7cm by 3cm, piece of card
8cm by 3cm, piece of card
I have finished doing all of my table of results and graphs, now I will try my prediction to see if I was right, but if I wasn't I will try to find a rule which will work.
From the results from section two, I don't expect to find a simple rule to link the length with the shorter side of the piece of card, and there is no relationship between the longer side and the length of the section either.
The equation for the volume is :
Vol. = x ( a  2x ) ( b  2x )
= x ( ab  2ax  2bx + 4x2 )
= x ( ab  ( 2a + 2b ) x + 4x2 )
The area of the card, A = ab, and
the perimeter P = 2a + 2b.
Therefore Vol. = x ( A  Px + 4x2 )
This proves that the Volume depends on both the Area, and the Perimeter of the rectangle. This also means that the volume will also depend on the shape of the rectangle. I think that the ratio of the two sides may be important, and I will investigate this.
There are three cases to consider :
a is less than b, so a/b is less than one ( a/b < 1 )
a = b. This is a square I did in problem one ( a/b = 1 )
a is greater than b, so a/b is greater than one ( a/b > 1 ). This is a mirror of the first case.
I will now draw a graph of x against a/b, and b/a.
From the graph it appears that the section of the graph for b/a less than one, is approximately a straight line. I will try and find a rule that fits this section of the graph.
The straight line part of the graph appears to cross the axis at about x = 0.75.n I will check this by taking another rectangle with a value of b/a that is very small. I will use a rectangle that is 100 X 3, and test values close to 0.75 only.
This proves that the line would cross the axis at approximately 0.75, and that the maximum value for a/b would also be 0.75.
If the line for the first part of the graph is straight, then it will have a simple straight line equation. ( y = mx + c ).
For this line, the gradient m = 0.5  0.75 =  0.25 = 0.25
1.0  0 1
and from the table above, the constant c = 0.75
The equation therefore is x = m ( a/b ) + c,
Which becomes, x =  0.25 ( a/b ) + 0.75
Multiply by 4 4x =  ( a/b ) + 3
3 is the short side i.e. a = 3,
So, 4x =  ( a/b ) + a = a  ( a/b )
Therefore, x = 1/4 [ a  ( a/b ) ]
Or in words, x = 1/4 [ short side  short side ]
long side
This is the equation of the straight line which approximates to the first part of the graph.
My second prediction was that the ratio of the two sides would effect the value of x. I have been able to prove an approximate relationship between the two sides, and x.
CALCULUS
I have now decided to use calculus to find a general rule, which would allow me to get the size of cut out which would give the biggest volume.
To work out the volume of the cubeoid, you need to use a general rule. This is how I got the rule:
Length = A2x
Width = B2x
Height = x
Vol. = (A2x)(B2x)x
= (AB2Ax2Bx+4x2)x
= ABx2Ax22Bx2+4x3
= 4x32Ax22Bx2+ABx
dy/dx = 12x24Ax4Bx+AB
= 12x2(4A+4B)x+AB
To work out the minimum or maximum values, the greadient needs to be "0".
To find x, you need to use = b ± \/(b24ac)
_________________
2a
Substitute as follows:
a = 12
b = 4(A+B)
c = AB
Therefore, x = +4(A+B) ± \/(4(A+B)2(4 . 12 . AB)
_______________________________________________
2 . 12
= 4(A+B) ± \/16(A2+2AB+B2)48AB
____________________________________________
24
= 4(A+B) ± \/16A2+32AB+16B248AB
_______________________________________________
24
= 4(A+B) ± \/16A216AB+16B2
________________________________________
24
= 4(A+B) ± 4 . \/A2AB+B2
________________________________________
24
= A+B ± \/A2AB+B2
__________________________
6
= 1/6 [ A+B ± \/A2AB+B2 ]

= 1/6 [ A+B ± \/A2AB+B2 ]
This is a general rule for any rectangle (including squares). Squares also have their own formula, as A = B :
x = 1/6 [ A+A ± \/A2AA+A2 ]
= 1/6 [ 2A ± \/A2 ]
= 1/6 [ 2A ± A ]
= 1/6 [ 3A ] or 1/6 [ A ]
= 1/2 A or 1/6 A
= MIN MAX
Therefore, the maximum volume is when x = 1/6 A
Conclusion
This proves by calculus the result that I obtained in the First Open Box problem. By substituting different ratios of sides I can also prove a relationship for any rectangle. For example I could use B = 2A, or B = 8A, or B = nA and obtain a result as follows :
x = 1/6 [ A+nA ± \/(A2  A.nA + (nA)2) ]
= 1/6 [ (1+n)A ± \/(A2  nA2 + n2A2) ]
= 1/6 [ (1+n)A ± \/A2 (1n+n2) ]
= 1/6 [ (1+n)A ± A\/(1n)2 ]
= 1/6 A [ 1+n ± (1n) ]
= 1/6 A [ 2 ] or 1/6 A [ 2n ]
= MIN MAX