for the highest value of an open box is X=L/6. I gained two values where L/2 and L/6.I released it could not be L/2 when comparing this with my graphs turning points. The relationship is cemented by the writing on the table on the previous page as well. Here is my working to show how I mathematical discovered this relationship. v=y(l-2y)2 v=y(l-2y)(l-2y) v=(yl-2y2)(l-2y) v=yl2 - 2ly2 - 2ly2 + 4y3 v=4x3 -4y2l + yl2 I then differentiated the x values and this is my results: dy/dx = 12y2 - 8yl + l2 I then made a quadratic equation out of this: (6y - l)(2y-l) = 0 This gave me the following answers: 6y - l = 0 or 2y - l = 0 6y = l 2y = l y = l/6 y = l/2 y=l/6 I will prove this now by putting the formula to use. With l=6 which is my first tested table and in which I went to 5 decimal places. Y=5/6 Y=0.833333333 recurring This proves that the relationship for the formula is correct. Graphs ...read more.

This student written piece of work is one of many that can be found in our GCSE Open Box Problem section.

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