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In this investigation, a box without a lid must be made from a sheet of card, as shown below.
Identical squares must be cut out of each corner and the dotted lines folded along to form the sides of the box.
The goal of the investigation is to find out a relationship between the size of the initial piece of card, the size of the identical corner squares and the volume of the resulting open box. This will allow me to say what size corner square will produce the box of the largest volume, for any given rectangular sheet of card.
To begin with, I am going to investigate the size of the corner square that must be cut out to make an open box of the largest volume, for any sized square sheet of card. Once I have found the formula that allows me to find this out easily, I will progress to using an initial piece of card that is rectangular in shape.
The formula used to obtain the volume of a box is
VOLUME = Length * Width * Height
(where * is multiplication)
To show a simple example of how this formula works with the open box, I will first of all use a initial piece of card that is 20cm by 20 cm, and a corner square (from now on called 'cut-off') of 2cm by 2cm.
Once the cut-offs are taken away, the net will look like this.
From this we can see that when the dotted lines are folded along, there is a height of 2cm, a length of 16cm and a width of 16cm.
Since Volume = Length * Width * Height, the volume of this open box is 16*16*2 = 512cm3
We can also see relationships between the cut-off and the dimensions of the net from this example.
1. The size of the cut-off is the same size as the height of the box
2. The initial sizes of the length and width have decreased by 2 times the size of the cut-off
So, when the initial piece of card is square, the formula for the volume can be written:
Volume = (Length-2*Cut-off) * (Width-2*Cut-off) * Cut-off
V = (L-2C) * (W-2C) * C
V = (20-2*2) * (20-2*2) * 2
V = 16*16*2
V= 512cm3
And since the length and width are always equal to each other in a square, the formula can be simplified:
Volume = (Side-2*Cut-off)2 * Cut-off
V = (S-2C)2 *C
(where S is either the length or the width)
Using this theory I created a spreadsheet that allowed me to make tables of results quickly and efficiently.
This spreadsheet is set up so that when the side length and cut-off are entered in the top left, the resulting volume of a box with those attributes is calculated instantly in the table beneath, where the V = (S-2C)2 * C formula has been incorporated.
The cut-off plus value is added on to the initial cut-off 10 times, so that the greatest volume can be found quickly.
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When all the data is calculated for one size cut-off, I look at the volume column the find the largest, then find which cut-off produced the largest volume, then enter this value of cut of back into the top left, and choose a more precise cut-off plus. Doing this a number of times allows me to get a very accurate and precise value for the cut-off that gives the largest volume for a certain size square.
A sheet with the formulas used by the spreadsheet is included at the end of this investigation to proove that it is genuine.
As you can see in the table on the right, as the side length of the square increases, as does the maximum cut-off and the maximum volume. Therefore, the side length is proportional to the maximum cut-off and the maximum volume. However, since I only want to find a rule for the maximum cut-off, I can discount the volume for the moment. Also apparent is that most of the maximum cut-off values are recurring decimals. Since my spreadsheet does not recognise a recurring decimal (they are always rounded up after a certain amount of decimal places) I will convert these into fractions using the method shown on the next page.
If r = 0.833333333...
10r = 8.33333333...
10r - r = 8.33333333... - 0.833333333...
9r = 7.5
18r = 15
r = 15/18
r = 5/6
Using the same method, I transferred all of the maximum cut-off values I had obtained, from decimals into fractions.
Side Length
Maximum Cut-off
5
5/6
6
1/1
7
7/6
8
4/3
9
3/2
10
5/3
11
11/6
Then I put all the fractions over a common denominator to compare them easier.
Side Length
Maximum Cut-off
5
5/6
6
6/6
7
7/6
8
8/6
9
9/6
10
10/6
11
11/6
Here I can see a definite pattern between the side length and the maximum cut-off. The maximum cut-off is equal to the side length divided by 6. To test this theory, I predict that with a side length of 12, the cut off that will produce the maximum volume will be 12/6, or 2.
My prediction has been proved right in practice. So now I know that for a square, the cut-off that gives the greatest volume of box is equal to the length of one of the sides divided by 6. Now that I know this, I can rewrite my original formula to give the largest volume possible for a square, by substituting "Side/6" where I see "Cut-off". So:
Largest Volume = (Side - 2 * Side/6)2 * Side/6
LV = (S - 2S/6)2 * S/6
To test this I will use a side length of nine, which I already know has a largest volume of 54cm3
LV = (9 - 2 * 9/6)2 * 9/6
LV = (9 - 3)2 * 9/6
LV = (6)2 * 9/6
LV = 36 * 9/6
LV = 54cm3
This proves that the formula is correct and completes the first part of my investigation. I now need to investigate the size of cut-off that makes a box of the largest volume when the initial piece of card is rectangular.
