This test has proved that cut-outs of length 6cm will give the box with the biggest possible volume. I have proved this because when I took cut-outs of length 5.9cm and 6.1cm the volumes achieved were smaller, so therefore 6cm must be the optimum value.
Conclusion
This result has proved my hypothesis, in that to quickly work out what size the square cut-outs will need to be in order to make a box with the biggest volume you take the size of the sheet of square card you are using and divide it by 6. This can be simply put into the algebraic formula:
X = L
6
X represents that size of the square cut-outs while L represents the length of the square sheet of card.
36cm using algebra
I am now going to prove that this formula is accurate by using it to work out the required cut-out size for a piece of card of length 36cm. I know that the answer should come out at 6cm, so if it does then this formula must be correct and accurate.
V = L x B x H
= x(36 – 2x)(36 – 2x)
= x(1296 – 72x – 72x + 4x²)
= x(4x² - 144x +1296)
V = 4x³ - 144x² + 1296x
I am now going to find the gradient through the use of differentiation. At the biggest value the gradient should be 0.
Gradient = 12x² - 288x + 1296
= 12x² - 288x + 1296 = 0
12x² - 288x + 1296 = 0
12
x² - 24x + 108 = 0
(x – 18)(x – 6) = 0
Either x – 18 = 0 or x – 6 = 0
+18 +18 +6 +6
x = 18 x = 6
x cannot be equal to 18 because if both the cut-outs were 18cm then there would be no cart left so x must be equal to 6cm.
This result has shown that my formula is correct. However, to make sure I am also going to test the formula on square sheets of card 24cm and 30cm in length.
24cm using algebra
I am now going to test my formula using a sheet of card 24cm square.
V = L x B x H
= x(24 – 2x)(24 – 2x)
= x(576 – 48x – 48x +4x²)
= x(4x² - 96x + 576)
V = 4x³ - 96x² + 576x
I am now going to find the gradient using differentiation. The gradient should be 0 at the biggest value.
Gradient = 12x² - 192x + 576
= 12x² - 192x + 576 = 0
12x² - 192x + 576 = 0
12
x² - 16x + 48 = 0
(x – 4)(x – 12) = 0
Either x – 12 = 0 or x – 4 = 0
+12 +12 +4 +4
x = 12 x = 4
12 cannot be equal to x because two cut-outs of 12cm would mean there was no card left so therefore x must be equal to 4cm.
30cm using algebra
I am now going to test my formula using a sheet of card 30cm square.
V = L x B x H
= x(30 – 2x)(30 – 2x)
= x(900 – 60x – 60x + 4x²)
= x(4x² - 120x + 900)
V = 4x³ - 120x² +900x
I am now going to differentiate to find the gradient because the gradient will be 0 at the biggest volume.
Gradient = 12x² - 240x + 900
= 12x² - 240x + 900 = 0
12x² - 240x + 900 = 0
12
x² - 20x + 75 = 0
(x – 5)(x – 15) = 0
Either x – 5 = 0 or x – 15 = 0
+5 +5 +15 +15
x = 5 x = 15
x cannot be equal to 15cm because two cut-outs of 15cm would mean there was no card left so x must be equal to 5.
Calculus and Algebra
I am now going to prove, through the use of calculus and algebra, that in the formula x = L
6
That x stands for the size of each of the four cu-outs and that L stands for the length of one side of the card.
V = L x B x H
= x(L – 2x)(L – 2x)
= x(L² - 2Lx – 2Lx + 4x²)
= x(L² - 4Lx + 4x²)
V = 4x³ - 4Lx² + L²x
At the biggest volume the gradient is always equal to 0 so I am going to find the gradient using differentiation and put it equal to 0.
Gradient = 12x² - 8Lx + L²
= 12x² - 8Lx + L² = 0
I am now going to factorise this formula.
-12Lx – Lx = -13Lx
-3Lx – 4Lx = -7Lx
-6Lx – 2Lx = -8Lx
(6x – L)(2x – L) = 0
Either 6x – L = 0 or 2x – L = 0
+L +L +L +L
6x = L 2x = L
6 6 2 2
x = L x = L
6 2
L
2 cannot be right because if you cut off half the card twice then there will be no card left so L must be the correct rule.
6
I have proved using algebra and calculus that the cut-out required to achieve the biggest volume is the length of one side of the card divided by 6.
