## Poll

 Pass the dunce cap, I don't know how to solve it. 2 votes (28.57%) I know, but I'm on 24-hour delay. 1 vote (14.28%) I don't understand the question. 1 vote (14.28%) It depends on the exact surface area. No votes (0%) Hannah Minx hasn't made a video in years. 1 vote (14.28%) I support a ban on leaf blowers. 2 votes (28.57%) When I die, I want my body eaten by vultures. 1 vote (14.28%) Super bowl prop bet challenge deadline reminder. No votes (0%) I like under 7.5 punts in SB 53 No votes (0%) Why the Roman numerals with the Super Bowl number? 3 votes (42.85%)

7 members have voted

Wizard
Joined: Oct 14, 2009
• Posts: 25947
February 1st, 2019 at 7:00:37 AM permalink
This was asked in the 27 chips thread, but I think it's worthy of its own thread.

What should be the ratio of the height to the radius of the base of a dunce cap to maximize the ratio of volume to surface area of the dunce cap (thus not including the base of the cone)?

As usual, I request anyone who has won a beer to not submit an answer for 24 hours, but they may correct other responses or ask for clarification of my wording.
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
charliepatrick
Joined: Jun 17, 2011
• Posts: 2930
February 1st, 2019 at 9:01:10 AM permalink
Perhaps I'm misunderstanding the problem, so could you clarify the problem as the logic below suggests an infinitely tall hat.

Let's assume the radius of the cone is 1, so the question is what height maximizes V/A.
The volume is proportional to h (V = Pi r^2 h/3 = (Pi r^2 / 3) * h = constant*h for a fixed r.
The area is propprtional to l (the length of the slant) = Pi r l = constant*l for a fixed r.
So the ratio of volume to area is the same as h to l (where l = SQRT(h^2+1) ).
This means an infinitely tall hat is best.

Another way of looking at it is consider a hat of height h then one of height 2h.
The volume is twice as much (as all we've done is stretch the cone in one direction).
The area is proportional to the distance from the cone's base to the tip.
Consider a triangle with these three points
B : the cone's base (x=1, y=0)
A = cone's tip at height h (x=0, y=h).
C = cone's tip at height 2h (x=0, y=2h).
AB = l (the original side length of smaller cone)
BC = h (the bigger cone is h higher)
AC = new side length of larger cone.
Since this forms a triangle AC is less than AB+BC; as l>h this is also less than 2l.
So the area of the large cone is less than twice the area of the small cone.

So doubling the height doubles the volume but not the area.
Thus it increases the ratio of volume to area.

So continue doulbing the height (for ever).
RogerKint
Joined: Dec 5, 2011
• Posts: 1915
February 1st, 2019 at 9:14:20 AM permalink
10:10?
100% risk of ruin
Wizard
Joined: Oct 14, 2009
• Posts: 25947
February 1st, 2019 at 11:14:26 AM permalink
I think I may have asked the problem badly. Give me a moment.
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
Wizard
Joined: Oct 14, 2009
• Posts: 25947
February 1st, 2019 at 3:26:29 PM permalink
Well, this problem is harder than it seems. I wasn't satisfied with a formula for the lateral surface area of cone so derived it from scratch, which took about two hours. Harder than it seems.

That was the hard part. Let's just say that I have an answer and it doesn't involve an infinitely high length. Sorry, Charlie, only good tasting tuna get to be Star-Kist, I'll have to try to find your flaw another time, I have wasted enough time on this for one day.

BTW, I think I asked the problem just fine.
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
unJon
Joined: Jul 1, 2018
• Posts: 4337
February 1st, 2019 at 4:00:48 PM permalink

Don’t give me the beer because I couldn’t remember formula for lateral surface area of a cone so looked it up. Learned it’s like the formula for surface area of a pyramid where the base perimeter is 2rpi. Pretty cool.

A = pi * r * L
Where L is the slant length so L = sqrt(r^2 + h^2). Because L is the hypotenuse of the right triangle formed by the height and radius of the cone.

