V = 162.5625
This means that the value X = 2.17 gives the largest volume.
Size of card – 14cm by 14cm
X must be 0<X<7: This is because if X is 0 there would not be a side to fold and if X is 7 then there would be nothing left.
X = 1 X = 2 X = 3
V = (14-(2x1)) ²x1 V = (14-(2x2)) ²x2 V = (14-(2x3)) ²x3
V = 144 V = 200 V = 192
X = 4 X = 5 X = 6
V = (14-(2x4)) ²x4 V = (14-(2x5)) ²x5 V = (14-(2x6)) ²x6
V = 144 V = 45 V = 24
X = 7
Not possible due to reasons mentioned above. If seven was cut from each corner then there would be nothing left to make the box with.
Graph
To show the cut off compared to the volume.
The cut off that gives the largest volume is 2. I will now look for the cut off that will leave the largest volume for the box. I will be looking for the cut off to 1d.p. between 1.5 – 2.5
X = 1.5 X = 1.6 X = 1.7
V = (14-(2x1.5)) ²x1.5 V = (14-(2x1.6)) ²x1.6 V = (14-(2x1.7)) ²x1.7
V = 181.5 V = 186.624 V = 191.012
X = 1.8 X = 1.9 X = 2.1
V = (14-(2x1.8)) ²x1.8 V = (14-(2x1.9)) ²x1.9 V = (14-(2x2.1)) ²x2.1
V = 194.688 V = 197.676 V = 201.684
X = 2.2 X = 2.3 X = 2.4
V = (14-(2x2.2)) ²x2.2 V = (14-(2x2.3)) ²x2.3 V = (14-(2x2.4)) ²x2.4
V = 202.752 V = 203.228 V = 203.136
X = 2.5
V = (14-(2x2.5)) ²x2.5
V = 202.5
The cut off that will leave the largest volume so far is 2.3. I will now look for the cut off to 2d.p. between 2.25 – 2.35
X = 2.25 X = 2.26 X = 2.27
V = (14-(2x2.25)) ²x2.25 V = (14-(2x2.26)) ²x2.26 V = (14-(2x2.27)) ²x2.27
V = 203.0625 V = 203.107104 V = 203.107104
X = 2.28 X = 2.29 X = 2.31
V = (14-(2x2.28)) ²x2.28 V = (14-(2x2.29)) ²x2.29 V = (14-(2x2.31)) ²x2.31
V = 203.179008 V = 203.206356 V = 203.243964
X = 2.32 X = 2.33 X = 2.34
V = (14-(2x2.32)) ²x2.32 V = (14-(2x2.33)) ²x2.33 V = (14-(2x2.34)) ²x2.34
V = 203.254272 V = 203.258948 V = 203.258016
X = 2.35
V = (14-(2x2.35)) ²x2.35
V = 203.2515
This means that the value X = 2.33 give the largest volume.
Highest values of the cut off
I will put all the cut off values that give the maximum volume in table to find a general formula.
Total of the last column : 29.9983
Average (total/amount of numbers) : 5.99966
The answer to 1 s.f. : 6
Therefore if we take the side of any square and divide it by 6 we get the cut off that would leave the highest volume.
E.g.
If the size of card is 14cm.
Then the size of cut off would be 14/6 = 2.33 (2.d.p)
Gradient
I will plot a graph and find the gradient of the line.
Graph
(See appendix 1)
Gradient = L/><
L = 13 – 11 = 2
>< = 2.17 – 1.83 = 0.34
Gradient = 2/0.34
= 5.8823
= 6 (1 s.f.)
Both of the above formulas say that if you divide the size of the card by 6 you will get the size of the cut off.
Testing
Size of card – 20cm by 20cm
Prediction
Using the formula that I worked out using the two methods above I will predict what the answer should be to see if it matches the actual answer. If it does then my formula is correct. If not then it is wrong.
X = L/6
X = 20/6
X = 3.33…
Working out
I will now do all the working out to see if my prediction and formula were right.
X must be 0<X<10: This is because if X is 0 there would not be a side to fold and if X is 10 then there would be nothing left.
X = 1 X = 2 X = 3
V = (20-(2x1)) ²x1 V = (20-(2x2)) ²x2 V = (20-(2x3)) ²x3
V = 324 V = 512 V = 588
X = 4 X = 5 X = 6
V = (20-(2x4)) ²x4 V = (20-(2x5)) ²x5 V = (20-(2x6)) ²x6
V = 576 V = 500 V = 384
X = 7 X = 8 X = 9
V = (20-(2x7)) ²x7 V = (20-(2x8)) ²x8 V = (20-(2x9)) ²x9
V = 252 V = 128 V = 36
X = 10
Not possible due to reasons mentioned above. If ten was cut from each corner then there would be nothing left to make the box with.
The cut off that gives the largest volume is 3. I will now look for the cut off that will leave the largest volume for the box. I will be looking for the cut off to 1d.p. between 2.5 – 3.5
X = 2.5 X = 2.6 X = 2.7
V = (20-(2x2.5)) ²x2.5 V = (20-(2x2.6)) ²x2.6 V = (20-(2x2.7)) ²x2.7
V = 562.5 V = 569.504 V = 575.532
X = 2.8 X = 2.9 X = 3.1
V = (20-(2x2.8)) ²x2.8 V = (20-(2x2.9)) ²x2.9 V = (20-(2x3.1)) ²x3.1
V = 580.608 V = 584.756 V = 590.364
X = 3.2 X = 3.3 X = 3.4
V = (20-(2x3.2)) ²x3.2 V = (20-(2x3.3)) ²x3.3 V = (20-(2x3.4)) ²x3.4
V = 591.872 V = 592.548 V = 592.416
X = 3.5
V = (20-(2x3.5)) ²x3.5
V = 591.5
The cut off that will leave the largest volume so far is 3.3. I will now look for the cut off to 2d.p. between 3.25 – 3.35
X = 3.25 X = 3.26 X = 3.27
V = (20-(2x3.25)) ²x3.25 V = (20-(2x3.26)) ²x3.26 V = (20-(2x3.27)) ²x3.27
V = 592.3125 V = 592.375904 V = 592.431132
X = 3.28 X = 3.29 X = 3.31
V = (20-(2x3.28)) ²x3.28 V = (20-(2x3.29)) ²x3.29 V = (20-(2x3.31)) ²x3.31
V = 592.478208 V = 592.517156 V = 592.570764
X = 3.32 X = 3.33 X = 3.34
V = (20-(2x3.32)) ²x3.32 V = (20-(2x3.33)) ²x3.33 V = (20-(2x3.34)) ²x3.34
V = 592.585472 V = 592.592148 V = 592.590816
X = 3.35
V = (20-(2x3.35)) ²x3.35
V = 592.5815
This means that the value X = 2.33 give the largest volume. This also means that me prediction was right and therefore the formula I used was correct. However to prove that it works for any number I will use differentiation.
Differentiation
I will use differentiation to prove that X = L/6
Differentiation is used to find the maximum and minimum of a point on a curve, of the gradient of a curve.
The gradient should always be equal to 0 (see below)
V = Xn This is the formula for differentiation.
dV/dX This is the symbol for differentiation.
dV/dX = nX n – 1 Multiply the power by the coefficient of X (the n umber in front of X), then minus 1 from the power.
