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  • Level: GCSE
  • Subject: Maths
  • Word count: 2831

The Open Box Problem

Extracts from this document...

Introduction

The 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, which is then folded along the dotted lines to make the box. The main aim is to find out the size of the square cut which makes the volume of the box as large as possible for any rectangular sheet of card. I am going to begin by looking at which size of cut out square will give the largest volume for a square sheet of card. I will now look at a square with a side of 6 cm. The letter 'x' represents the length of the square cut out to form the box. The letter 'l' is used for the length of each side - 2x, this is going to be the length of the open box created. The letter 'h' is going to be the height of the open box created, and 'v' is the volume of the open box created. I will use the formula Volume=Length�Width�Height or v=l�h to find out the volume of the box. I will investigate the volume of boxes, where x is between 0.5 and 2.5, at intervals of 0.5. I cannot go higher than 2.5 to 3, because then the card would be cut in half, and would not form an open box. I have used the information above to construct the spreadsheet below. The number in bold is the maximum volume. I intend to use spreadsheets to find the volumes given by different values of x, so I can find which value of x gives the maximum volume. ...read more.

Middle

+ 4x� - x�( 2l + 2w ) This is now ready for differentiation When dv/dx = 0 dv/dx = lw + 12x� - 2(2l + 2w) 0 = 12x� - (4l + 4w) + lw This is now a quadratic equation, we can use this formula: X = -b ? ??b� - 4ac 2a when a = Co-efficient of x� b = Co-efficient of x c = constant in this case a = 12, b = - (4l + 4w) and c = lw I will now put our quadratic into this formula: X = (4l - 4w) ? ?-(4l + 4w)� - 4 � 12 �lw 2�12 I will now simplify: X = 4(l + w) ? ?-(4l +4w)(4l +4w) - 48lw 24 Now I will multiply the two adjacent brackets together: X = 4(l + w) ? ?16l� + 16w� +32lw - 48lw 24 Now simplify: X = 4(l + w) ? ?16l� + 16w� - 16lw 24 Now factorise: X = 4(l + w) ? ?16(w� + l� - lw) 24 I can move the 16 out of the square root to give 4: X = 4(l + w) ? 4 ?(w� + l� - lw) 24 I can now simplify by dividing by common factor 4 X = l+w ? ?w� + l� - lw 6 I cannot simplify this anymore, using this expression I should be able to find the value of x for any rectangle, by substituting l for the length of that rectangle, and w for the width. I will now make some predictions with this formula, for the maximum volumes of different size rectangles, and I will then prove the predictions by using spreadsheets and the formula for quadratic expressions. ...read more.

Conclusion

These results confirm this formula to be correct: X = l+w ? ?w� + l� - lw 6 This does in fact give the value for the length of x, or square cut out in each corner which gives the maximum volume of the open box. I have confirmed the answers given by this formula for three different rectangles, through the use of spreadsheets and quadratic formulas. However this formula does not only apply tot the rectangle, it can also be used for the square, which is the first half of my investigation. In relation to the square it can be simplified, as the length and width or l and w are the same. So therefore: X = 2l ? ?2l� - l� 6 I will now use this for the square of length 10cm, I have already found out in the previous half of my investigation, that the length of square cut out for this sized square is 1.667cm. X = 20 ? ?200 - 100 6 x = 20? ?100 6 x = 20 +10 = 5.000 6 x = 20 - 10 = 1.667 6 This is the same answer as I have got before. Therefore the length of the square cut out in each corner of any rectangle should be l+w ? ?w� + l� - lw 6 This is where l is the length of the rectangle, and w is the width of the rectangle, this will always give the maximum volume for the open box created. The length of the square cut out in the corner of any square should be 2l ? ?2l� - l� 6 This is where l is the length of the square, this will always give the maximum volume for the open box created. This concludes my investigation. ...read more.

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