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unJon
unJon 
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February 1st, 2019 at 6:13:27 PM permalink
Quote: ThatDonGuy

Quote: Wizard

So a simple ratio of radius to lateral length of 0.5 gets us to a ratio of volume to surface area of 0.1443, much higher than the extremes of a very squat or tall dunce cap.

However, maybe you can do better than a ratio of 0.5.


To earn the beer, I don't want to see a trial and error answer but a full solution.



You didn't ask for the lateral length to base radius ratio; you asked for the height to base radius ratio.

In your three examples:

Height to Radius ratio .00999949999 has Ratio of Volume to Lateral Surface Area of 0.003333167

Height to Radius ratio 0.1425 has Ratio of Volume to Lateral Surface Area of 0.046552229

Height to Radius ratio 1.732 has Ratio of Volume to Lateral Surface Area of 0.144337567.

I have a feeling you either (a) asked the wrong problem, (b) misinterpreted "height," or (c) solved the wrong problem.

I agree with this.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Wizard
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Wizard
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February 1st, 2019 at 6:57:30 PM permalink
Okay, let me try this again. I think the way I phrased the original question is fine, but my last post was confusing.



If we put in a very small radius we get a very big ratio of height to radius. In the following example, a height to radius ratio of almost 1,000 results in a volume to surface area ratio of 0.000333.

lateral length 1
radius 0.001
height 0.9999995
volume 1.0472E-06
lateral surface area 0.003141593
Ratio volume to surface area 0.000333333
Ratio height to radius 999.9995


I need to show only one counter-example to disprove the infinitely tall dunce cap answer. Putting in a radius of 0.5 compared to a lateral length of 1 gives us a volume to surface ratio of 0.144, much more than the 0.000333 of a extremely tall dunce cap.

lateral length 1
radius 0.5
height 0.866025404
volume 0.226724921
lateral surface area 1.570796327
Ratio volume to surface area 0.144337567
Ratio height to radius 1.732050808


It's not whether you win or lose; it's whether or not you had a good bet.
ThatDonGuy
ThatDonGuy
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February 1st, 2019 at 7:51:39 PM permalink
Quote: Wizard

Okay, let me try this again. I think the way I phrased the original question is fine, but my last post was confusing.



If we put in a very small radius we get a very big ratio of height to radius. In the following example, a height to radius ratio of almost 1,000 results in a volume to surface area ratio of 0.000333.

lateral length 1
radius 0.001
height 0.9999995
volume 1.0472E-06
lateral surface area 0.003141593
Ratio volume to surface area 0.000333333
Ratio height to radius 999.9995


I need to show only one counter-example to disprove the infinitely tall dunce cap answer. Putting in a radius of 0.5 compared to a lateral length of 1 gives us a volume to surface ratio of 0.144, much more than the 0.000333 of a extremely tall dunce cap.

lateral length 1
radius 0.5
height 0.866025404
volume 0.226724921
lateral surface area 1.570796327
Ratio volume to surface area 0.144337567
Ratio height to radius 1.732050808




The solution may depend on the specific height and radius values separately.
I got "infinite height" based on a fixed radius - i.e. for a radius of, say, 1, the volume/LSA ratio increases as the height increases.

However, compare these two cones, one being twice the size (in terms of linear distance) of the other:
radius12
heightsqrt(99)2 sqrt(99)
lateral length1020
volumePI * sqrt(99)/3PI * 8 sqrt(99)/3
lateral surface area10 PI40 PI
height/radius ratiosqrt(99)sqrt(99)
volume/LSA ratiosqrt(99)/30sqrt(99)/15

The volumes vary as the linear distances cubed, and the areas vary as the linear distances squared, so the ratios vary as the linear distances, even though the height to radius ratio is a constant.

The answer may be different for each radius.



If the problem becomes finding the height/radius ratio for a given lateral distance that has the highest volume/lateral surface area ratio:

Let d be the (constant) lateral distance, r the radius, and h the height
h = sqrt(d2 - r2)
Volume v = PI/3 * r2 * h = PI/3 * r2 sqrt(d2 - r2)
Lateral Surface Area a = PI * r * d
v/a = r sqrt(d2 - r2) / (3d)
The first derivative of v/a with respect to r = sqrt(d2 - r2) - r / sqrt(d2 - r2)
This is zero when d2 = 2 r2 -> r = sqrt(2)/2 d = h -> h/r = 1
Rather than trying to calculate the second derivative of v/a to show that the value is a minimum when r = sqrt(2)/2 d, note that there is only one zero for the first derivative, so it is either a maximum, a minimum, or an inflection point
r = sqrt(2)/2 d -> v/a = 2/3 d
r = 1/2 d < sqrt(2)/2 d -> v/a = sqrt(3)/12 d < 2/3 d
r = 3/4 d > sqrt(2)/2 d -> v/a = sqrt(7)/16 d < 2/3 d
Since v/a < 2/3 d for values of r both less than and greater than sqrt(2)/2 d, v/a is a maximum at that point.

Last edited by: ThatDonGuy on Feb 1, 2019
Wizard
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Wizard
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February 1st, 2019 at 9:56:25 PM permalink
Okay, Don makes a good point. Much as the dimensions for the optimal sized can, maximizing volume to surface face, changes depending on the surface area, so does it matter in this problem.