Therefore firstly you multiply out the brackets:
V = (L-2X) (L-2X) X
= (L 2 – 2LX – 4X 2 ) X
= L 2 X 1-1 – 4LX 2-1 – 4X 3-1
dV/dX = L 2 – 8LX + 12X 2 = 0
To find the maximum dV/dX = 0
Rearrange the formula into the form of a quadratic equation:
12X 2 – 8LX + L 2 = 0
Using the quadratic formula I will solve the equation.
-b+/_√b 2 – 4ac
The quadratic formula
2a
aX 2 + bX + c = 0 Formula to decide what a, b and c stand for.
12X 2 – 8LX +L 2 = 0
Therefore:
a = 12 b = -8L c = L 2
+8L+/_√ (-8L)2 – 4 x 12 x L2
Substituting a, b and c into the formula
2 x 12
+8L+/_√64L2 – 48L2
Simplifying the formula
24
+8L+/_√16L2
Simplifying the formula
24
+8L+/_4L
Simplifying the formula
24
This now means that x can be either one of two values:
X = 8L + 4L X = 8L – 4L
Or
24 24
= 12L = 4L
24 24
X = L X = L
2 6
This cannot be the formula because L/2 = 0, there would be no card left to make the box with.
However this means that my formula I previously found was correct, X = L/6. So if I divide the length of any square by 6 it will give me the cut off that will leave the maximum volume.
Plan 2:
Now I will use trial and improvement to experiment with different sizes of a rectangular box. By doing this I will find out the size of cut off that will leave me with the largest volume inside the box. To find out the volume I will need to know the size of the cut off side and the width length.
x = length off the square cut off
L = original length off the width
The formula that I will use to work out the volume is:
Volume = (2L - 2X) (L - 2X) X for the 1:2 ratio
Volume = (3L - 2X) (L - 2X) X for the 1:3 ratio
Volume = (4L - 2X) (L - 2X) X for the 1:4 ratio
To find the different sizes of card I will use ratio. I will use three different sizes: 1:2, 1:3, 1:4. After this I will then look at the ratio, 1: n. I will determine the size of x that will give the highest volume to 2d.p. After finding the highest value of X I will prove that my answer if right by using differentiation. Finally I will try and find a rule that allows me to find the highest value of X for a piece of rectangular card and check that it works with any size of card.
Trail and improvement
Size of card – 1:2
Size of width 7
X must be 0<X<3.5: This is because if X is 0 there would not be a side to fold and if X is 3.5 then there would be nothing left.
X = 1 X = 2 X = 3
V = (2x7-2x1) (7-2x1) x1 V = (2x7-2x2) (7-2x2) x2 V = (2x7-2x3) (7-2x3) x3
V = 60 V = 60 V = 24
Graph
To show the cut off compared to the volume.
The cut off that gives the largest volume lies between 1 and 2. I will now look for the cut off that will leave the largest volume for the box. I will be looking for the cut off to 1d.p. between 1 – 2.
X = 1.1
V = (2x7-2x1.1) (7-2x1.1) x1.1
V = 62.304
X = 1.2
V = (2x7-2x1.2) (7-2x1.2) x1.2
V = 64.032
X = 1.3
V = (2x7-2x1.3) (7-2x1.3) x1.3
V = 65.2
X = 1.4
V = (2x7-2x1.4) (7-2x1.4) x1.4
V = 65.856
X = 1.5
V = (2x7-2x1.5) (7-2x1.5) x1.5
V = 66
X = 1.6
V = (2x7-2x1.6) (7-2x1.6) x1.6
V = 65.664
X = 1.7
V = (2x7-2x1.7) (7-2x1.7) x1.7
V = 64.872
X = 1.8
V = (2x7-2x1.8) (7-2x1.8) x1.8
V = 63.648
X = 1.9
V = (2x7-2x1.9) (7-2x1.9) x1.9
V = 62.016
The cut off that will leave the largest volume so far is 1.5. I will now look for the cut off to 2d.p. between 1.45 – 1.55
X = 1.45
V = (2x7-2x1.45) (7-2x1.45) x1.45
V = 65.9895
X = 1.46
V = (2x7-2x1.46) (7-2x1.46) x1.46
V = 66.001344
X = 1.47
V = (2x7-2x1.47) (7-2x1.47) x1.47
V = 66.008292
X = 1.48
V = (2x7-2x1.48) (7-2x1.48) x1.48
V = 66.010368
X = 1.49
V = (2x7-2x1.49) (7-2x1.49) x1.49
V = 66.007596
X = 1.51
V = (2x7-2x1.51) (7-2x1.51) x1.51
V = 65.987604
X = 1.52
V = (2x7-2x1.52) (7-2x1.52) x1.52
V = 65.970432
X = 1.53
V = (2x7-2x1.53) (7-2x1.53) x1.53
V = 65.948508
X = 1.54
V = (2x7-2x1.54) (7-2x1.54) x1.54
V = 65.921856
X = 1.55
V = (2x7-2x1.55) (7-2x1.55) x1.55
V = 65.8905
This means that the value X = 1.48 give the largest volume.
Size of width 8
X must be 0<X<4: This is because if X is 0 there would not be a side to fold and if X is 4 then there would be nothing left.
X = 1 X = 2 X = 3
V = (2x8-2x1) (8-2x1) x1 V = (2x8-2x2) (8-2x2) x2 V = (2x8-2x3) (8-2x3) x3
V = 84 V = 96 V = 60
X = 4
Not possible due to the reasons mentioned above. If four was cut from each corner then there would be nothing left to make the sides with.
Graph
To show the cut off compared to the volume.
The cut off that gives the largest volume is 2. I will now look for the cut off that will leave the largest volume for the box. I will be looking for the cut off to 1d.p. between 1.5 – 2.5
X = 1.5
V = (2x8-2x1.5) (8-2x1.5) x1.5
V = 97.5
X = 1.6
V = (2x8-2x1.6) (8-2x1.6) x1.6
V = 98.304
X = 1.7
V = (2x8-2x1.7) (8-2x1.7) x1.7
V = 98.532
X = 1.8
V = (2x8-2x1.8) (8-2x1.8) x1.8
V = 98.208
X = 1.9
V = (2x8-2x1.9) (8-2x1.9) x1.9
V = 97.356
X = 2.1
V = (2x8-2x2.1) (8-2x2.1) x2.1
V = 94.164
X = 2.2
V = (2x8-2x2.2) (8-2x2.2) x2.2
V = 91.872
X = 2.3
V = (2x8-2x2.3) (8-2x2.3) x2.3
V = 89.148
X = 2.4
V = (2x8-2x2.4) (8-2x2.4) x2.4
V = 86.016
X = 2.5
V = (2x8-2x2.5) (8-2x2.5) x2.5
V = 82.5
The cut off that will leave the largest volume so far is 1.7. I will now look for the cut off to 2d.p. between 1.65 – 1.75
X = 1.65
V = (2x8-2x1.65) (8-2x1.65) x1.65
V = 98.4885
X = 1.66
V = (2x8-2x1.66) (8-2x1.66) x1.66
V = 98.508384
X = 1.67
V = (2x8-2x1.67) (8-2x1.67) x1.67
V = 98.522652
X = 1.68
V = (2x8-2x1.68) (8-2x1.68) x1.68
V = 98.531328
X = 1.69
V = (2x8-2x1.69) (8-2x1.69) x1.69
V = 98.534436
X = 1.71
V = (2x8-2x1.71) (8-2x1.71) x1.71
V = 98.524044
X = 1.72
V = (2x8-2x1.72) (8-2x1.72) x1.72
V = 98.510592
X = 1.73
V = (2x8-2x1.73) (8-2x1.73) x1.73
V = 98.491668
X = 1.74
V = (2x8-2x1.74) (8-2x1.74) x1.74
V = 98.467296
X = 1.75
V = (2x8-2x1.75) (8-2x1.75) x1.75
V = 98.4375
This means that the value X = 1.69 give the largest volume.
