The maximum volume I found in this table was at 3.33cm with its volume at 592.59215cm
Size of square cut = 30 by 30 (cm)
From this result I can tell the highest volume is the length cut is from 5cm – 6cm. I am going to find out where the highest volume by using trail and improvement and breaking the two numbers down into 1 decimal place.
The table shows us that as the cut is increased so did the volume, but when it reached 5cm, the volume decreased. Therefore the maximum volume was at 5cm where it had reached 2000cm. I should put the table in 2 decimal place to find an accurate result.
The maximum volume I found in this table was at 5cm with its volume at 2000cm
Size of square cut = 40 by 40 (cm)
From this result I can tell the highest volume is the length cut is from 6cm – 7cm. I am going to find out where the highest volume by using trail and improvement and breaking the two numbers down into 1 decimal place.
The table shows us that as the cut is increased so did the volume, but when it reached 6.7cm, the volume decreased. Therefore the maximum volume was at 6.7cm where it had reached 4740.652cm. I should put the table in 2 decimal place to find an accurate result.
The maximum volume I found in this table was at 6.67cm with its volume at 4740.7399cm
I made a table below to show the maximum volume of each square more clearly.
I created another table to find the relationship between the 4 squares.
From each of the four squares I found out the relationship from each of them with the highest volume divided by the same size. I found out at the relationship is 1/6.
Differentiation
I am checking my results using differentiation. Differentiation is a useful way of finding the gradient at any point. If y = xn then dy/dx = nxn-1.
If the relationship is 1/6 then I choose a square divisible by 6. In this case I choose a 6 by 6 square.
Volume = y
Volume = (6 - 2x)2x
= (36 + 4x2 - 24x) x
= 36 + 4x3 - 24x2
Then dy/dx = 36 + 12x2 - 48x
Factorising = 12 (x2- 4x +3) -3 * -1 = 3
-3 + -1 = -4
= 12((x - 1) (x - 3)) = 0
x = 1 or x = 3
x = 3 isn’t possible because if u cut 3cm from the side it would cut the entire square leaving nothing.
If I differentiate again I can tell if the gradient at the turning point is a maximum or minimum.
dy/dx if I differentiate again it will be D dy/dx / dx = d2y/dx2
If dy/dx = 12x2 – 48x + 36 d2y/dx2 = 24x – 48
So d2y/dx2 < 0 so it is at its maximum when x = 1
(24 * 1) – 48
= 24 – 48 = -24
I choose another square divisible by 6. I choose a 12 by 12 square.
Volume = y
Volume = (12 - 2x)2x
= (144 + 4x2 - 48x) x
= 144 + 4x3 - 48x2
Then dy/dx = 144 + 12x2 - 96x
Factorising = 12 (x2- 8x +12) -6 * -2 = 12
-6 + -2 = -8
= 12((x - 2) (x - 6)) = 0
x = 2 or x = 6
x = 6 isn’t possible because if u cut 6cm from the side it would cut the entire square leaving nothing.
If I differentiate again I can tell if the gradient at the turning point is a maximum or minimum.
dy/dx if I differentiate again it will be D dy/dx / dx = d2y/dx2
If dy/dx = 12x2 – 96x + 144 d2y/dx2 = 24x – 96
So d2y/dx2 < 0 so it is at its maximum when x = 2
(24 * 2) – 96
= 48 – 96 = -48
I choose another square divisible by 6. I choose an 18 by 18 square.
Volume = y
Volume = (18 - 2x)2x
= (324 + 4x2 - 72x) x
= 144 + 4x3 - 72x2
Then dy/dx = 324 + 12x2 - 144x
Factorising = 12 (x2- 12x +27) -9 * -3 = 27
-9 + -3 = -12
= 12((x - 3) (x - 9)) = 0
x = 3 or x = 9
x = 9 isn’t possible because if u cut 9cm from the side it would cut the entire square leaving nothing.
If I differentiate again I can tell if the gradient at the turning point is a maximum or minimum.
dy/dx if I differentiate again it will be D dy/dx / dx = d2y/dx2
If dy/dx = 12x2 – 144x + 324 d2y/dx2 = 24x – 144
So d2y/dx2 < 0 so it is at its maximum when x = 3
(24 * 3) – 144
= 72 – 144 = -72
Further Investigation
To find the length of the cut-out corner squares that gives the maximum open box volume for rectangular pieces of card of different sizes. The length of the cut-out corner squares will be to 3 significant figures of accuracy.
Method- I will investigate the size of the corner cut-out squares that give the largest open box volume for rectangular shaped pieces of card that have the width to length ratio of 1:2, those being 12 x 24, 24 x 48 and 48 x 96. I will then produce a formula for the maximum open box volume, for all rectangles that have width to length ratios of 1:2.
Size of square cut = 10 by 20 (cm)
From this result I can tell the highest volume is the length cut is from 2cm – 3cm. I am going to find out where the highest volume by using trail and improvement and breaking the two numbers down into 1 decimal place.
The table shows us that as the cut is increased so did the volume, but when it reached 2.1cm, the volume decreased. Therefore the maximum volume was at 2.1cm where it had reached 192.444cm. I should put the table in 2 decimal place to find an accurate result.
The maximum volume I found in this table was at 2.11cm with its volume at 192.4497cm
Size of square cut = 20 by 40 (cm)
From this result I can tell the highest volume is the length cut is from 4cm – 5cm. I am going to find out where the highest volume by using trail and improvement and breaking the two numbers down into 1 decimal place.
The table shows us that as the cut is increased so did the volume, but when it reached 4.2cm, the volume decreased. Therefore the maximum volume was at 4.2cm where it had reached 1539.552cm. I should put the table in 2 decimal place to find an accurate result.
The maximum volume I found in this table was at 6.34cm with its volume at 5196.152cm
Size of square cut = 30 by 60 (cm)
From this result I can tell the highest volume is the length cut is from 6cm – 7cm. I am going to find out where the highest volume by using trail and improvement and breaking the two numbers down into 1 decimal place.
The table shows us that as the cut is increased so did the volume, but when it reached 6.3cm, the volume decreased. Therefore the maximum volume was at 6.3cm where it had reached 5195.988cm. I should put the table in 2 decimal place to find an accurate result.
The maximum volume I found in this table was at 6.34cm with its volume at 5196.152cm
Size of square cut = 40 by 80 (cm)
From this result I can tell the highest volume is the length cut is from 8cm – 9cm. I am going to find out where the highest volume by using trail and improvement and breaking the two numbers down into 1 decimal place.
The table shows us that as the cut is increased so did the volume, but when it reached 8.5cm, the volume decreased. Therefore the maximum volume was at 8.5cm where it had reached 12316.5cm. I should put the table in 2 decimal place to find an accurate result.
The maximum volume I found in this table was at 8.46cm with its volume at 12316.8cm
I made a table below to show the maximum volume of each square more clearly.
I created another table to find the relationship between the 4 squares.
From each of the four squares I found out the relationship from each of them with the highest volume divided by the same size. I found out at the relationship is 1/5. I found the relationship by finding the difference of the Size of square cut divided by the Cut for volume maximum.
