Paper Bags ...
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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
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where the side is described parametrically by
and we have used the shorthand
h
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where the side is described parametrically by
w
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equation for the functional
.
’ ,
h
A
w
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’ 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.
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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).
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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
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…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…
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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)
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a=0.8
a=4.1
The quickest path in this family of curves gives a time:
T3 = 0.5898350922T3 = 0.5898350922
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T4
= 0.5806014444T4
= 0.5806014444
The general family of circles through A and B leads to a
T5
= 0.5798158652T5
= 0.5798158652
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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
.
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=
what is the best Cycloid?
It turns out it is the one with zero radient at x=1.
Ts
= 0.5610910637Ts
= 0.5610910637
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=
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
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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
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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
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We finall et to the ex ression for the maximum volume of a
bag of width w,
’
h w
...
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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