As it can clearly be seen from the table above that biggest volume has found to be 37.92cm3 not 37.81cm3 as shown in the previous table. This isn’t the highest possible volume; therefore I am now going to use calculus to find the maximum possible volume.
V = x × ((8 - 2x) × (8 - 2x))
V = x × (64 - 16x - 16x + 4x2)
V = x × (64 - 32x + 4x2)
V = 64x - 32x2 + 4x3
If we rearrange this, it gives us:
4x3 - 32x2 + 64x
Now, to find the maximum volume I will differentiate the following to find the slope at where it’s at its maximum.
dv = 12x2 - 64x + 64
dx
= 12x2 - 64x + 64
3
= 3x2 – 16x + 16
The equation above tells us the slope of a curve at any given point, so now I am going to find out where the maximum volume lies on the curve.
Using the quadratic formula -b±√b2 – 4ac :
2a
= 16±√162 - 4(3) (16)
2(3)
= 16±√256 - 192
6
= 16±√64
6
= 16 + 8 OR = 16 - 8
6 6
= 4 = 4
3
The first value is out of range because it’s greater than half the width of the box. So the second value must give the maximum height for the volume.
Now I am going to substitute it back into the equation.
4 × ((8 - 2 × (4) × (8 - 2 × (4)))
3 3 3
= 4 × 28.4
3
= 37.9259 (4dp)
This is the highest possible volume for an 8 by 8cm square sheet of card. This value could have also been found using the first method, but it would have taken lot more time.
Part 2
I will be investigating the size of the cut out square which makes an open box of the largest volume for any rectangular square sheet of card.
The volume for any sized rectangular sheet of card will be gained by using the following formula.
Volume = height × area of base
V = x × ((L - 2x) × (W - 2x))
X = height of cut out square
L= length of rectangle
W = width of rectangle
I am going to use the following formula to work out the volume of an 10cm by 8cm rectangular sheet of card. I am going to keep the size of card the same while changing the height of the cut out square.
V = x × ((10 - 2x) × (8 - 2x))
From the table above it can clearly be seen that the biggest volume gained from a 10 by 8 piece of rectangular sheet is 52.5cm (2dp), this volume lies between the height of 1.25cm and 1.75cm. Therefore I am going to further investigate the height between these two points to gain the highest possible volume.
As it can clearly be seen from the table above that biggest volume has found to be 52.505cm3 (3dp) not 52.5cm3 as shown in the previous table. This isn’t the highest possible volume; therefore I am now going to use calculus to find the maximum possible volume.
V = x × ((10 - 2x) × (8 - 2x))
V = x × (80 - 20x - 16x + 4x2)
V = x × (80 - 36x + 4x2)
V = 80x - 36x2 + 4x3
If we rearrange this, it gives us:
4x3 - 36x2 + 80x
Now, to find the maximum volume I will differentiate the following to find the slope at where it’s at its maximum.
dv = 12x2 - 72x + 80
dx
= 12x2 - 72x + 80
4
= 3x2 – 18x + 20
The equation above tells us the slope of a curve at any given point, so now I am going to find out where the maximum volume lies on the curve.
Using the quadratic formula -b±√b2 - 4ac :
2a
= 18±√182 - 4(3) (20)
2(3)
= 18±√324 - 240
6
= 18±√80
6
= 18 +√80 OR = 18 -√80
6 6
= 4.49 (2dp) = 1.51 (2dp)
The first value is out of range because it’s greater than half the width of the box. So the second value must give the maximum height for the volume.
Now I am going to substitute it back into the equation.
1.51 × ((10 - 2 × (1.51) × (8 - 2 × (1.51)))
= 1.51 × ((6.98) × (4.98))
= 1.51 × 34.77
= 52.50 (2dp)
This is the highest possible volume for a 10 by 8cm rectangular sheet of card. This value could have also been found using the first method, but it would have taken lot more time.