V = pi * r^2 * h / 3

Let’s make life easier by setting r to 1.

Maximize V/A = (pi * h) / 3(pi * sqrt(1 + h^2))

Take the derivative d/dh = 1/3(1 + h^2)^3/2

Hmmm. This derivative would seem to go to zero at h being infinite.

So I did something wrong too I guess.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
charliepatrick
Joined: Jun 17, 2011
• Posts: 2930
February 1st, 2019 at 4:20:22 PM permalink
Quote: unJon

...A =....V = ....Hmmm....so I did something wrong too I guess.

I agree with the same formula.
If you put "Area of Cone" into google you get the formula when clkcing laternal surface and realising that the value is L.

Another way of looking at it is to consider the surface area created by dragging a circle, albeit getting smaller, The average size for the circle will be 1/2, so it's circumference is Pi (2 Pi r where r = 1/2). The distance the circle covers is along the outside edge, i.e. L. So the area is Pi L.
unJon
Joined: Jul 1, 2018
• Posts: 4337
February 1st, 2019 at 4:41:40 PM permalink
Quote: charliepatrick

I agree with the same formula.

If you put "Area of Cone" into google you get the formula when clkcing laternal surface and realising that the value is L.

Another way of looking at it is to consider the surface area created by dragging a circle, albeit getting smaller, The average size for the circle will be 1/2, so it's circumference is Pi (2 Pi r where r = 1/2). The distance the circle covers is along the outside edge, i.e. L. So the area is Pi L.

Right. Or:
as h goes to zero, the area of the cone goes to the area of the circle and L goes to r
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Wizard
Joined: Oct 14, 2009
• Posts: 25947
February 1st, 2019 at 5:32:08 PM permalink
Let's say the lateral distance (from tip to any point on the circumference) of the dunce cap is 1. Then those who argue optimal ratio is found at an infinite height would say that the height should also be 1.

Let's put in a radius of 0.01 and see what we get:

 lateral length 1 radius 0.01 height 0.99995 volume 0.000104715 lateral surface area 0.0314159 Ratio volume to surface area 0.00333317

 lateral length 1 radius 0.99 height 0.141067 volume 0.144786 lateral surface area 3.11018 Ratio volume to surface area 0.0465522

Let's split the difference and put in a radius of 0.5

 lateral length 1 radius 0.5 height 0.866025 volume 0.226725 lateral surface area 1.5708 Ratio volume to surface area 0.144338

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.
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 5944
February 1st, 2019 at 5:56:27 PM permalink
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.

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.

unJon
Joined: Jul 1, 2018
• Posts: 4337
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.

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
Joined: Oct 14, 2009
• Posts: 25947
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 1 volume 1.0472e-06 lateral surface area 0.00314159 Ratio volume to surface area 0.000333333 Ratio height to radius 1000

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 0.866025 volume 0.226725 lateral surface area 1.5708 Ratio volume to surface area 0.144338 Ratio height to radius 1.73205

“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 5944
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 1 volume 1.0472e-06 lateral surface area 0.00314159 Ratio volume to surface area 0.000333333 Ratio height to radius 1000

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 0.866025 volume 0.226725 lateral surface area 1.5708 Ratio volume to surface area 0.144338 Ratio height to radius 1.73205

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:
 radius 1 2 height sqrt(99) 2 sqrt(99) lateral length 10 20 volume PI * sqrt(99)/3 PI * 8 sqrt(99)/3 lateral surface area 10 PI 40 PI height/radius ratio sqrt(99) sqrt(99) volume/LSA ratio sqrt(99)/30 sqrt(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.

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
Joined: Oct 14, 2009
• Posts: 25947
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.
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
unJon
Joined: Jul 1, 2018
• Posts: 4337
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 isn’t 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
Joined: Oct 14, 2009
• Posts: 25947
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?
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
unJon
Joined: Jul 1, 2018
• Posts: 4337
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
Joined: Jun 22, 2011
• Posts: 5944
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
Joined: Jun 17, 2011
• Posts: 2930
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
Joined: Jan 17, 2019
• Posts: 45
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
OnceDear is a Dear!
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 5944
February 5th, 2019 at 4:02:15 PM permalink
I have found (so far) three different solutions, depending on how the problem is interpreted.