Size of width 9
X must be 0<X<4.5: This is because if X is 0 there would not be a side to fold and if X is 4.5 then there would be nothing left.
X = 1 X = 2 X = 3
V = (2x9-2x1) (9-2x1) x1 V = (2x9-2x2) (9-2x2) x2 V = (2x9-2x3) (9-2x3) x3
V = 112 V = 140 V = 108
X = 4
V = (2x9-2x3) (9-2x3) x3
V = 40
Graph
To show the cut off compared to the volume.
The cut off that gives the largest volume is 2. I will now look for the cut off that will leave the largest volume for the box. I will be looking for the cut off to 1d.p. between 1.5 – 2.5
X = 1.5
V = (2x9-2x1.5) (9-2x1.5) x1.5
V = 135
X = 1.6
V = (2x9-2x1.6) (9-2x1.6) x1.6
V = 137.344
X = 1.7
V = (2x9-2x1.7) (9-2x1.7) x1.7
V = 138.992
X = 1.8
V = (2x9-2x1.8) (9-2x1.8) x1.8
V = 139.968
X = 1.9
V = (2x9-2x1.9) (9-2x1.9) x1.9
V = 140.296
X = 2.1
V = (2x9-2x2.1) (9-2x2.1) x2.1
V = 139.104
X = 2.2
V = (2x9-2x2.2) (9-2x2.2) x2.2
V = 137.632
X = 2.3
V = (2x9-2x2.3) (9-2x2.3) x2.3
V = 135.608
X = 2.4
V = (2x9-2x2.4) (9-2x2.4) x2.4
V = 133.056
X = 2.5
V = (2x9-2x2.5) (9-2x2.5) x2.5
V = 130
The cut off that will leave the largest volume so far is 1.9. I will now look for the cut off to 2d.p. between 1.85 – 1.95
X = 1.85
V = (2x9-2x1.85) (9-2x1.85) x1.85
V = 140.2115
X = 1.86
V = (2x9-2x1.86) (9-2x1.86) x1.86
V = 140.241024
X = 1.87
V = (2x9-2x1.87) (9-2x1.87) x1.87
V = 140.264212
X = 1.88
V = (2x9-2x1.88) (9-2x1.88) x1.88
V = 140.281088
X = 1.89
V = (2x9-2x1.89) (9-2x1.89) x1.89
V = 140.291676
X = 1.91
V = (2x9-2x1.91) (9-2x1.91) x1.91
V = 140.294084
X = 1.92
V = (2x9-2x1.92) (9-2x1.92) x1.92
V = 140.285952
X = 1.93
V = (2x9-2x1.93) (9-2x1.93) x1.93
V = 140.271628
X = 1.94
V = (2x9-2x1.94) (9-2x1.94) x1.94
V = 140.251136
X = 1.95
V = (2x9-2x1.95) (9-2x1.95) x1.95
V = 140.2245
This means that the value X = 1.91 give the largest volume.
Highest values of the cut off
I will put all the cut off values that give the maximum volume in table to find a general formula.
Total of the last column : 14.17549973
Average (total/amount of numbers) : 4.725166575
The answer to 2d.p. : 4.73
Therefore if we take the width of any rectangular card and divide it by 4.73 we get the cut off that would leave the highest volume.
E.g.
If the size of card is 8 by 16cm.
Then the size of cut off would be 8/4.73 = 1.69 (2.d.p)
Gradient
I will plot a graph and find the gradient of the line.
Graph
(See appendix 2)
Gradient = L/><
L = 8 – 7 = 1
>< = 1.69 – 1.48 = 0.21
Gradient = 1/0.21
= 4.76190
= 4.76 (2d.p.)
Both of the above formulas say different answers so I will now use differentiation to find an exact formula that will give the size of cut off that will leave the largest volume.
Differentiation
I will use differentiation to find the formula.
Differentiation is used to find the maximum and minimum of a point on a curve, of the gradient of a curve.
The gradient should always be equal to 0 (see below)
V = Xn This is the formula for differentiation.
dV/dX This is the symbol for differentiation.
dV/dX = nX n – 1 Multiply the power by the coefficient of X (the n umber in front of X), then minus 1 from the power.
Therefore firstly you multiply out the brackets:
V = (2L-2X) (L-2X) X
= (2L 2 – 6LX + 4X 2 ) X
= 2L 2 X – 6LX 2 + 4X 3
dV/dX = 2L 2 – 12LX +12X 2 = 0
To find the maximum dV/dX = 0
Using the quadratic formula I will solve the equation.
-b+/_√b 2 – 4ac
The quadratic formula
2a
aX 2 + bX + c = 0 Formula to decide what a, b and c stand for.
2X 2 – 12LX +2L 2 = 0
Therefore:
a = 12 b = -12L c = 2L 2
+12L+/_√ (-12L)2 – 4 x 12 x 2L2
Substituting a, b and c into the formula
2 x 12
+12L+/_√144L2 – 96L2
Simplifying the formula
24
+12L+/_√48L2
Simplifying the formula
24
+12L+/_6.9282…L
Simplifying the formula
24
This now means that x can be either one of two values:
X = 12L + 6.9282…L X = 12L – 6.9282…L
Or
24 24
= 18.92820…L = 5.071796…L
24 24
= 0.78867…L = 0.21132L
= 1 = 1
The reciprocal
0.78867L 0.21132L
X = 1.267L X = 4.732L
X = L X = L
The reciprocal
1.267 4.732
This cannot be the formula because L/1.267 = >L/2, this means that you will be cutting off more card than there actually is and therefore there would be no card left to make the box with.
However this means that my formula I previously found was incorrect. So instead if I divide the length of any rectangle with the ratio 1:2 by the 4.732 it will give me the cut off that will leave the maximum volume.
Testing
Size of card – 6cm by 12cm
Prediction
Using the formula that I worked out using differentiation above I will predict what the answer should be to see if it matches the actual answer. If it does then my formula is correct. If not then it is wrong.
X = L/4.73
X = 6/4.73
X = 1.268….
Working out
I will now do all the working out to see if my prediction and formula were right.
X must be 0<X<3: This is because if X is 0 there would not be a side to fold and if X is 3 then there would be nothing left.
X = 1 X = 2
V = (2x6-2x1) (6-2x1) x1 V = (2x6-2x2) (6-2x2) x2
V = 40 V = 32
X = 3
Not possible due to the reasons mentioned above. If four was cut from each corner then there would be nothing left to make the sides with.