Rectangle
Length 2 times Width
I am now going to investigate the size of the cut-out square which will make an open box of the biggest volume for a rectangular piece of card.
I am first going to use a rectangle where the length is twice the size of the width. This rectangle has a length of 20 cm and a width of 10cm.
V = L x B x H
= x(20 – 2x)(10 – 2x)
= x(200 – 40x – 20x +4x²)
= x(200 – 60x +4x²)
V = 4x³ - 60x² +200x
I am now going to differentiate to find the gradient which will be 0 at the biggest volume.
Gradient = 12x² - 120x +200
= 12x² - 120x + 200 = 0
12x² - 120x +200 = 0
4
3x² - 30x +50 = 0
-25x – 6x = -31x
-75x – 2x = -77x
-50x – 3x = -53x
-x – 150x = -151x
-30x – 5x = -35x
-15x – 10x = -25x
This doesn’t factorise normally so I will have to make use of the quadratic formula to factorise it instead.
x = -b b² - 4ac
2a
a = 3
b = -30
c = 50
4ac = 600
x = --30 900 – 600
6
x = 30 300
6
Either x = 7.89 or x = 2.11
You cannot cut 7.89cm out twice because there isn’t enough card so the answer must be 2.11cm.
30cm by 15cm
I am now going to use a rectangle which has a length of 30cm and a width of 15cm.
V = L x B x H
= x(30 - 2x)(15 – 2x)
= x(450 – 60x – 30x + 4x²)
= x(4x² - 90x + 450)
V = 4x³ - 90x² + 450x
I am now going to differentiate to find the gradient which will be 0 at the biggest volume.
Gradient = 12x² - 180x + 450
= 12x² - 180x + 450 = 0
12x² - 180x + 450 = 0
6
2x² - 30x +75 = 0
-50x – 3x = -53x
-6x – 25x = -31x
This formula will not factorise in the normal way so I will have to use the quadratic formula to factorise it instead.
x = -b b² - 4ac
2a
a = 2
b = -30
c = 75
4ac = 600
x = --30 900 – 600
4
x = 30 300
4
Either x = 11.83 or x = 3.17
You cannot cut out 11.83cm twice from this sheet of card because there would be no card left so the answer must be 3.17cm.
10cm by 5cm
I am now going to use a rectangle that has a length of 10cm and a width of 5cm.
V = L x B x H
= x(10 – 2x)(5 – 2x)
= x(50 – 2x – 10x + 4x²)
= x(4x² - 30x + 50)
V = 4x³ - 30x² + 50x
I am now going to differentiate to find the gradient which will be 0 at the biggest volume.
Gradient = 12x² - 60x +50
= 12x² - 60x +50 = 0
12x² - 60x + 50 = 0
2
6x² - 30x + 25 = 0
This formula will not factorise normally so I will have to use the quadratic formula to factorise it instead.
x = -b b² - 4ac
2a
a = 6
b = -30
c = 25
4ac = 600
x = --30 900 – 600
12
X = 30 300
12
Either x = 3.94 or x = 1.06
x cannot be equal to 3.94cm because two cut-outs of that size would mean that there was no card left so x must be equal to 1.06cm.
I am now going to tabulate my results to see if there is any sort of pattern beginning to emerge.
There is a pattern emerging. If you divide the length by approximately 5 then you get the size of the cut-outs required to achieve the biggest volume with that particular piece of card.
This can be summed up in the formula: x W
5
Algebra and Calculus
I am now going to prove, through the use of algebra and calculus that this formula is correct and accurate.
V = L x B x H
= x(2w – 2x)(w – 2x)
= x(2w² - 4wx – 2wx + 4x²)
= x(4x² - 6wx + 2w²)
V = 4x³ - 6wx² + 2w²x
I am now going to differentiate to find the gradient which will be 0 at the biggest value.
Gradient = 12x² - 12wx + 2w²
= 12x² - 12wx + 2w² = 0
12x² - 12wx + 2w² = 0
2
6x² - 6wx + w² = 0
This formula will not factorise in the normal way so I am going to have to use the quadratic formula to factorise it instead.
x = -b b² - 4ac
2a
a = 6
b = -6w
c = w²
4ac = 24w²
x = --6w 36w² - 24w²
12
x = 6w 12w²
12
12 x w²
12 x w
12 w
x = 6w 12 w
12
Either x = 9.464101615w or x = 2.535898385w
- 12
x is always equal to the value you get when you subtract b² - 4ac from –b in the quadratic formula. Therefore x must be equal to 2.535898385w
12
x = w
4.732
Length 3 times Width
I am now going to consider a rectangle which has a length 3 times the size of the width.
