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=l²h 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.
This shows that the maximum volume is obtained when x is between 1 and 1.5, I will now make another three spreadsheets, to find the maximum volume correct to three decimal places, as I believe that this first spreadsheet does not give an accurate result.
These show, that for a square of width 6cm, that the maximum volume is obtained when x=1.000.
I will now repeat the spreadsheets above for a square of width 12cm, to find the maximum volume for this size of card, and to attempt to find any patterns in maximum volumes.
These show that the maximum volume for a square of width 12cm is obtained when x=2.
Width of square Value of x to give maximum volume
6 1
2 2
I have noticed that so far, my results would suggest that the maximum volume if given when x is 1/6th of the width of the square sheet of card. To prove this, I will have to find the maximum volume for another square, and see whether that obeys with what I have predicted.
Therefore a square of width 10cm should give a maximum volume when:
X=10=1.667
6
This is the spreadsheet for the square of length 10cm, this will give me what value of x gives the maximum volume for this size sheet of card.
The maximum volume is in fact obtained when x=1.667, which is what I predicted. This proves that the maximum volume is in fact obtained when x=1/6th of the width of the square.
It is possible to confirm my results for the maximum area through using differentiation. Differentiation will tell us the turning points on a graph, where the gradient = 0, one of the two numbers we will get will be the maximum value (which I have already found through spreadsheets) and the other will be the minimum value, this maximum will be the same as the values I have already come to, proving that they are correct.
I will now use differentiation for the square of width 6.
x(6 - 2x)² = V
x(6 - 2x)(6 - 2x) = V
x(4x² - 24x + 36) = V
4x³ - 24x² +36x = dv
dx
at maximum or minimum dv = 0
dx
2x² - 48x + 36 = 0
simplify by dividing by 12
x² - 4x + 3 = 0
(x - 1)(x - 3) = 0
x is therefore either 1 or 3
Since through my spreadsheets I have ...
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I will now use differentiation for the square of width 6.
x(6 - 2x)² = V
x(6 - 2x)(6 - 2x) = V
x(4x² - 24x + 36) = V
4x³ - 24x² +36x = dv
dx
at maximum or minimum dv = 0
dx
2x² - 48x + 36 = 0
simplify by dividing by 12
x² - 4x + 3 = 0
(x - 1)(x - 3) = 0
x is therefore either 1 or 3
Since through my spreadsheets I have already determined that the answer is 1, it must be 1 here, not 3, which is the minimum because at 3 the volume is 0, because as I have already stated, the box would be cut in half, and would not form a box.
The graph below shows that the maximum turning point is at 1, and that the minimum turning point is at 3.
All this proves that my answer for the square of side 6cm is correct, I will now repeat the differentiation for the other two sizes that I have chosen.
Square of length 12cm
x(12 - 2x)² = V
x(12 - 2x)(12 - 2x) = V
x(4x² - 48x + 144) = V
4x³ - 48x² + 144x = dv
dx
at maximum or minimum dv = 0
dx
2x² - 96x + 144 = 0
simplify by dividing all by 12
x² - 8x + 12 = 0
(x - 6)(x - 2) = o
x = 6 or 2, but must be 2, as 6 would give no box. This answer is the same as what I got earlier, using the spreadsheets.
Square of length 10cm
x(10 - 2x)² = V
x(10 - 2x)(10 - 2x) = V
x(4x² - 40x + 100) = V
4x³ - 40x² + 100x = V
simplify by dividing all by 4
x³ - 10x² + 100x = dv
dx
at maximum or minimum dv = 0
dx
3x² - 20x + 25 = 0
(3x - 5)(x - 5) = 0
x = 5 or 5, x must be 5 or 1.667, as 5 would not get a box, this volume is the
3 3
same as when I worked the volume out using spreadsheets.
This proves all my answers for the value of x which gives the maximum volume. I will now put the expression for the volume into general form to prove my finding that the maximum volume is given when x = 1/6th of the length of the card.
V = x(l [length] - 2x)²
V = x(l - 2x) ( l - 2x)
V = x(l² + 4x² - 2lx - 2lx)
V = x(l² + 4x² - 4lx)
V = 4x³ - 4lx² + l²x
dv = 4x³ - 4lx² + l²x
dx
at maximum or minimum dv = 0
dx
0 = 12x² - 8lx + l²
0 = (6x - 1)(2x - 1)
6x = L 1/2 x = L
x = L/6 x = L/2
x = L/6 or L/2
The x = L/6 proves my finding that the maximum volume is obtained when x is 1/6th of L. The L/2 can be ignored, because this always gives the minimum volume.
I will now move onto the second part of my investigation, which is investigating what size cut out will provide the largest volume for an open box made out of any rectangle.
I will begin this part of the investigation by attempting to find a general formula for the volume of the rectangle, and using this to find out the maximum volume for an open box made out of any sized rectangular sheet of card.
The volume of an open box can be expressed like this:
V = x (L - 2x) (W - 2x)
V stands for volume, L stands for the length of the piece of card, W for the width, and x is the length of the square cutout in each corner.
If I multiply the two brackets together I get this:
V = x (4x² + lw - 2wx - 2lx)
If I now multiply the bracket out I get:
V = 4x³ + 2lx² - 2wx² +lwx
I can simplify this by factorizing several parts:
V = x( lw ) + 4x³ - x²( 2l + 2w )
This is now ready for differentiation
When dv/dx = 0
dv/dx = lw + 12x² - 2(2l + 2w)
0 = 12x² - (4l + 4w) + lw
This is now a quadratic equation, we can use this formula:
X = -b ? ??b² - 4ac
2a
when a = Co-efficient of x²
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)² - 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) ? ?16l² + 16w² +32lw - 48lw
24
Now simplify:
X = 4(l + w) ? ?16l² + 16w² - 16lw
24
Now factorise:
X = 4(l + w) ? ?16(w² + l² - lw)
24
I can move the 16 out of the square root to give 4:
X = 4(l + w) ? 4 ?(w² + l² - lw)
24
I can now simplify by dividing by common factor 4
X = l+w ? ?w² + l² - 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.
