Part of The Student Room Group

  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  • Level: GCSE
  • Subject: Maths
  • Word count: 2232
    • View my saved documents

    The Open Box Problem

    Extracts from this document...

    Introduction

    Mathematics GCSE The Open Box Problem Tiers F, I and H Introduction An open box is to be made from a sheet of card. Identical squares are cut off the four corners of the card as shown in figure 1. Figure 1: The card is then folded along the dotted lines to make the box. The main aim of this activity is to determine the size of the square cut out which makes the volume of the box as large as possible for any given rectangular sheet of card. 1. For any sized square sheet of card, investigate the size of the cut out square which makes an open box of the largest volume. 2. For any sized rectangular sheet of card, investigate the size of the cut out square which makes an open box of the largest volume. Question 1 I began work on question 1, which was to investigate the correct cut out size to satisfy the largest possible volume of the open box on any sized square piece of card. I started by using a square piece of card measuring 10 x 10. To work out the volume for the open box with the variable being 'x' I made a formula to help: V=x(l-2x)(l-2x) Here are my results: size of cut out 'x' (cm) volume (cm�) 1 64 2 72 3 48 4 16 The highest volume is 72cm� with a cut out size of 2cm. ...read more.

    Middle

    To work out the maximum volume for a square piece of card substituting 'x' with 1/6L: V=(1/6L)L�-4(1/6L)�L+4(1/6L)� V=1/6L�-4/36L�+4/216L� V=1/6L�-1/9L�+1/54L� V=9/54L�-6/54L�+1/54L� V=4/54L� V=2/27L� Maximum volume=2/27L� This formula can be used to work out the maximum volume of any open box from a square shaped piece of card without needing to work out anything to do with the cut out size. Question 2 I them moved on to question 2, which was to investigate the correct cut out size to satisfy the largest possible volume of the open box on any sized rectangular piece of card. I had already found the cut out size for a rectangle of ratio 1:1 in the first question, as a square is also a type of rectangle. I first tried the formula for ratio 1:2 rectangles: To work out the volume of the open box from a rectangular piece of card the formula is: V=(KW-2X)(W-2X)xX I investigated the ideal cut out size for a rectangle 16 x 8: size of cut out 'x' (cm) volume (cm�) 1 84 1.5 97.5 2 96 2.5 82.5 3 60 3.5 31.5 4 0 I then homed in between 1.5 and 2cm for 'X': size of cut out 'x' (cm) volume (cm�) 1.5 97.5 1.6 98.304 1.7 98.532 1.8 98.208 1.9 97.356 2 96 I then homed in once more with values of 'X' to two decimal places: size of cut out 'x' (cm) ...read more.

    Conclusion

    volume (cm�) 1.5 157.5 1.6 159.744 1.7 161.092 1.8 161.568 1.9 161.196 2 160 I then homed in again between 1.8 and 1.9cm for 'x': size of cut out 'x' (cm) volume (cm�) 1.8 161.568 1.81 161.568564 1.82 161.560672 1.83 161.544348 1.84 161.519616 1.85 161.4865 1.86 161.445024 1.87 161.395212 1.88 161.337088 1.89 161.270676 1.9 161.196 The highest volume for this rectangular sheet of card is 161.568564cm� when 'X' is to two decimal places being 1.81cm. I then divided 8 by 1.81 and got the answer 4.419889503. This is very similar to the answer for rectangular sheet 18x6 so this is proof that for rectangles with ratio 1:3 the cut out size is about 1/4 when rounded down from 4.4.... I now have results for rectangles with 3 different ratios and I have created this table: ratio cut out size if width=1 fraction 1:1 0.16666 1/6 1:2 0.21111 about 1/5 1:3 0.22666 about 1/4 I investigated some formulae to find the cut out size easily with less calculation. This rectangular problem can be solved algebraically: Area=(KW-2X)(W-2X) Volume=(KW-2X)(W-2X)xX =(KW�-2KWX-2XW-4X�)xX =KW�X-2KWX�-2X�W-4X� Gradient=KW�-4KWX-4XW+12X� 0=12X�-4WX(K+L)+KW� A=12 B=-4W(K+1) C=KW� General Formula: X=-B�� B�-4AC 2A X=4W(K+1) � -(4W(K+1))�-48KW� 24 X=4W(K+1) � 16W�(K+1)(K+1)-48KW� 24 X=4W(K+1) � 16W�(K�+2K+1)-48KW� 24 X=4W(K+1) � 16K�W�+32KW�+16W�-48KW� 24 X=4W(K+1) � 16K�W�-16KW�+16W� 24 X=4W(K+1) � 16W�(K�-K+1) 24 X=W(K+1) � 4W K�-K+1 24 X=W(K+1) � W K�-K+1 6 X=cut out size W=Width K=Integer that is multiplied by 'W' to get the length. This formula can be used to calculate the maximum cut out section of any size rectangular sheet of card, in order to obtain the maximum volume. ...read more.