V = L x B x H
= x(3w – 2x)(w – 2x)
= x(3w² - 6wx – 2wx +4x²)
= x(4x² - 8wx +3w²)
V = 4x³ - 8wx² + 3w²x
I am now going to differentiate to find the gradient which will be 0 at the biggest value.
Gradient = 12x² - 16wx + 3w²
= 12x² - 16wx + 3w² = 0
This formula will not factorise in the normal way so I will have to use the quadratic formula instead.
x = -b b² - 4ac
2a
a = 12
b = -16w
c = 3w²
4ac = 144w²
x = --16w 256w² - 144w²
24
x = 16w 112w²
24
112 x w²
112 x w
112 w
x = 16w 112 w
24
Either x = 26.58300524w or x = 5.416994756w
- 24
x = w
4.431
Length 4 times Width
I am now going to consider a rectangle which has a length 4 times bigger than the width.
V = L x B x H
= x(4w – 2x)(w – 2x)
= x(4w² - 8wx – 2wx + 4x²)
= x(4x² - 10wx + 4w²)
V = 4x³ - 10wx² + 4w²x
I am now going to find the gradient using differentiation. The gradient will be 0 at the biggest volume.
Gradient = 12x² - 20wx + 4w²
= 12x² - 20wx + 4w² = 0
This formula will not factorise in the normal way so I am going to have to use the quadratic formula to factorise it instead.
x = -b b² - 4ac
2a
a = 12
b = -20w
c = 4w²
4ac = 192w²
x = --20w 400w² - 192w²
24
x = 20w 208w²
24
208 x w²
208 x w
208 w
x = 20w 208 w
24
I am going to use the answer that I got when I took the square root of 208 from 20 because the other answer is always too much.
x = 5.577794898w
24
x = w
4.303
Length 5 times Width
I am now going to consider a rectangle which has a length 5 times bigger than the width.
V = L x B x H
= x(5w – 2x)(w – 2x)
= x(5w² - 10wx – 2wx + 4x²)
= x(5w² - 12wx + 4 x²)
V = 4x² - 12wx² + 5w²x
I am now going to differentiate to find the gradient which will be 0 at the biggest value.
Gradient = 12x² - 24wx + 5w²
= 12x² - 24wx + 5w² = 0
This formula will not factorise in the normal way so I will have to use the quadratic formula to factorise it instead.
x = -b b² - 4ac
2a
a = 12
b = -24w
c = 5w²
4ac = 240w²
x = --24w 576w² - 240w²
24
x = 24w 336w²
24
336 x w²
336 x w
336 w
X = 24w 336 w
24
Either x = 42.33030278w or x = 5.66969722w
- 24
x = w
4.233
I am now going to put my results into a table to see if there is a pattern beginning to emerge.
There is pattern beginning to emerge here. All the numbers begin with 4 and the denominator gradually gets smaller each time.
I believe that the denominator will never be less than 4 no matter what the size of the rectangle is. I am going to test my hypothesis using a rectangle which has a length 145 times bigger than the width.
Length 145 times Width
V = L x B x H
= x(145w – 2x)(w – 2x)
= x(145w² - 290wx – 2wx + 4x²)
= x(4x² - 292wx + 145w²)
V = 4x³ - 292wx² + 145w²x
I am now going to differentiate to find the gradient which will be 0 at the biggest value.
Gradient = 12x² - 584w +145w²
= 12x² - 584w +145w² = 0
This formula will not factorise in the normal way so I am going to have to use the quadratic formula to factorise it instead.
x = -b b² - 4ac
2a
a = 12
b = -584w
c = 145w²
4ac = 6960w²
x = --584w 341056w² - 6960w²
24
x = 584w 334096w²
24
334096 x w²
334096 x w
334096 w
x = 584w 334096 w
24
x = 5.98961947w
24
x = w
4.007
This result leads me to believe that my hypothesis was correct and that the denominator will never be lower than 4.
Length 1.5 times Width
I am now going to consider a rectangle which has a length 1.5 times bigger than the width.
V = L x B x H
= x(1.5w – 2x)(w -2x)
= x(1.5w² = 3wx = 2wx + 4x²)
= x(4x² - 5wx +1.5w²)
V = 4x³ - 5wx² + 1.5w²x
I am now going to differentiate to find the gradient which will be 0 at the biggest value.