The formula for the volume of a rectangle is:
Volume = Cut-off * (Length - 2 * Cut-off) * (Width - 2 * Cut-off)
V = C(L-2C)(W-2C)
Multiplying out the brackets:
V = C (L - 2C)(W - 2C)
V = C (LW - 2CL - 2CW + 4C2)
V = CLW - 2C2L - 2C2W + 4C3
The central part of this formula simplifies to:
V = CLW - 2C2L - 2C2W + 4C3
V = CLW - 2(L+W)C2 + 4C3
If I were to draw a graph of Cut-off against Volume using this formula, I would find a maximum and a minimum value. I want to find the maximum value for the Cut-off, but I don't want to have to draw a graph every time I want to find it. Differentiation is a method used to calculate the slope of a graph without actually drawing the graph, so I will use this.
The maximum (and minimum) values are found when the slope of the graph is equal to zero. At the point where there is no slope, it will no longer continue in the same direction, and go back the opposite way, leaving behind a maximum or minimum.
So, If I differentiate the equation I get the slope, and wherever the slope is equal to zero is either a maximum or minimum value.
The differentiation operator is d/dC(V) (where C is the cut-off and V is the volume). d/dC(V) is the differential of V, and shows the slope of the graph of V.
Differentiating the formula:
V = CLW - 2(L+W)C2 + 4C3
d/dC(V) = d/dC [CLW - 2(L+W)C2 + 4C3]
d/dC(V) = d/dC(CLW) - d/dC[2(L+W)C2] + d/dC(4C3)
d/dC(V) = LW*d/dC(C) - 2(L+W)*d/dC(C2) + 4*d/dC(C3)
d/dC(V) = LW*1 - 2(L+W)*2C + 4*3C2
d/dC(V) = LW - 4(L+W)C + 12C2
Since we want the slope, d/dC(V), to be zero:
LW - 4(L+W)C + 12C2 = 0
This can be solved using the common quadratic formula, giving the maximum or minimum value of C. The common quadratic formula is:
X = [-b +- sqrt(b2 - 4ac)] / 2a
Substituting my values in the formula and simplifying:
C = [-[-4(L+W)] +- sqrt[(-4(L+W))2 - 4*12*LW]] / 2*12
C = [4(L+W) +- sqrt[16(L+W)2 - 48LW]] / 24
C = [(L+W) +- sqrt[(L+W)2 - 3LW]] / 6
C = [(L+W) +- sqrt(L2+2LW+W2 - 3LW)] / 6
C = [(L+W) +- sqrt(L2-LW+W2)] / 6
This will produce two values for the Cut-off, one giving the largest volume possible and one giving the smallest. I will now test this formula by entering some random data and checking it with a spreadsheet.
Test
I have a rectangle of length 35cm and width 20cm. I want to find the cut-out that will give the open box of the largest possible volume.
C = [(35+20) +- sqrt(352-35*20+202)] / 6
C = [55+- sqrt(1225-700+400)] / 6
C = (55 +- sqrt925) / 6
C = (55 +- 30.4138) / 6
C = 85.4138 / 6 or C = 24.5862 / 6
C = 14.2356 or C = 4.0977
Substituting these results back into my formula for the volume, I can find which gives the largest volume. (see next page)
Using C = 14.2356
V = 14.2356 * (35 - 2*14.2356) * (20 - 2*14.3256)
V = 14.2356 * 6.5288 * -8.4717
V = -787.3715cm3
(since the volume cannot be negative, the optimum cut-off must be 4.0977)
Using C = 4.0977
V = 4.0977 * (35 - 2*4.0977) * (20 - 2*4.0977)
V = 4.0977 * 26.8046 * 11.8046
V = 1296.5843cm3
So, the largest volume possible for a box of initial dimensions 35cm by 20cm, is 1296.5843cm3 (to 4 decimal places). This occurs when the cut-off is 4.0977cm (to 4 decimal places). To prove that this is correct, I will change my original spreadsheet slightly to account for the use of rectangles instead of squares. These formula are again printed at the back of the project.
As you can see in the picture above, the maximum volume is attained when the cut-off is 4.0977, proving that my results were correct.
This concludes my project. The formula to work out the optimum cut-off from a rectangle is C = [(L+W) +- sqrt(L2-LW+W2)] / 6 This also works for a square, but the simpler formula for a square is S/6
Open Box Problem. (2020, Jun 02). Retrieved from https://studymoose.com/open-box-problem-new-essay
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