    The above preview is unformatted text

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

    Found what you're looking for?

    • Start learning 29% faster today
    • 150,000+ documents available
    • Just £6.99 a month

    or


    Not the one? Search for your essay title...
    • Join over 1.2 million students every month
    • Accelerate your learning by 29%
    • Unlimited access from just £6.99 per month

    See related essaysSee related essays

    Related GCSE Open Box Problem essays

    1. Investigation: The open box problem.

      �x1.75 V = (11-(2x1.76)) �x1.76 V = (11-(2x1.77)) �x1.77 V = 98.4375 V = 98.472704 V = 98.503332 X = 1.78 X = 1.79 X = 1.81 V = (11-(2x1.78)) �x1.78 V = (11-(2x1.79)) �x1.79 V = (11-(2x1.81)) �x1.81 V = 98.529408 V = 98.550956 V = 98.580564 X = 1.82 X = 1.83 X = 1.84 V = (11-(2x1.82))

    2. Maximum box investigation

      which is impossible when the only numbers left to try were between 3 and 3.05. So, instead I decided to try numbers between 2 and 3. This proved to be more successful and I found that if the corner square length was 2.5 then the volume of the box would be 250 cm�.

    1. Open Box Problem

      74.07407407 1.66666 6.66668 6.66668 1.66666 74.07407407 1.66667 6.66666 6.66666 1.66667 74.07407407 1.66668 6.66664 6.66664 1.66668 74.07407407 1.66669 6.66662 6.66662 1.66669 74.07407406 1.6667 6.6666 6.6666 1.6667 74.07407405 1.66671 6.66658 6.66658 1.66671 74.07407404 1.66672 6.66656 6.66656 1.66672 74.07407402 1.6668 6.6664 6.6664 1.6668 74.07407372 1.6669 6.6662 6.6662 1.6669 74.07407299 1.667 6.666 6.666 1.667

    2. Maths Courseowrk - Open Box

      These are the values when l=10 X L V=x(10-2x) 2 1 10 64 1.5 10 73.5 2 10 72 2.5 10 62.5 3 10 48 3.5 10 31.5 4 10 16 4.5 10 4.5 1.75 10 73.9375 Here is a graph when l=10 These are the values when l=20 X L v=x(20-2x)

    1. Tbe Open Box Problem

      - "L/6 (L - 2L/6) L/6 This is: V = (L� - 2L�/6 - 2L�/6 - 4L�)L/6 I will multiply the lot by L/6 V = L�/6 - 2L�/36 - 2L�/36 + 4L�/216 Now I have put the denominator as 216 V = (4L� + 36L� - 12L� - 12L�)/216

    2. The Open Box Problem

      18.9 6.9 2.53 2.52 332.54323 18.96 6.96 2.52 2.51 332.5258 18.98 6.98 2.51 2.49 332.4658 19.02 7.02 2.49 Graph comparing the length of Small Side to the Volume for a rectangular shaped piece of card with dimensions 12 x 24 Rectangular piece of card with dimensions 24 x 48 Small

    1. THE OPEN BOX PROBLEM

      6.62 1.69 74.063 1.7 6.6 6.6 1.7 74.052 As you can see by the table above, the largest volume is achieved when the cut-out size of each corner of the box is 1.67cm. I also made a graph to prove that the maximum cut-out size is around 1.67cm.

    2. The open box problem

      3.96 3.96 1.02 15.99523 1.03 3.94 3.94 1.03 15.98931 1.04 3.92 3.92 1.04 15.98106 1.05 3.9 3.9 1.05 15.9705 You can see in each case that a 1.00 cm size cut-out gives the biggest volume of the open box. At first I used whole integers as my cut-out size, I

    • Over 160,000 pieces
      of student written work
    • Annotated by
      experienced teachers
    • Ideas and feedback to
      improve your own work