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Maths Open Box Problem
How should I go about writing up the maths open box problem? I've carried out the investigation on loads of different bits of paper so now I've got to write it up. Has anybody got any suggestions for the best way to write it up?
Matt45
How should I go about writing up the maths open box problem? I've carried out the investigation on loads of different bits of paper so now I've got to write it up. Has anybody got any suggestions for the best way to write it up?
You know, half the people here are not even doing A levels. I, as an example, am doing the IB curriculum. I might be able to help you if you gave me the whole problem.
It's the GCSE 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.
The card is then folded along the dotted lines to make a box.
The aim of this investigation is to determine the size of the square cut which makes the volume of the box as large as possible for any given square or rectangular sheet of card.
Investigation Part 1 is for square pieces of card and investigation part 2 is for rectangular pieces of card.
You have to draw graphs showing the results and then form a general equation to work out the size that needs to be cut off the corner in order to form the largest volume open box when only given the length and width.
Hope that explains it more clearly.
An open box is to be made from a sheet of card. Identical squares are cut off the four corners of the card.
The card is then folded along the dotted lines to make a box.
The aim of this investigation is to determine the size of the square cut which makes the volume of the box as large as possible for any given square or rectangular sheet of card.
Investigation Part 1 is for square pieces of card and investigation part 2 is for rectangular pieces of card.
You have to draw graphs showing the results and then form a general equation to work out the size that needs to be cut off the corner in order to form the largest volume open box when only given the length and width.
Hope that explains it more clearly.
to give the greatest volume, would the side length of the square cut out be equal to 1/3 of the side length of the square piece piece of card?
to write up: what conclusions you can draw
how they relate to your hypothesis
where these conclusions come from (give proper figures and graph references)
say how you know your results/conclusions are accurate
to write up: what conclusions you can draw
how they relate to your hypothesis
where these conclusions come from (give proper figures and graph references)
say how you know your results/conclusions are accurate
Right...that confused me since I worked it out as the following:
Taking the piece of card to be a rectangle of width 'c' and length 'b'. (It helps to draw a diagram). Taking a square of length 'a' from each side means the remaining length is b-2a and the remaining width is c-2a.
The volume is now a(c-2a)(b-2a), and, when expanded, gives:
abc-2a²c - 2a²b + 4a³
Now, b and c are constants, only 'a' is variable. Using differentiation to find the maximum volume possible (this is an AS/A-level technique)...remembering b and c are constant:
dV/da = bc - 4ca - 4ba + 12a².
This is max (or min) when dV/da = 0.
bc - 4ca - 4ba + 12a² = 0.
Since 'b' and 'c' are constant, we have a quadratic in 'a' so we can solve:
12a² - (4b+4c)a + bc = 0
Taking the coefficients '12', '-(4b+4c)' and 'bc' to plug into the quadratic formula known for GCSE gives:
b+c plus or minus the square root of [(b-c)² + bc] all over 6. Clearly now, whatever value of 'b' and 'c' can take, the value of 'a' to maximise will be completely different.
Hope this helps....maybe not.
Kirk
Taking the piece of card to be a rectangle of width 'c' and length 'b'. (It helps to draw a diagram). Taking a square of length 'a' from each side means the remaining length is b-2a and the remaining width is c-2a.
The volume is now a(c-2a)(b-2a), and, when expanded, gives:
abc-2a²c - 2a²b + 4a³
Now, b and c are constants, only 'a' is variable. Using differentiation to find the maximum volume possible (this is an AS/A-level technique)...remembering b and c are constant:
dV/da = bc - 4ca - 4ba + 12a².
This is max (or min) when dV/da = 0.
bc - 4ca - 4ba + 12a² = 0.
Since 'b' and 'c' are constant, we have a quadratic in 'a' so we can solve:
12a² - (4b+4c)a + bc = 0
Taking the coefficients '12', '-(4b+4c)' and 'bc' to plug into the quadratic formula known for GCSE gives:
b+c plus or minus the square root of [(b-c)² + bc] all over 6. Clearly now, whatever value of 'b' and 'c' can take, the value of 'a' to maximise will be completely different.
Hope this helps....maybe not.
Kirk
Just a bit of a flaw: calculus doesn't exist within the boundary conditions of GCSE.
GeneralGrievous
Just a bit of a flaw: calculus doesn't exist within the boundary conditions of GCSE.
It says in the syllabus that students will be expected to look outside the curriculum in order to get full marks - or something along those lines.
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