I will now use this formula to find the value of x which gives the maximum volume for a rectangle of width 12cm, and length 10cm.
X = 10 + 12 ? ?12² + 10² - 10×12
6
x = 22 ? ?144 + 100 - 120
6
x = 22? ?124
6
x = 22 + 11.13552873 = 5.52
6
x = 22 - 11.13552873 = 1.8107, or 1.811 rounded up to three decimal places
6
x must be 1.8107, because 5.52 is over twice the length, so would therefore not form an open box. I will now find the value of x which gives the maximum volume for a rectangle of width 14cm and length 10cm.
X = 14 + 10 ? ?14² + 10² - 14×10
6
x = 24 ? ?196 + 100 - 140
6
x = 24 + 12.489996 = 6.082
6
x = 24 - 12.489996 = 1.918
6
Again, x must be 1.918, because 6.082 is over twice the length, so it would not form an open box. I will now ifnd which value of x gives the maximum volume for a rectangle of width 16cm and length 10cm.
X = 16 + 10? ?16² + 10² - 16×10
6
x = 26 ? ?256 + 100 - 160
6
x = 40 = 6.667
6
x = 12 = 2.000
6
Again, x must be 2.000, because it cannot be 6.667 because that is over twice the length, and would therefore not make an open box.
I will now produce a spreadsheet for each of these three rectangles, 10 by 12, 10 by 14 and 10 by 16. I will do this to prove that the values for x which give maximum volume which I have already found, are correct. I will do each spreadsheet to three decimal places for a degree of accuracy, and they will be very similar to those that I have done for the squares (above).
Spreadsheets for rectangle of width 12cm and length 10cm:
These show that the value for x which gives the maximum volume of the open box is 1.811 correct to three decimal places, which is what I predicted.
Here are the spreadsheets for the rectangle of width 14cm, and length 10cm.
These show that the maximum volume of the open box is obtained when x, the square cut out in each corner, is 1.918cm long. This is what I predicted in my formula.
Below are the spreadsheets for the rectangle of width 16cm and length 10cm:
These show that the maximum volume for the open box is obtained when x, the length of the square cut out in each corner is 2.000 cm, which is what I have already predicted.
I can use the formula:
X = -b ? ??b² - 4ac
2a
when a = Co-efficient of x²
b = Co-efficient of x
c = constant
to prove my results.
I can use in the formula
V = x (l - 2x)(w I 2x)
For the volume, differentiate this, and then put the results into the above formula to prove my results for x.
I will now do this for the rectangle of width 12cm and length 10cm.
V = x (10 - 2x)(12 - 2x)
V = x (4x² - 44x + 120)
V = 4x³ - 44x² + 120x
Now differentiate:
DV/DX = 4x³ - 44x² + 120x
When DV/DX = 0
0 = 12x² - 88x + 120
simplify by dividing by common factor four:
0 = 3x² - 22x + 30
This is now ready for the formula:
X = 22 ???22² - 4×3×30
2×3
x = 22 ??124
6
x = 22 + 11.13552873 = 5.5225
6
X = 22 - 11.13552873 = 1.8107 or 1.811 rounded up to three decimal places
6
The answer must be 1.811, because 5.5225 is over twice the length, and would therefore not form a box. This confirms what I have predicted with my formula, and proven once with the use of spreadsheets for this rectangle.
I will now do the same for the rectangle of width 14cm and length 10cm
V = (10 - 2x)(14 - 2x)
V = (4x² - 48x + 140)
V = 4x³ - 48x² + 140x
Now differentiate
DV/DX = 4x³ - 48x² + 140x
When DV/DX = 0
0 = 12x² - 96x + 140
Divide by common factor 4
0 = 3x² - 24x + 35
This is now ready for the formula:
X = 24 ??-24² - 4×3×35
2×3
X = 24 ??576 - 4×3×35
6
x = 24 ??156
6
x = 24 + 12.489996 = 6.08
6
x = 24 - 12.489996 = 1.918
6
The answer must be 1.918, as 6.08 is over twice the length and would therefore not form an open box. 1.918 is the same answer that I got through the use of the formula above and through the spreadsheets.
I will now do the same for the rectangle of width 16cm and length 10cm.
V = x(10 - 2x)(16 - 2x)
V = x(4x² - 52x + 160)
V = 4x³ - 52x² + 160x
Now differentiate:
DV/DX = 4x³ - 52x² + 160x
When DV/DX = 0
0 = 12x² - 104x + 160
Divide by common factor 4
0 = 3x² - 26x + 40
This is now ready for the formula:
X = 26??26² - 4×3×40
2×3
x = 26??676 - 480
6
x = 26??196
6
x = 26 + 14 = 6.667
6
x = 26 - 14 = 2.000
6
X must be 2.000, as 6.667 is over twice the length of the rectangle, and would therefore not form an open box. 2.000 is the same as the answer I got both through my use of my formula above and through the spreadsheets.
I have put the information from the spreadsheet for the rectangle into a graph on the following page, this shows that the maximum turning point is at 2.000, this helps prove that my answers are correct, and that the differentiation I have used gives the maximum turning point as 2.000 which is correct.
These results confirm this formula to be correct:
X = l+w ? ?w² + l² - 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 ? ?2l² - l²
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 ? ?w² + l² - 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 ? ?2l² - l²
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.