Let r = the radius of the base, and h = the height
Let V = the volume = PI/3 r2 h
Let A = the lateral surface area = PI r sqrt(r2 + h2)
Let R = the desired ratio = V/A = rh / (3 sqrt(r2 + h2))

Let x = h/r, where r is a constant; h = rx
R = x r2 / (3 sqrt(r2 + x2r2)
= x r / (3 sqrt(1 + x2))
= r / 3 * 1 / sqrt(1 + x2)
dR/dx = r / 3 * (sqrt(1 + x2) - x^2 / sqrt(1 + x2)) / (x2+1)
Note that 1 > 0 -> x2 + 1 > x2 -> (x2 + 1) / sqrt(x2 + 1) > x2 / sqrt(x2 + 1)
-> sqrt(x2 + 1) > x2 / sqrt(x2 + 1) -> sqrt(x2 + 1) - x2 / sqrt(x2 + 1) > 0
Therefore, dR/dx > 0 for all positive x, which means R increases as x increases; since r is fixed, x increases as h increases, so R increases as h increases
There is no h/r ratio where volume/lateral surface area is a maximum as it increases as h increases

Let d = the lateral distance = a constant; since d is a constant, d2 is also constant
d2 = h2 + r2 ->h = sqrt(d2 - r2)
R = hr / (3d) = 1 / (3d) * r sqrt(d2 - r2)
dR/dr = 1 / (3d) * (sqrt(d2 - r2) - r2 / sqrt(d2 - r2))
This is zero when sqrt(d2 - r2) = r2 / sqrt(d2 - r2)
-> d2 = 2 r2 -> h2 = d2 - r2 = r2 -> h = r
The volume/lateral surface area is a maximum when r = h (= sqrt(2)/2 x the lateral distance)

I think this is what netzer discovered

Let d = 3V / PI = r2 h; since V is a constant, this is also a constant
h = d / r2
R = hr / (3 sqrt(r2 + h2) = d / r2 * r / (3 sqrt(r2 + (d2 / r4))
= d / 3r * 1 / (sqrt(r2 + (d2 / r4))
= d / 3r * 1 / (sqrt(r6 / r4 + d2 / r4))
= d / 3 * r / sqrt(r6 + d2)
dR/dr = d / 3 * (sqrt(r6 + d2) - r * 3 r5 / sqrt(r6 + d2)) / (r6 + d2)
This is zero when sqrt(r6 + d2) = 3 r6 / sqrt(r6 + d2))
-> r6 + d2 = 3 r6 ->d2 = 2 r6 -> d = sqrt(2) r3
h = d / r2 = sqrt(2) r
The volume/lateral surface area is a maximum when h = sqrt(2) r

Coming up next: what if the lateral surface area is a maximum (i.e. you have a certain area of material you can use to make a conical cup; how much water can it hold?)
unJon
Joined: Jul 1, 2018
• Posts: 4337
February 5th, 2019 at 4:06:41 PM permalink
Quote: ThatDonGuy

I have found (so far) three different solutions, depending on how the problem is interpreted.

Let r = the radius of the base, and h = the height
Let V = the volume = PI/3 r2 h
Let A = the lateral surface area = PI r sqrt(r2 + h2)
Let R = the desired ratio = V/A = rh / (3 sqrt(r2 + h2))