The cut off that gives the largest volume is 1. I will now look for the cut off that will leave the largest volume for the box. I will be looking for the cut off to 1d.p. between 0.5 – 1.5
X = 0.5
V = (2x6-2x0.5) (6-2x0.5) x0.5
V = 27.5
X = 0.6
V = (2x6-2x0.6) (6-2x0.6) x0.6
V = 31.104
X = 0.7
V = (2x6-2x0.7) (6-2x0.7) x0.7
V = 34.132
X = 0.8
V = (2x6-2x0.8) (6-2x0.8) x0.8
V = 36.608
X = 0.9
V = (2x6-2x0.9) (7-2x0.9) x0.9
V = 38.556
X = 1.1
V = (2x6-2x1.1) (6-2x1.1) x1.1
V = 40.964
X = 1.2
V = (2x6-2x1.2) (6-2x1.2) x1.2
V = 41.472
X = 1.3
V = (2x6-2x1.3) (6-2x1.3) x1.3
V = 41.548
X = 1.4
V = (2x6-2x1.4) (6-2x1.4) x1.4
V = 41.216
X = 1.5
V = (2x6-2x1.5) (6-2x1.5) x1.5
V = 40.5
The cut off that will leave the largest volume so far is 1.3 I will now look for the cut off to 2d.p. between 1.25 – 1.35
X = 1.25
V = (2x6-2x1.25) (6-2x1.25) x1.25
V = 41.5625
X = 1.26
V = (2x6-2x1.26) (6-2x1.26) x1.26
V = 41.567904
X = 1.27
V = (2x6-2x1.27) (6-2x1.27) x1.27
V = 41.569132
X = 1.28
V = (2x6-2x1.28) (6-2x1.28) x1.28
V = 41.566208
X = 1.29
V = (2x6-2x1.29) (6-2x1.29) x1.29
V = 41.559156
X = 1.31
V = (2x6-2x1.31) (6-2x1.31) x1.31
V = 41.532764
X = 1.32
V = (2x6-2x1.32) (6-2x1.32) x1.32
V = 41.513472
X = 1.33
V = (2x6-2x1.33) (6-2x1.33) x1.33
V = 41.490148
X = 1.34
V = (2x6-2x1.34) (6-2x1.34) x1.34
V = 41.462816
X = 1.35
V = (2x6-2x1.35) (6-2x1.35) x1.35
V = 41.4315
This means that the value X = 1.27 gives the largest volume. This also means that me prediction was right and therefore the formula I used was correct.
Size of card – 1:3
Size of width 7
X must be 0<X<3.5: This is because if X is 0 there would not be a side to fold and if X is 3.5 then there would be nothing left.
X = 1 X = 2 X = 3
V = (3x7-2x1) (7-2x1) x1 V = (3x7-2x2) (7-2x2) x2 V = (3x7-2x3) (7-2x3) x3
V = 95 V = 102 V = 45
Graph
To show the cut off compared to the volume.
The cut off that gives the largest volume is 2. I will now look for the cut off that will leave the largest volume for the box. I will be looking for the cut off to 1d.p. between 1.5 – 2.5.
X = 1.5
V = (3x7-2x1.5) (7-2x1.5) x1.5
V = 108
X = 1.6
V = (3x7-2x1.6) (7-2x1.6) x1.6
V = 108.224
X = 1.7
V = (3x7-2x1.7) (7-2x1.7) x1.7
V = 107.712
X = 1.8
V = (3x7-2x1.8) (7-2x1.8) x1.8
V = 106.488
X = 1.9
V = (3x7-2x1.9) (7-2x1.9) x1.9
V = 104.576
X = 2.1
V = (3x7-2x0.9) (7-2x0.9) x0.9
V = 98.784
X = 2.2
V = (3x7-2x2.2) (7-2x2.2) x2.2
V = 94.952
X = 2.3
V = (3x7-2x2.3) (7-2x2.3) x2.3
V = 90.528
X = 2.4
V = (3x7-2x2.4) (7-2x2.4) x2.4
V = 85.536
X = 2.5
V = (3x7-2x2.5) (7-2x2.5) x2.5
V = 80
The cut off that will leave the largest volume so far is 1.6 I will now look for the cut off to 2d.p. between 1.55 – 1.65
X = 1.55
V = (3x7-2x1.55) (7-2x1.55) x1.55
V = 108.2055
X = 1.56
V = (3x7-2x1.56) (7-2x1.56) x1.56
V = 108.224064
X = 1.57
V = (3x7-2x1.57) (7-2x1.57) x1.57
V = 108.235172
X = 1.58
V = (3x7-2x1.58) (7-2x1.58) x1.58
V = 108.238848
X = 1.59
V = (3x7-2x1.59) (7-2x1.59) x1.59
V = 108.235116
X = 1.61
V = (3x7-2x1.61) (7-2x1.61) x1.61
V = 108.205524
X = 1.62
V = (3x7-2x1.62) (7-2x1.62) x1.62
V = 108.179712
X = 1.63
V = (3x7-2x1.63) (7-2x1.63) x1.63
V = 108.146588
X = 1.64
V = (3x7-2x1.64) (7-2x1.64) x1.64
V = 108.106176
X = 1.65
V = (3x7-2x1.65) (7-2x1.65) x1.65
V = 108.0585
This means that the value X = 1.81 give the largest volume.
Size of width 8
X must be 0<X<4: This is because if X is 0 there would not be a side to fold and if X is 4 then there would be nothing left.
X = 1 X = 2 X = 3
V = (3x8-2x1) (8-2x1) x1 V = (3x8-2x2) (8-2x2) x2 V = (3x8-2x3) (8-2x3) x3
V = 132 V = 160 V = 108
X = 4
Not possible due to the reasons mentioned above. If four was cut from each corner then there would be nothing left to make the sides with.
Graph
To show the cut off compared to the volume.
The cut off that gives the largest volume is 2. I will now look for the cut off that will leave the largest volume for the box. I will be looking for the cut off to 1d.p. between 1.5 – 2.5
X = 1.5
V = (3x8-2x1.5) (8-2x1.5) x1.5
V = 157.5
X = 1.6
V = (3x8-2x1.6) (8-2x1.6) x1.6
V = 159.744
X = 1.7
V = (3x8-2x1.7) (8-2x1.7) x1.7
V = 161.092
X = 1.8
V = (3x8-2x1.8) (8-2x1.8) x1.8
V = 161.568
X = 1.9
V = (3x8-2x1.9) (8-2x1.9) x1.9
V = 161.196
X = 2.1
V = (3x8-2x2.1) (8-2x2.1) x2.1
V = 158.004
X = 2.2
V = (3x8-2x2.2) (8-2x2.2) x2.2
V = 155.232
X = 2.3
V = (3x8-2x2.3) (8-2x2.3) x2.3
V = 151.708
X = 2.4
V = (3x8-2x2.4) (8-2x2.4) x2.4
V = 147.456
X = 2.5
V = (3x8-2x2.5) (8-2x2.5) x2.5
V = 142.5
The cut off that will leave the largest volume so far is 1.8. I will now look for the cut off to 2d.p. between 1.75 – 1.85
X = 1.75
V = (3x8-2x1.75) (8-2x1.75) x1.75
V = 161.4375
X = 1.76
V = (3x8-2x1.76) (8-2x1.76) x1.76
V = 161.480704
X = 1.77
V = (3x8-2x1.77) (8-2x1.77) x1.77
V = 161.515332
X = 1.78
V = (3x8-2x1.78) (8-2x1.78) x1.78
V = 161.541408
X = 1.79
V = (3x8-2x1.79) (8-2x1.79) x1.79
V = 161.558956
X = 1.81
V = (3x8-2x1.81) (8-2x1.81) x1.81
V = 161.568564
X = 1.82
V = (3x8-2x1.82) (8-2x1.82) x1.82
V = 161.560672
X = 1.83
V = (3x8-2x1.83) (8-2x1.83) x1.83
V = 161.544348
X = 1.84
V = (3x8-2x1.84) (8-2x1.84) x1.84
V = 161.519616
X = 1.85
V = (3x8-2x1.85) (8-2x1.85) x1.85
V = 161.4865
This means that the value X = 1.81 give the largest volume.
