This chart shows the volume compared to the 3 different sizes of original card (10cm², 20cm² and 30cm²) and the size of the squares cut out.
For each size of A (original piece of card) the volume increases to a maximum point decreases there after.
Firstly I will find a general rule for the volume in terms of X and A.
To find a general rule for the volume of any box, I will be using the following letters in place of numeric values.
- X = Length of the side of one of the small squares removed from each corner. This length becomes the height of the box.
- W = The width of one side of the original piece of card once the squares have been removed from each corner. This becomes the width of the box.
- L = The length of one side of the original piece of card once the squares have been removed from each corner. This becomes the length of the box.
- Y = In a square W = L. Y is used to represent both L and W as separate letters are not needed.
- A = The length of one side of the original square.
- V = Volume.
e.g., In this example the original piece of card is 10cm by 10cm.
X = 2
Y = 8
A = 10
10cm
8cm
10cm
2cm
2cm 8cm
Y = A – 2X
The volume of the box = X x Y x Y (length x width x height)
Substitute Y with A – 2X in the equation V = X x Y x Y.
X (A – 2X) ² = V
Check:
2 (10 – 2x2) ²
2(10 – 4) ²
2 (6) ²
2 x 36
72
I have produced a graph to show how the volume changes according to the size of the squares cut from each corner of the original. A 10cm x 10cm piece of card was used. 1.1>X>2.2 because I know from previous calculations that the maximum point lies between X = 1cm and X = 2cm
X = Length of one side of the squares cut out
A = length of one side of the original square before the corners are cut out.
This graph shows that the volume reaches its maximum when X = 1.6
I noted down the maximum volume for each size of original card:
A V
10 cm x 10cm = 1.6cm³
20 cm x 20 cm = 3.3cm³
30cm x 30cm = 5cm³
I divided A by V and each time the answer was 6. This means that X = A/6
AIM 2:
Aim two follows a similar procedure using rectangular pieces of card instead of squares.
A square is being cut from each corner. Aim 2 is to find out what fraction of the whole original rectangle needs to be cut from each corner to make the largest possible volume.
In the square investigation A was used in place of W and L because W = L. In a rectangle the length and width differ so A is not used.
The formula found for the square:
X (A – 2X) ² = V
Is also relevant for the rectangle but A is split into W and L so the formula becomes:
X (W – 2X) (L – 2X) = V
As no immediate connection between X and V is obvious other means of calculation are necessary.
Calculus is used to find the relationship between X and V using a graph to show the maximum volume. At the maximum volume the gradient or dV/dX = 0
In this example dV/dX = X (W – 2X) (L – 2X)
This formula need to be differentiated with respect to X as X is the value we are trying to find.
Before I differentiated I put L in terms of W so the formula is simpler and doesn’t contain too many different letters, which could be confusing.
The following example demonstrates the route to answering the aim. This process can be carried out using any size rectangle but I will be using a rectangle in which L = 2W.
Therefore X (W – 2X) (2W – 2X) will be differentiated with respect to X.
Before differentiation can occur the formula must be expanded then simplified - getting rid of the brackets.
EXPAND:
V = (WX – 2X²) (2W – 2X)
V = 2W²X – 2WX² - 4X²W + 4X³
SIMPLIFY:
2W²X – 6WX² + 4X³
DIFFRENTIATE:
dV/dX must equal 0 for the maximum volume.
dV/dX = 2W² - 12WX + 12X²
This can be simplified by dividing by 2
dV/dX = W² - 6WX + 6X²
I recognized this formula as quadratic so I used the general formula:
To find out what X equaled.
A = 6
B = - 6
C = W
In a rectangle 10cm x 20cm the size of X needed to make the maximum volume is 2.1.
The area of this rectangle is 200cm². 2.1² is cut out from each corner. 2.1² = 4.41
2.205% is cut out from each corner to make the largest possible volume of box.