Let x = h/r, where r is a constant; h = rx
R = x r2 / (3 sqrt(r2 + x2r2)
= x r / (3 sqrt(1 + x2))
= r / 3 * 1 / sqrt(1 + x2)
dR/dx = r / 3 * (sqrt(1 + x2) - x^2 / sqrt(1 + x2)) / (x2+1)
Note that 1 > 0 -> x2 + 1 > x2 -> (x2 + 1) / sqrt(x2 + 1) > x2 / sqrt(x2 + 1)
-> sqrt(x2 + 1) > x2 / sqrt(x2 + 1) -> sqrt(x2 + 1) - x2 / sqrt(x2 + 1) > 0
Therefore, dR/dx > 0 for all positive x, which means R increases as x increases; since r is fixed, x increases as h increases, so R increases as h increases
There is no h/r ratio where volume/lateral surface area is a maximum as it increases as h increases

Let d = the lateral distance = a constant; since d is a constant, d2 is also constant
d2 = h2 + r2 ->h = sqrt(d2 - r2)
R = hr / (3d) = 1 / (3d) * r sqrt(d2 - r2)
dR/dr = 1 / (3d) * (sqrt(d2 - r2) - r2 / sqrt(d2 - r2))
This is zero when sqrt(d2 - r2) = r2 / sqrt(d2 - r2)
-> d2 = 2 r2 -> h2 = d2 - r2 = r2 -> h = r
The volume/lateral surface area is a maximum when r = h (= sqrt(2)/2 x the lateral distance)

I think this is what netzer discovered

Let d = 3V / PI = r2 h; since V is a constant, this is also a constant
h = d / r2
R = hr / (3 sqrt(r2 + h2) = d / r2 * r / (3 sqrt(r2 + (d2 / r4))
= d / 3r * 1 / (sqrt(r2 + (d2 / r4))
= d / 3r * 1 / (sqrt(r6 / r4 + d2 / r4))
= d / 3 * r / sqrt(r6 + d2)
dR/dr = d / 3 * (sqrt(r6 + d2) - r * 3 r5 / sqrt(r6 + d2)) / (r6 + d2)
This is zero when sqrt(r6 + d2) = 3 r6 / sqrt(r6 + d2))
-> r6 + d2 = 3 r6 ->d2 = 2 r6 -> d = sqrt(2) r3
h = d / r2 = sqrt(2) r
The volume/lateral surface area is a maximum when h = sqrt(2) r

Coming up next: what if the lateral surface area is a maximum (i.e. you have a certain area of material you can use to make a conical cup; how much water can it hold?)

I think netzer discovered fixed lateral area (ie a set amount of construction paper) not fixed volume. My intuition is that the answer is the same.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
cmlotito
Joined: Jun 3, 2013
• Posts: 371
February 5th, 2019 at 5:45:02 PM permalink
We should ask the governor of Virginia. He appears to be an expert in the matter.
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 5944
February 5th, 2019 at 6:19:16 PM permalink
Quote: unJon

Quote: ThatDonGuy

Coming up next: what if the lateral surface area is a maximum (i.e. you have a certain area of material you can use to make a conical cup; how much water can it hold?)

I think netzer discovered fixed lateral area (ie a set amount of construction paper) not fixed volume. My intuition is that the answer is the same.

Let r = the radius of the base, and h = the height
Let V = the volume = PI/3 r^2 h
Let A = the lateral surface area = PI r sqrt(r^2 + h^2)
Let R = the desired ratio = V/A = rh / (3 sqrt(r^2 + h^2))

Let A be a constant
A = PI r sqrt(r^2 + h^2)
A / (PI r) = sqrt(r^2 + h^2)
A^2 / PI^2 * 1 / r^2 = r^2 + h^2
h^2 = A^2 / PI^2 * 1 / r^2 - r^2
h = sqrt(A^2 / PI^2 * 1 / r^2 - r^2)

R = rh / (3 sqrt(r^2 + h^2))
= 1/3 * r sqrt (A^2 / PI^2 * 1 / r^2 - r^2) / sqrt(r^2 + A^2 / PI^2 * 1 / r^2 - r^2)
= 1/3 * r sqrt (A^2 / PI^2 * 1 / r^2 - r^2) / sqrt(A^2 / PI^2 * 1 / r^2)
= PI/(3A) * r^2 sqrt (A^2 / PI^2 * 1 / r^2 - r^2)
= PI/(3A) * sqrt (A^2 / PI^2 * r^2 - r^6)