Size of width 9
X must be 0<X<4.5: This is because if X is 0 there would not be a side to fold and if X is 4.5 then there would be nothing left.
X = 1 X = 2 X = 3
V = (3x9-2x1) (9-2x1) x1 V = (3x9-2x2) (9-2x2) x2 V = (3x9-2x3) (9-2x3) x3
V = 175 V = 230 V = 189
X = 4
V = (3x9-2x3) (9-2x3) x3
V = 76
Graph
To show the cut off compared to the volume.
The cut off that gives the largest volume is 2. I will now look for the cut off that will leave the largest volume for the box. I will be looking for the cut off to 1d.p. between 1.5 – 2.5
X = 1.5
V = (3x9-2x1.5) (9-2x1.5) x1.5
V = 21.6
X = 1.6
V = (3x9-2x1.6) (9-2x1.6) x1.6
V = 220.864
X = 1.7
V = (3x9-2x1.7) (9-2x1.7) x1.7
V = 224.672
X = 1.8
V = (3x9-2x1.8) (9-2x1.8) x1.8
V = 227.448
X = 1.9
V = (3x9-2x1.9) (9-2x1.9) x1.9
V = 229.216
X = 2.1
V = (3x9-2x2.1) (9-2x2.1) x2.1
V = 229.824
X = 2.2
V = (3x9-2x2.2) (9-2x2.2) x2.2
V = 228.712
X = 2.3
V = (3x9-2x2.3) (9-2x2.3) x2.3
V = 226.688
X = 2.4
V = (3x9-2x2.4) (9-2x2.4) x2.4
V = 223.776
X = 2.5
V = (3x9-2x2.5) (9-2x2.5) x2.5
V = 220
The cut off that will leave the largest volume so far is 2.1. I will now look for the cut off to 2d.p. between 2.05 – 2.15.
X = 2.05
V = (3x9-2x2.05) (9-2x2.05) x2.05
V = 230.0305
X = 2.06
V = (3x9-2x2.06) (9-2x2.06) x2.06
V = 230.008064
X = 2.07
V = (3x9-2x2.07) (9-2x2.07) x2.07
V = 229.976172
X = 2.08
V = (3x9-2x2.08) (9-2x2.08) x2.08
V = 229.934848
X = 2.09
V = (3x9-2x2.09) (9-2x2.09) x2.09
V = 229.884116
X = 2.11
V = (3x9-2x2.11) (9-2x2.11) x2.11
V = 229.754524
X = 2.12
V = (3x9-2x2.12) (9-2x2.12) x2.12
V = 229.6757712
X = 2.13
V = (3x9-2x2.13) (9-2x2.13) x2.13
V = 229.587588
X = 2.14
V = (3x9-2x2.14) (9-2x2.14) x2.14
V = 229.490176
X = 2.15
V = (3x9-2x2.15) (9-2x2.15) x2.15
V = 229.3835
This means that the value X = 2 gives the largest volume.
Highest values of the cut off
I will put all the cut off values that give the maximum volume in table to find a general formula.
Total of the last column : 13.35026925
Average (total/amount of numbers) : 4.45008975
The answer to 2d.p. : 4.45
Therefore if we take the width of any rectangular card and divide it by 4.45 we get the rough cut off that would leave the highest volume.
E.g.
If the size of card is 8 by 24cm.
Then the size of cut off would be 8/4.45 = 1.8(2.d.p)
Gradient
I will plot a graph and find the gradient of the line.
Graph
(See appendix 3)
Gradient = L/><
L = 8 – 7 = 1
>< = 1.81 – 1.58 = 0.0.23
Gradient = 1/0.23
= 4.347826
= 4.35 (2d.p.)
Both of the above formulas say different formulas therefore I will now use differentiation to find the exact formula.
Differentiation
I will use differentiation to find the exact formula that will give me the cut off that will leave the maximum volume.
Differentiation is used to find the maximum and minimum of a point on a curve, of the gradient of a curve.
The gradient should always be equal to 0 (see below)
V = Xn This is the formula for differentiation.
dV/dX This is the symbol for differentiation.
dV/dX = nX n – 1 Multiply the power by the coefficient of X (the n umber in front of X), then minus 1 from the power.
Therefore firstly you multiply out the brackets:
V = (3L-2X) (L-2X) X
= (3L 2 – 6LX – 2LX + 4X 2 ) X
= 3L 2 X – 8LX 2 + 4X 3
dV/dX = 3L 2 – 16LX + 12X 2 = 0
To find the maximum dV/dX = 0
Using the quadratic formula I will solve the equation.
-b+/_√b 2 – 4ac
The quadratic formula
2a
aX 2 + bX + c = 0 Formula to decide what a, b and c stand for.
3X 2 – 16LX + 12L 2 = 0
Therefore:
a = 12 b = -16L c = 3L 2
+16L+/_√ (-16L)2 – 4 x 12 x 3L2
Substituting a, b and c into the formula
2 x 12
+16L+/_√256L2 – 144L2
Simplifying the formula
24
+16L+/_√112L2
Simplifying the formula
24
+16L+/_10.583…L
Simplifying the formula
24
This now means that x can be either one of two values:
X = 16L + 10.583…L X = 16L – 10.583…L
Or
24 24
= 26.583…L = 5.417…L
24 24
= 1.10762…L = 0.22573…L
= 1 = 1
The reciprocal
1.10762L 0.22573L
X = 0.90L X = 4.43L
X = L X = L
The reciprocal
0.90 4.43
This cannot be the formula because L/0.90 = >L/2, this means that you will be cutting off more card than there actually is and therefore there would be no card left to make the box with.
However this means that my formula I previously found was nearly correct, X = L/4.43. So if I divide the length of any rectangle with the ratio 1:3 by the 4.43 it will give me the cut off that will leave the maximum volume.
Testing
Size of card – 6cm by 12cm
Prediction
Using the formula that I worked out using the two methods above I will predict what the answer should be to see if it matches the actual answer. If it does then my formula is correct. If not then it is wrong.
X = L/4.43
X = 6/4.43
X = 1.354….
Working out
I will now do all the working out to see if my prediction and formula were right.
