16
V = 21952+x³-28x²-784x
16
Right is the graph of the equation.
If we substitute x for +4 and –4, it will give us the volume of the possibilities. –4 gives a higher volume; when P is the longer side. Therefore I will always use the 320mm side for the perimeter of the cross-section in section 1 of this investigation.
Task 2
Task 2 is to investigate the maximum attainable area for a 3 sided cross-section, I begin with a 3-sided shape because this is the least number of sides a 2 dimensional shape can have.
From now on, let No. of sides of cross-section = n.
A =½bh
Let us for the moment assume b = 20mm to reduce the number of variables.
A = 10h
As h is the perpendicular of b, using pythagoras theorem, we know that
h² = a²-(20-x)² = c²-x²
Where x is the distance from the vertex of a and b to
where the angle bisector of the angle between a and c meets line b.
As we know the perimeter to be 320, we can substitute c for 300 – a.
h² = a²-(20-x)² = (300-a)²-x²
Which clearly shows that a>h and 300-a>h. This means
that h is greatest when a = 300-a = 300/2 = 150.
If two sides should be the same size for optimum area, we can calculate the optimum
Length of the third one. First some limitations:
2a+b = 320
a+b>a
2a>b
h² = a²-(a-x)² = (320-2a)²-x²
this again shows that a>h and 320-2a>h. This means that h is
greatest when a = 320-2a = 320/3.
The triangle with the greatest area has all sides the same.
Therefore all its angles are 60°.
Therefore it is equilateral.
Task 3
Task 3 is the same as task 2 but n = 4.
Left is a rhombus. The top and bottom sides are parallel and equal, call them a. The sides on the left and right are also parallel and equal. Call them b.d is the base of the triangle formed of d, b and c. c is the perpendicular of a. A is the area of the shape. θ is the angle between b and a
A = ac.
c = √(b²-d²)
when d = 0, b=c
when b = c, θ=90
b>c
Therefore, the area is greatest when θ=90 as that is when b = c
and c cannot be greater than b.
Therefore a quadrilateral with 4 90° angles has the greatest area.
The sides must also be the same to attain the greatest area, for:
The average length of a side is 80mm
Area of a square = 80² = 6400mm²
Let 2x = the deviation in lengths
Area = (80-x)(80+x)
= 80²+80x-80x-x²
= 6400-x²
6400 > 6400-x²
The suggestion does seem to be that regular shapes have greater areas than irregular ones. I have as of yet been unable to prove it. It should be possible to do this using intergration, however, attempting to intergrate anything more complex than a straight line gives me a headache. I can also sense the outline of a possiblity to do it algebraeically, I tried to expand the formulae for calculating the area of various shapes in order to apply them to each other, creating a ‘formulae of everything’. It was a hard lesson in how concentrated some of these formulae are. Herons formula, in the terms of a, b and c, is a whole page of numbers. I abandoned my efforts as I foresaw no good coming from them.
The graph shows the volumes of the first 22 possible prisms with regular cross sections. As the number of sides tend toward ∞, the volume tends towards 2000cm³, roughly.
Task 3
To find a formula for the volume of a prism with a regular cross-section.
All regular shapes can be divided into identical right angled triangles like the square below (except circles).
Let the perimeter of the shape = P
Let the no. of sides = n
Let the area of the cross-section = A
Let th volume of the tube = V
Let the length = L
The number of triangles is equal to 2n.
One angle is equal to half the interior angle of the shape (180(n-2))/2n
One side, the adjacent side to the known angle, is equal to half the side of the shape P/2n.
The opposite side of the known angle can be found with trigonometry (P/2n)tan(180(n-2))/2n)
Knowing two sides and the angle between them, which is the right angle, one can calculate the area of a triangle ½(P/2n)(P/2n)tan(180(n-2)/2n)sin90
As sin 90 = 1 we can dispense from it ½(P/2n)(P/2n)tan(180(n-2)/2n)
Multiply this by the number of triangles (½(P/2n)(P/2n)tan(180(n-2)/2n))*2n
A = ((P/4n)(P/2n)tan(180(n-2))/2n)*2n
A = ((P²/8n²)tan(180(n-2)/2n))*2n
A = (P²/4n)tan(180(n-2)/2n)
A = (P²/4n)tan(180/n)
Multiply by the length to get the volume
AL = V
V = P²L
4n tan(180/n)
The graph of this equation will be as the one on the last page.
One cannot, as implied earlier, use this formula to find the volume of a circular prism, which has an infinite number of sides, this does not pose a problem however as one can simply make an equation in Л.
Let r be the radius
The Perimeter is the circumference
The circumference = 2 Лr r = P/(2 Л)
The area is Лr² A = Л P/(2 Л)²
Multiplied by length V= Л P/(2 Л)²L
The maximum volume of a prism made out of a piece of card 240 by 320mm is therefore 1955695.941mm³
Section 2
To ascertain how to create a tube with the greatest possible volume using a piece of card with an area (B) of 1200cm².
1200 = PL
From section one, we know that the cross-section with the greatest area is a circle. So we can make an equation for the volume like so:
P = 1200/L
Radius = P/2Л
Area = radius² Л
Area = ((1200/L)/2Л)² Л
V = ((1200/L)/2Л)² ЛL
This equation gives an exponential curve as shown below. L is x on this graph, V is Y, so it clearly shows that the smaller L is, the greater the volume.
As P ∞, V ∞
A Cylinder where P = ∞ would not be a 3 dimensional objest, and so have a Volume of 0, it would be an infinite sheet, a plane. This means that one can make a prism out of a card of any area with an unlimited volume. It is therefore impossible to give a definite answer to the task, as it is always possible to increase the volume of the piece of card simply by shortening the Length and increasing the Perimeter of the cross-section.
