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I am doing an investigation into how get the maximum volume from a cuboid using a square with smaller squares cut out from each corner to then

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Introduction

C/W Max Box Coursework John Tovey 11M 31/10/05 Introduction I am doing an investigation into how get the maximum volume from a cuboid using a square with smaller squares cut out from each corner to then fold it up into a cuboid. Cut out the red squares and fold inwards on the blue lines to get a cuboid. To get the maximum volume from the cuboid you need to work out the sizes of the squares you want to cut out from each corner. The formula I used to work out the volume for each cuboid was height x width x length. Height is the width or length of the cut out square. Width is the length of the square minus 2H, (2H is the width of the cut out square times two). Width and length are the same. How I got started Firstly I chose a size of square to use, I chose 20cm. After this I went through all of the possible lengths for the cut out squares to be. ...read more.

Middle

This is because it would be impossible to receive a negative volume. The Y axis of my graph will be the volume of the cuboid in cm^3 and the X axis will be the size of the cut out square. The method I used was accurate but very time consuming. It would be easier and quicker to miss out all of the number crunching and create an equation using the Volume = Length x height x width. X = size of the square cut out. V=LHW Length = 20 - 2x H = X W = 20 - 2x V=(20-2x)x(20-2x) V=(20-2x)(20-2x)X V=(400-40x-40x+4x^2)x V=(400-80x+4x^2)x V=400x-80x^2+4x^3 Equation - V= 4x^3-80x+400x To be 100% sure that the equation was correct I took measurements of two differently sized squares, 25cm and 30cm. Once I had found the maximum volume of the squares when the original size was 25cm and 30cm I moved onto working with rectangles. The first rectangle I worked out had sides measuring 20cm and 15cm. ...read more.

Conclusion

General solution To find out the general solution I am going to use a square of 10cm each side. This is because these numbers seem the easiest to use. I went back to the equation and worked it through with algebra instead of numbers, starting off with the equation (10-2x)(10-2x)X. I worked this all through until it was at the stage of 12x^3-80x+100. The I worked it through but with letters for the width, length and height. The final equation I got after putting it all through was - 4(L+W)?????L+W)-48LW 24 The whole equation from start to finish is : L W H (10-2x) (10-2x) X LW -(L+W)H -(L+W)H 4H^2 H (100- 20x -20x -4x^2) x 4H^3 2(L+W)H^2 (LW)H 4x^3 -40x^2 +100x dy ax = (n x n-1) a=12 b=-80 c=100 12x^2 - 80x^1 +100 -b???b^2-4ac 12x^2-4(L+W)x+LWx 2a a= 12 b= -4(L+W) c=LW I put this: 12x^2-4(L+W)x+LWx into this -b???b^2-4ac 2a I got: 4(L+W)???-4(L+W) - (4x12xLW) 24 Simplified down to: 4(L+W)???-4(L+W) - 48LW 24 Final general solution - 4(L+W)???-4(L+W) - 48LW 24 ?? ?? ?? ?? John Tovey 11M Maths coursework ...read more.

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