X must be 0<X<3: This is because if X is 0 there would not be a side to fold and if X is 3 then there would be nothing left.
X = 1 X = 2
V = (3x6-2x1) (6-2x1) x1 V = (3x6-2x2) (6-2x2) x2
V = 64 V = 56
X = 3
Not possible due to the reasons mentioned above. If four was cut from each corner then there would be nothing left to make the sides with.
The cut off that gives the largest volume is 1. I will now look for the cut off that will leave the largest volume for the box. I will be looking for the cut off to 1d.p. between 0.5 – 1.5
X = 0.5
V = (3x6-2x0.5) (6-2x0.5) x0.5
V = 42.5
X = 0.6
V = (3x6-2x0.6) (6-2x0.6) x0.6
V = 48.384
X = 0.7
V = (3x6-2x0.7) (6-2x0.7) x0.7
V = 53.452
X = 0.8
V = (3x6-2x0.8) (6-2x0.8) x0.8
V = 57.728
X = 0.9
V = (3x6-2x0.9) (7-2x0.9) x0.9
V = 61.236
X = 1.1
V = (3x6-2x1.1) (6-2x1.1) x1.1
V = 66.044
X = 1.2
V = (3x6-2x1.2) (6-2x1.2) x1.2
V = 67.392
X = 1.3
V = (3x6-2x1.3) (6-2x1.3) x1.3
V = 68.068
X = 1.4
V = (3x6-2x1.4) (6-2x1.4) x1.4
V = 68.096
X = 1.5
V = (3x6-2x1.5) (6-2x1.5) x1.5
V = 67.5
The cut off that will leave the largest volume so far is 1.4 I will now look for the cut off to 2d.p. between 1.35 – 1.45
X = 1.35
V = (3x6-2x1.35) (6-2x1.35) x1.35
V = 68.1615
X = 1.36
V = (3x6-2x1.36) (6-2x1.36) x1.36
V = 68.161024
X = 1.37
V = (3x6-2x1.37) (6-2x1.37) x1.37
V = 68.154212
X = 1.38
V = (3x6-2x1.38) (6-2x1.38) x1.38
V = 68.141088
X = 1.39
V = (3x6-2x1.39) (6-2x1.39) x1.39
V = 68.121676
X = 1.41
V = (3x6-2x1.41) (6-2x1.41) x1.41
V = 68.064084
X = 1.42
V = (3x6-2x1.42) (6-2x1.42) x1.42
V = 68.025952
X = 1.43
V = (3x6-2x1.43) (6-2x1.43) x1.43
V = 67.981628
X = 1.44
V = (3x6-2x1.44) (6-2x1.44) x1.44
V = 67.931136
X = 1.45
V = (3x6-2x1.45) (6-2x1.45) x1.45
V = 67.8745
This means that the value X = 1.35 gives the largest volume. This also means that me prediction was right and therefore the formula I used was correct. However this also proves that the two methods I used before differentiation were wrong and unreliable compared with differentiation.
Size of card – 1:4
Size of width 7
X must be 0<X<3.5: This is because if X is 0 there would not be a side to fold and if X is 3.5 then there would be nothing left.
X = 1 X = 2 X = 3
V = (4x7-2x1) (7-2x1) x1 V = (4x7-2x2) (7-2x2) x2 V = (4x7-2x3) (7-2x3) x3
V = 130 V = 144 V = 66
Graph
To show the cut off compared to the volume.
The cut off that gives the largest volume is 2. I will now look for the cut off that will leave the largest volume for the box. I will be looking for the cut off to 1d.p. between 1.5 – 2.5.
X = 1.5
V = (4x7-2x1.5) (7-2x1.5) x1.5
V = 150
X = 1.6
V = (4x7-2x1.6) (7-2x1.6) x1.6
V = 150.784
X = 1.7
V = (4x7-2x1.7) (7-2x1.7) x1.7
V = 150.552
X = 1.8
V = (4x7-2x1.8) (7-2x1.8) x1.8
V = 149.328
X = 1.9
V = (4x7-2x1.9) (7-2x1.9) x1.9
V = 147.136
X = 2.1
V = (4x7-2x0.9) (7-2x0.9) x0.9
V = 139.944
X = 2.2
V = (4x7-2x2.2) (7-2x2.2) x2.2
V = 134.992
X = 2.3
V = (4x7-2x2.3) (7-2x2.3) x2.3
V = 129.168
X = 2.4
V = (4x7-2x2.4) (7-2x2.4) x2.4
V = 122.496
X = 2.5
V = (4x7-2x2.5) (7-2x2.5) x2.5
V = 115
The cut off that will leave the largest volume so far is 1.6 I will now look for the cut off to 2d.p. between 1.55 – 1.65
X = 1.55
V = (4x7-2x1.55) (7-2x1.55) x1.55
V = 150.5205
X = 1.56
V = (4x7-2x1.56) (7-2x1.56) x1.56
V = 150.593664
X = 1.57
V = (4x7-2x1.57) (7-2x1.57) x1.57
V = 150.656572
X = 1.58
V = (4x7-2x1.58) (7-2x1.58) x1.58
V = 150.709248
X = 1.59
V = (4x7-2x1.59) (7-2x1.59) x1.59
V = 150.751716
X = 1.61
V = (4x7-2x1.61) (7-2x1.61) x1.61
V = 150.806124
X = 1.62
V = (4x7-2x1.62) (7-2x1.62) x1.62
V = 150.818112
X = 1.63
V = (4x7-2x1.63) (7-2x1.63) x1.63
V = 150.819988
X = 1.64
V = (4x7-2x1.64) (7-2x1.64) x1.64
V = 150.811776
X = 1.65
V = (4x7-2x1.65) (7-2x1.65) x1.65
V = 150.7935
This means that the value X = 1.63 give the largest volume.
Size of width 8
X must be 0<X<4: This is because if X is 0 there would not be a side to fold and if X is 4 then there would be nothing left.
X = 1 X = 2 X = 3
V = (4x8-2x1) (8-2x1) x1 V = (4x8-2x2) (8-2x2) x2 V = (4x8-2x3) (8-2x3) x3
V = 180 V = 224 V = 156
X = 4
Not possible due to the reasons mentioned above. If four was cut from each corner then there would be nothing left to make the sides with.
Graph
To show the cut off compared to the volume.