Gradient = 12x² - 10wx + 1.5w²
= 12x² - 10wx + 1.5w² = 0
This formula will not factorise in the normal way so I am going to have to use the quadratic formula to factorise it instead.
x = -b b² - 4ac
2a
a = 12
b = -10w
c = 1.5w²
4ac = 72w²
x = --10w 100w² - 72w²
24
x = 10w 28w²
24
28 x w²
28 x w
28 w
x = 10w 28 w
24
x = 4.708497378w
24
x = w
5.097
Length 1.2 times Width
I am now going to investigate a rectangle which has a length 1.2 times bigger than its width.
V = L x B x H
= x(1.2w -2x)(w – 2x)
= x(1.2w² - 2.4wx – 2wx + 4x²)
= x(4x² - 4.4wx + 1.2w²)
V = 4x³ - 4.4wx² + 1.2w²x
I am now going to find the gradient through the use of differentiation. The gradient will be 0 at the biggest value.
Gradient = 12x² - 8.8wx + 1.2w²
= 12x² - 8.8wx +1.2w² = 0
This formula will not factorise in the normal way. This means that I am going to have to use factorisation to factorise it instead.
x = -b b² - 4ac
2a
a = 12
b = -8.8w
c = 1.2w²
4ac = 57.6w²
x = --8.8w 77.44w² - 57.6w²
24
x = 8.8w 19.84w²
24
19.84 x w²
19.84 x w
19.84 w
x = 4.34578851w
24
x = w
5.523
Length 1.1 times Width
I am now going to consider a rectangle which has a length 1.1 times bigger than the width.
V = L x B x H
= x(1.1w – 2x)(w – 2x)
= x(1.1w² - 2.2wx – 2wx +4x²)
= x(4x² - 4.2wx + 1.1w²)
V = 4x³ - 4.2wx² + 1.1w²x
I am now going to find the gradient through the use of differentiation. The gradient will be 0 at the biggest value.
Gradient = 12x² - 8.4wx + 1.1w²
= 12x² - 8.4wx + 1.1w² = 0
This formula will not factorise in the normal way. This means that I am going to have to use the quadratic formula to factorise it instead.
x = -b b² - 4ac
2a
a = 12
b = -8.4w
c = 1.1w²
4ac = 52.8w²
x = --8.4w 70.56w² - 52.8w²
24
x = 8.4w 17.76w²
24
17.76 x w²
17.76 x w
17.76 w
x = 8.4w 17.76 w
24
x = 4.214261501w
24
x = w
5.695
Conclusion
For a piece of card that is square in shape I came up with the formula:
x = L
6
x stands for the size of the cut-outs, while L stands for the length of one side of the piece of card.
After that I considered a rectangular piece of card which had a length that was twice the size of the width. For that I came up with the formula:
x w
5
I then perfected and refined this formula through the use of algebra and calculus. The new formula I came up with is:
x = w
4.752
I then considered a rectangular piece of card which had a length 3 times the size of the width. For this piece of card I came up with the formula:
x = w
4.431
After that I considered a rectangular piece of card which had a length 4 times bigger than the width. For this particular piece of card I came up with the formula:
x = w
4.303
Finally, I investigated a rectangular piece of card which had a length 5 times bigger than the width. For this piece of card I came up with the formula:
x = w
4.233
I then came up with the hypothesis that the denominator, that is the number on the bottom of the formula, will never be less than four no matter what the dimensions of the rectangle are. I decided to test this hypothesis by taking a rectangle which had a length 145 times bigger than the width. This investigation resulted in the formula:
x = w
4.007
This result proved beyond reasonable doubt that my hypothesis was correct and accurate.
I then decided to look at rectangles where the length was less than 2 times the size of the width to investigate what happened to the denominator.
The first rectangle I decided to investigate was one where the length was 1.5 times bigger than the width. The formula this investigation resulted in was:
x = w
5.097
I then investigated a rectangle where the length was 1.2 times bigger than the width. This resulted in the formula:
x = w
5.523
Finally, I investigated a rectangle where the length was 1.1 times bigger than the width. This investigation resulted in the formula:
x = w
5.695
This has led me to conclude that as the ratio of length to width gets smaller, the rectangle becomes closer and closer in its dimensions to a square. Therefore, the denominator will become closer and closer to 6, which is the denominator for a square.