Let me introduce a new rule -- The lateral distance (from tip of dunce cap to any point on the circumference) is 1.
It's not whether you win or lose; it's whether or not you had a good bet.
unJon
unJon 
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February 2nd, 2019 at 4:47:12 AM permalink
Quote: Wizard

Okay, Don makes a good point. Much as the dimensions for the optimal sized can, maximizing volume to surface face, changes depending on the surface area, so does it matter in this problem.

Let me introduce a new rule -- The lateral distance (from tip of dunce cap to any point on the circumference) is 1.



My guess is that making L=1 isnít critical so much as saying L must be a constant.

A = pi * r * L
V = pi * r^2 * h / 3
h= sqrt(L^2 - r^2)
L=1

V/A = pi*r^2 *sqrt(1-r^2) / 3(pi*r)
d/dr = (1 - 2r^2) / (3 - sqrt(1-r^2)) = 0
r = 1/sqrt(2)

Which makes r = h, which also makes sense.
Last edited by: unJon on Feb 2, 2019
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Wizard
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February 2nd, 2019 at 6:17:58 AM permalink
Quote: unJon

A = pi * r * L
V = pi * r^2 * h / 3
h= sqrt(L^2 - r^2)
L=1

V/A = pi*r^2 *sqrt(1-r^2) / 3(pi*r)
d/dr = (1 - 2r^2) / (3 - sqrt(1-r^2)) = 0
r = 1/sqrt(2)

Which makes r = h, which also makes sense.



I'm getting a different answer. What is your volume to surface area ratio?
It's not whether you win or lose; it's whether or not you had a good bet.
unJon
unJon 
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February 2nd, 2019 at 6:39:27 AM permalink
Quote: Wizard

Quote: unJon

A = pi * r * L
V = pi * r^2 * h / 3
h= sqrt(L^2 - r^2)
L=1

V/A = pi*r^2 *sqrt(1-r^2) / 3(pi*r)
d/dr = (1 - 2r^2) / (3 - sqrt(1-r^2)) = 0
r = 1/sqrt(2)

Which makes r = h, which also makes sense.



I'm getting a different answer. What is your volume to surface area ratio?





Hmm. Not at computer so let me do by hand.
V/A = pi * 1/2 * sqrt(1/2) / 3pi*(1/sqrt(2))
I think that all cancels down to 1/6.

Can you beat 1/6?
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
ThatDonGuy
ThatDonGuy
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February 2nd, 2019 at 8:00:46 AM permalink
Quote: Wizard

I'm getting a different answer. What is your volume to surface area ratio?



I get what unJon gets: r = h = sqrt(2)/2

Volume = PI r2 h / 3 = PI * sqrt(2) / 12
Lateral Surface Area = PI r d (where d = lateral distance) = PI * sqrt(2) / 2
Ratio = 1/6
Numbers, and maximum, confirmed by spreadsheet

If you want a more rigorous proof of it being a maximum at r = sqrt(2) / 2, besides the one in one of my previous responses:

df / dr = sqrt(1 - r2) - r / sqrt(1 - r2)

d2f / dr2 = -2 / sqrt(1 - r2) - (sqrt(1- r2) + r2 / sqrt(1 - r2)) / (1 - r2)

For r = sqrt(2)/2, r2 = 1/2 -> sqrt(1 - r2) = sqrt(2)/2
Substitute x for both r and sqrt(1 - r^2):

d2f / dr2 = -2 / x - (x + x^2 / x) / x^2 = -4 / x = -2 sqrt(2) < 0, so f(r) is a maximum at r = sqrt(2) / 2

Last edited by: ThatDonGuy on Feb 2, 2019
charliepatrick
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February 2nd, 2019 at 9:47:05 AM permalink

Area = Pi R L; since Pi and L are constant it's K1 * R.
Volume = Pr R^2 H / 3, or K2 * R * R * H
Sol Volume/Area = ( K2 * R * R * H )/( K1 * R ) = K3 * R * H

However we know R^2+H^2 = 1 and want to maximize R*H.
Given 0<R<1 and 0<H<1 maximizing R*H is the same as maximizing R*R*H*H = (R^2)*(1-R^2).
This happens when R^2 = 1/2 (consider points on a unit circle and where x*y is maximum or also consider proof by symmetry). It also means H^2 = 1/2, and that R=H.

Now back to the formulae.
Area = Pi R
Volume = Pi R * R * R / 3 (replacing H by R as they have the same value).
Dividing Volume/Area = R * R / 3 = 1/2 * 1/3 = 1/6.
netzer
netzer
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February 3rd, 2019 at 10:10:23 AM permalink
I can suggest a more practical statement of the problem that produces several solutions with an Internet search.

What is the optimal shape of a conical paper cup from the point of view of how much water it will hold in relation to the amount of paper used in its construction?

"Cone optimization" is a good search term.

In this setting, intuition suggests that the shape should be independent of actual dimensions and that an infinitely high cup is out of the question.

A video on this problem may be found on YouTube at watch?v=UCFJPxVl3bE This, like the other solutions I found, do the calculation for particular volumes, they arrive at h/r = SQRT(2) numerically. I repeated this solution with a symbolic volume and found h/r = SQRT(2) exactly.

The Wizard might be interested in this video since it includes a derivation of the surface area of a cone as a sector of a circle.

I won't display the calculations since the expressions are humongous and difficult to typeset. I had to use symbolab to simplify the expressions. In the past I have used Derive and Maple. Mathematica is similar.
Last edited by: netzer on Feb 3, 2019
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