The cut off that gives the largest volume is 2. I will now look for the cut off that will leave the largest volume for the box. I will be looking for the cut off to 1d.p. between 1.5 – 2.5
X = 1.5
V = (4x8-2x1.5) (8-2x1.5) x1.5
V = 217.5
X = 1.6
V = (4x8-2x1.6) (8-2x1.6) x1.6
V = 221.184
X = 1.7
V = (4x8-2x1.7) (8-2x1.7) x1.7
V = 223.652
X = 1.8
V = (4x8-2x1.8) (8-2x1.8) x1.8
V = 224.928
X = 1.9
V = (4x8-2x1.9) (8-2x1.9) x1.9
V = 225.036
X = 2.1
V = (4x8-2x2.1) (8-2x2.1) x2.1
V = 221.844
X = 2.2
V = (4x8-2x2.2) (8-2x2.2) x2.2
V = 218.592
X = 2.3
V = (4x8-2x2.3) (8-2x2.3) x2.3
V = 214.268
X = 2.4
V = (4x8-2x2.4) (8-2x2.4) x2.4
V = 208.896
X = 2.5
V = (4x8-2x2.5) (8-2x2.5) x2.5
V = 202.5
The cut off that will leave the largest volume so far is 1.9. I will now look for the cut off to 2d.p. between 1.85 – 1.95
X = 1.85
V = (4x8-2x1.85) (8-2x1.85) x1.85
V = 225.1265
X = 1.86
V = (4x8-2x1.86) (8-2x1.86) x1.86
V = 225.131424
X = 1.87
V = (4x8-2x1.87) (8-2x1.87) x1.87
V = 225.124812
X = 1.88
V = (4x8-2x1.88) (8-2x1.88) x1.88
V = 225.106688
X = 1.89
V = (4x8-2x1.89) (8-2x1.89) x1.89
V = 225.077076
X = 1.91
V = (4x8-2x1.91) (8-2x1.91) x1.91
V = 224.983484
X = 1.92
V = (4x8-2x1.92) (8-2x1.92) x1.92
V = 224.919552
X = 1.93
V = (4x8-2x1.93) (8-2x1.93) x1.93
V = 224.844228
X = 1.94
V = (4x8-2x1.94) (8-2x1.94) x1.94
V = 224.757536
X = 1.95
V = (4x8-2x1.95) (8-2x1.95) x1.95
V = 224.6595
This means that the value X = 1.86 give the largest volume.
Size of width 9
X must be 0<X<4.5: This is because if X is 0 there would not be a side to fold and if X is 4.5 then there would be nothing left.
X = 1 X = 2 X = 3
V = (4x9-2x1) (9-2x1) x1 V = (4x9-2x2) (9-2x2) x2 V = (4x9-2x3) (9-2x3) x3
V = 238 V = 320 V = 270
X = 4
V = (4x9-2x3) (9-2x3) x3
V = 112
Graph
To show the cut off compared to the volume.
The cut off that gives the largest volume is 2. I will now look for the cut off that will leave the largest volume for the box. I will be looking for the cut off to 1d.p. between 1.5 – 2.5
X = 1.5
V = (4x9-2x1.5) (9-2x1.5) x1.5
V = 297
X = 1.6
V = (4x9-2x1.6) (9-2x1.6) x1.6
V = 304.384
X = 1.7
V = (4x9-2x1.7) (9-2x1.7) x1.7
V = 310.352
X = 1.8
V = (4x9-2x1.8) (9-2x1.8) x1.8
V = 314.928
X = 1.9
V = (4x9-2x1.9) (9-2x1.9) x1.9
V = 318.136
X = 2.1
V = (4x9-2x2.1) (9-2x2.1) x2.1
V = 320.544
X = 2.2
V = (4x9-2x2.2) (9-2x2.2) x2.2
V = 319.792
X = 2.3
V = (4x9-2x2.3) (9-2x2.3) x2.3
V = 317.768
X = 2.4
V = (4x9-2x2.4) (9-2x2.4) x2.4
V = 314.496
X = 2.5
V = (4x9-2x2.5) (9-2x2.5) x2.5
V = 310
The cut off that will leave the largest volume so far is 2.1. I will now look for the cut off to 2d.p. between 2.05 – 2.15.
X = 2.05
V = (4x9-2x2.05) (9-2x2.05) x2.05
V = 320.4355
X = 2.06
V = (4x9-2x2.06) (9-2x2.06) x2.06
V = 320.483264
X = 2.07
V = (4x9-2x2.07) (9-2x2.07) x2.07
V = 320.517972
X = 2.08
V = (4x9-2x2.08) (9-2x2.08) x2.08
V = 320.539648
X = 2.09
V = (4x9-2x2.09) (9-2x2.09) x2.09
V = 320.548316
X = 2.11
V = (4x9-2x2.11) (9-2x2.11) x2.11
V = 320.526724
X = 2.12
V = (4x9-2x2.12) (9-2x2.12) x2.12
V = 320.496512
X = 2.13
V = (4x9-2x2.13) (9-2x2.13) x2.13
V = 320.453388
X = 2.14
V = (4x9-2x2.14) (9-2x2.14) x2.14
V = 320.397376
X = 2.15
V = (4x9-2x2.15) (9-2x2.15) x2.15
V = 320.3285
This means that the value X = 2.09 gives the largest volume.
Highest values of the cut off
I will put all the cut off values that give the maximum volume in table to find a general formula.
Total of the last column : 12.90177389
Average (total/amount of numbers) : 4.300591298
The answer to 2d.p. : 4.3
Therefore if we take the width of any rectangular card and divide it by 4.3 we get the rough cut off that would leave the highest volume.
E.g.
If the size of card is 8 by 32cm.
Then the size of cut off would be 8/4.3 = 1.86(2.d.p)
Gradient
I will plot a graph and find the gradient of the line.
Graph
(See appendix 4)
Gradient = L/><
L = 8 – 7 = 1
>< = 1.86 – 1.63 = 0.23
Gradient = 1/0.23
= 4.347826
= 4.35 (2d.p.)
Both of the above formulas say different answers so I will now use differentiation to find an exact formula that will give the size of cut off that will leave the largest volume.
Differentiation
I will use differentiation to find the formula.
Differentiation is used to find the maximum and minimum of a point on a curve, of the gradient of a curve.
The gradient should always be equal to 0 (see below)
V = Xn This is the formula for differentiation.
dV/dX This is the symbol for differentiation.
dV/dX = nX n – 1 Multiply the power by the coefficient of X (the n umber in front of X), then minus 1 from the power.
Therefore firstly you multiply out the brackets:
V = (4L-2X) (L-2X) X
= (4L 2 – 8LX – 2LX + 4X 2 ) X
= 4L 2 X – 10LX 2 + 4X 3
dV/dX = 4L 2 – 20LX + 12X 2 = 0
To find the maximum dV/dX = 0
Using the quadratic formula I will solve the equation.
-b+/_√b 2 – 4ac
The quadratic formula
2a
aX 2 + bX + c = 0 Formula to decide what a, b and c stand for.
12X 2 – 20LX + 4L 2 = 0
Therefore:
a = 12 b = -20L c = 4L 2
+20L+/_√ (-20L)2 – 4 x 12 x 4L2
Substituting a, b and c into the formula
2 x 12
+20L+/_√400L2 – 192L2
Simplifying the formula
24
+20L+/_√208L2
Simplifying the formula
24
+20L+/_14.422…L
Simplifying the formula
24
This now means that x can be either one of two values:
X = 20L + 14.422…L X = 20L – 14.422…L
Or
24 24
= 34.42220…L = 5.577949…L
24 24
= 1.43425…L = 0.23241…L
= 1 = 1
The reciprocal
1.43425L 0.23241L
X = 0.698L X = 4.302L
X = L X = L
The reciprocal
0.698 4.302
This cannot be the formula because L/0.698 = >L/2, this means that you will be cutting off more card than there actually is and therefore there would be no card left to make the box with.
However this means that my formula I previously found was incorrect. So instead if I divide the length of any rectangle with the ratio 1:4 by the 4.3 it will give me the cut off that will leave the maximum volume.
