Paper Bags2

8/6/2019 Paper Bags2

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Paper Bags

...





We have to maximise the volume but keep various quantities

constant.

The lengths of the sides are constant (h, w)

The areas of the faces are constant (A)

h w



’

described by,

,

A

h w



where the side is described parametrically by

and we have used the shorthand

h



where the side is described parametrically by

w



A



equation for the functional

.

’ ,

h

A

w







’ Newton wasn’t ha but he su osedl solved

the problem in an evening after returning home

from work:‐

... e no s eep e a so ve ,

which was by four in the morning.”

... e no s eep e a so ve ,

which was by four in the morning.”

e sent t e so ut on to t e res ent o t e oya

Society, writing:‐

““

teased by foreigners about mathematical

things.”

teased by foreigners about mathematical

things.”

The Royal Society published Newton's solution

anonymously in the January 1697.



Five solutions were obtained, with Isaac Newton, Jacob Bernoulli,

Gottfried Leibniz and Guillaume de L'Hôpital all solving the problem

in

addition

to

Johann

Bernoulli.

Four solutions were published together in 1697: Leibniz’s, the two

Bernoullis’ and a Latin translation of Newton’s on a es 205‐223 of

Acta Eruditorum.

The solution by de L'Hôpital was not published until later (actually

nearly 300 years later in 1988).



Galileo had actuall studied the

problem in 1638.

He calculated the time taken from A

to B in a straight line.

A

Then

he

showed

you

would

reach

B

faster if you went along two line

se ments AC followed b CB where

C is a point on an arc of a circle. C

Galileo was correct but he then

made an error when he said that

u

to B would be an arc of a circle. B



…where is the magnitude of the acceleration due to gravity.

functional satisfies,

We can try and solve this second‐order nonlinear ODE, but

first we make a few uesses…



taken is then

T1

=

0.6330889378T1

=

0.6330889378

Using two straight lines you have a

choice of where to place the

intermediate point (xm , ym).

T2

=

0.5874903079T2

=

0.5874903079

which is for the point xm = 0.128, ym = 0.388

Johann Bernoulli attempted the problem)



a=0.8

a=4.1

The quickest path in this family of curves gives a time:

T3 = 0.5898350922T3 = 0.5898350922



T4

= 0.5806014444T4

= 0.5806014444

The general family of circles through A and B leads to a

T5

= 0.5798158652T5

= 0.5798158652



where is determined by the fact

that the curve must o throu h the

point (1,1)

so that

The minimum time is then == s

.s

.





=

what is the best Cycloid?

It turns out it is the one with zero radient at x=1.

Ts

= 0.5610910637Ts

= 0.5610910637



=

maximum volume of a closed bag.

B similarit the maximum volume of

a bag of width w must be h

A

and the maximum volume of an open

bag (half the closed bag) is then w



2

has burst at the bottom and so forms

a cylinder,

A lower bound could be formed by

creating a rectangular cross‐section

h‐x

2x

1‐2x





Then you can find the difference between the upper bound

the c linder and the measured value. The lost volume is

then estimated as

So the volume of an open bag is approximately

and for a closed ba it is



We finall et to the ex ression for the maximum volume of a

bag of width w,

’

h w

...



If we put w=x and h=1 we get

1x

and if we put w=1, h=x we get,

x 1

but all we did was turn the teabag through 90o



’



The paper bag problem is an “open” problem

There’s no million pound prize for it...

...but Tetley’s will sponsor your research.

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Paper Bags2