Testing
Size of card – 6cm by 32
Prediction
Using the formula that I worked out using differentiation above I will predict what the answer should be to see if it matches the actual answer. If it does then my formula is correct. If not then it is wrong.
X = L/4.3
X = 6/4.3
X = 1.3953488….
Working out
I will now do all the working out to see if my prediction and formula were right.
X must be 0<X<3: This is because if X is 0 there would not be a side to fold and if X is 3 then there would be nothing left.
X = 1 X = 2
V = (4x6-2x1) (6-2x1) x1 V = (4x6-2x2) (6-2x2) x2
V = 88 V = 80
X = 3
Not possible due to the reasons mentioned above. If four was cut from each corner then there would be nothing left to make the sides with.
The cut off that gives the largest volume is 1. I will now look for the cut off that will leave the largest volume for the box. I will be looking for the cut off to 1d.p. between 0.5 – 1.5
X = 0.5
V = (4x6-2x0.5) (6-2x0.5) x0.5
V = 57.5
X = 0.6
V = (4x6-2x0.6) (6-2x0.6) x0.6
V = 65.664
X = 0.7
V = (4x6-2x0.7) (6-2x0.7) x0.7
V = 72.772
X = 0.8
V = (4x6-2x0.8) (6-2x0.8) x0.8
V = 78.848
X = 0.9
V = (4x6-2x0.9) (7-2x0.9) x0.9
V = 83.916
X = 1.1
V = (4x6-2x1.1) (6-2x1.1) x1.1
V = 91.124
X = 1.2
V = (4x6-2x1.2) (6-2x1.2) x1.2
V = 93.312
X = 1.3
V = (4x6-2x1.3) (6-2x1.3) x1.3
V = 94.588
X = 1.4
V = (4x6-2x1.4) (6-2x1.4) x1.4
V = 94.976
X = 1.5
V = (4x6-2x1.5) (6-2x1.5) x1.5
V = 94.5
The cut off that will leave the largest volume so far is 1.4 I will now look for the cut off to 2d.p. between 1.35 – 1.45
X = 1.35
V = (4x6-2x1.35) (6-2x1.35) x1.35
V = 94.8915
X = 1.36
V = (4x6-2x1.36) (6-2x1.36) x1.36
V = 94.925824
X = 1.37
V = (4x6-2x1.37) (6-2x1.37) x1.37
V = 94.951412
X = 1.38
V = (4x6-2x1.38) (6-2x1.38) x1.38
V = 94.968288
X = 1.39
V = (4x6-2x1.39) (6-2x1.39) x1.39
V = 94.976476
X = 1.41
V = (4x6-2x1.41) (6-2x1.41) x1.41
V = 94.966884
X = 1.42
V = (4x6-2x1.42) (6-2x1.42) x1.42
V = 94.949152
X = 1.43
V = (4x6-2x1.43) (6-2x1.43) x1.43
V = 94.922828
X = 1.44
V = (4x6-2x1.44) (6-2x1.44) x1.44
V = 94.887936
X = 1.45
V = (4x6-2x1.45) (6-2x1.45) x1.45
V = 94.8445
This means that the value X = 1.39 gives the largest volume. This also means that me prediction was right and therefore the formula I used was correct. However to prove that it works for any number I will use differentiation.
Size of card – 1: n
I will now use differentiation to find a formula that would work with any size of card given.
Differentiation
Differentiation is used to find the maximum and minimum of a point on a curve, of the gradient of a curve.
The gradient should always be equal to 0 (see below)
V = Xn This is the formula for differentiation.
dV/dX This is the symbol for differentiation.
dV/dX = nX n – 1 Multiply the power by the coefficient of X (the n umber in front of X), then minus 1 from the power.
Therefore firstly you multiply out the brackets:
V = (L-2X) (nL-2X) X
= (nL 2 – 2nLX – 2LX + 4X 2) X
= nL 2 X – 2nLX2 – 2LX2 + 4X 3
dV/dX = nL 2 – 4nLX – 4LX + 12X 2 = 0
To find the maximum dV/dX = 0
Using the quadratic formula I will solve the equation.
-b+/_√b 2 – 4ac
The quadratic formula
2a
aX 2 + bX + c = 0 Formula to decide what a, b and c stand for.
12X 2 – (4nL + 4L) X + nL 2 = 0
Therefore:
a = 12 b = - (4nL + 4L) c = nL 2
+ (4nL + 4L) +/_√ (-(4nL + 4L))2 – 4 x 12 x nL2
2 x 12 Substituting a, b and c into the formula
+ (4nL + 4L) +/_√16L2(1 + n2 – n)
Simplifying the formula
24
+4L (1 + n) +/_4L√ (n2 – n +1)
Simplifying the formula
24
+4(L (1 + n) +/_L√n2 – n +1
Simplifying the formula
24 6
This now means that the formula that will give us the cut off that will leave the largest volume for any card is:
L (1 + n) +/- L√n2 – n +1
6
Through this we can prove that l/6 give the highest volume of any size of square card.
Testing n = 1 (testing for a square ratio 1:1)
L (1 + 1) +/-L√12 – 1 +1 Simplifying the formula
6
2L+/-L Simplifying the formula
6
This now means that x can be either one of two values:
X = 2L + L X = 2L – L
Or
6 6
= 3L = L
6 6
= L This cannot be the formula because this means that you will be cutting off more card than there actually is and therefore there would be no card left to make the box with.
This therefore means that my previous formula of L/6 was correct and will give the value of the cut off that will leave the largest volume.
Meanings of some of the mathematical words and symbols used in this investigation
Differentiation: This a method of calculations used to find the maximum or minimum of a point of a curve, also known as the gradient of a curve.
Trial and improvement: This is a method used to find a solution to a problem. It involves estimating number that might give the expected results and repeating this until the solution is found.
<: This is the sign for less than.
E.g. 2 < 4 is the same as saying 2 is less than 4.
>: This is the sign for more than.
E.g. 6 > 3 is the same as saying 6 is more than 3.
s.f.: This is short of significant figure. A significant figure is the first digit (*not zero) that you can reach, reading left to right. When approximating to a given number of significant figures, we use rounding off on the last significant number.
E.g. 2.3456 to 3s.f. = 2.35 however 0.067891 to 3s.f. = 0.0679.
d.p.: This is short for decimal place. A decimal place is the amount of numbers that come after the decimal point.
E.g. 2.3456 to 3.d.p = 2.346.
Gradient: The gradient of a line is a measurement of how ‘steep’ it is. The gradient of a line can be found by drawing right – angled triangles on the line.
Coefficient: Coefficient is the number that comes before a variable.
E.g. the coefficient of 5Y = 5.
Quadratic equation: A quadratic equation in x is an equation involving x to the power 2 but no higher powers of x. the general form of a quadratic equation is: ax² + bx + c = 0
E.g. 8x² - 4x + 6² = 0
When a quadratic equation has solutions and the equation is written in the general form, the solutions cab be found using the formula:
-b+/_√b 2 – 4ac
2a
Reciprocal: The reciprocal of a number x is the number 1/x.
E.g. the reciprocal of 3.4567 is 1/3.4567. The reciprocal of a number is the same as the multiplicative inverse of the number.