R' = PI/(3A) * ((2 A^2 / PI^2) r - 6 r^5) / (2 sqrt(A^2 / PI^2 * r^2 - r^6))
= PI/(3A) * 2r (A^2 / PI^2 - 3 r^4) / (2r sqrt(A^2 / PI^2 - r^4))
= PI/(3A) * (A^2 / PI^2 - 3 r^4) / sqrt(A^2 / PI^2 - r^4)
This = 0 when 3 r^4 = (A/PI)^2 -> r^2 = A / (PI sqrt(3))

h/r = sqrt(A^2 / PI^2 * 1 / r^2 - r^2) / r
h^2/r^2 = (A^2 / PI^2 * 1 / r^2 - r^2) / r^2
= A^2 / PI^2 * 1 / r^4 - 1
= A^2 / PI^2 * 1 / (A^2 / 3 PI^2) - 1
= 2
h / r = sqrt(2)

How about that - the volume/lateral surface area is a maximum when height/radius = sqrt(2)

unJon
Joined: Jul 1, 2018
• Posts: 4337
February 5th, 2019 at 6:34:23 PM permalink
Quote: ThatDonGuy

Let r = the radius of the base, and h = the height
Let V = the volume = PI/3 r^2 h
Let A = the lateral surface area = PI r sqrt(r^2 + h^2)
Let R = the desired ratio = V/A = rh / (3 sqrt(r^2 + h^2))

Let A be a constant
A = PI r sqrt(r^2 + h^2)
A / (PI r) = sqrt(r^2 + h^2)
A^2 / PI^2 * 1 / r^2 = r^2 + h^2
h^2 = A^2 / PI^2 * 1 / r^2 - r^2
h = sqrt(A^2 / PI^2 * 1 / r^2 - r^2)

R = rh / (3 sqrt(r^2 + h^2))
= 1/3 * r sqrt (A^2 / PI^2 * 1 / r^2 - r^2) / sqrt(r^2 + A^2 / PI^2 * 1 / r^2 - r^2)
= 1/3 * r sqrt (A^2 / PI^2 * 1 / r^2 - r^2) / sqrt(A^2 / PI^2 * 1 / r^2)
= PI/(3A) * r^2 sqrt (A^2 / PI^2 * 1 / r^2 - r^2)
= PI/(3A) * sqrt (A^2 / PI^2 * r^2 - r^6)

R' = PI/(3A) * ((2 A^2 / PI^2) r - 6 r^5) / (2 sqrt(A^2 / PI^2 * r^2 - r^6))
= PI/(3A) * 2r (A^2 / PI^2 - 3 r^4) / (2r sqrt(A^2 / PI^2 - r^4))
= PI/(3A) * (A^2 / PI^2 - 3 r^4) / sqrt(A^2 / PI^2 - r^4)
This = 0 when 3 r^4 = (A/PI)^2 -> r^2 = A / (PI sqrt(3))

h/r = sqrt(A^2 / PI^2 * 1 / r^2 - r^2) / r
h^2/r^2 = (A^2 / PI^2 * 1 / r^2 - r^2) / r^2
= A^2 / PI^2 * 1 / r^4 - 1
= A^2 / PI^2 * 1 / (A^2 / 3 PI^2) - 1
= 2
h / r = sqrt(2)

How about that - the volume/lateral surface area is a maximum when height/radius = sqrt(2)

Neat!
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Wizard
Joined: Oct 14, 2009
• Posts: 25947
February 5th, 2019 at 9:15:40 PM permalink
Thanks for the comments. I've been a bit busy with the Super Bowl but hope to write all this up properly soon.
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
netzer
Joined: Jan 17, 2019
• Posts: 45
February 6th, 2019 at 9:42:38 AM permalink
The maths gurus seem to be closing in on it. Now I'm wondering about the simplest possible solution. I think the answer may be in making the derivative that must be calculated and simplified as simple as possible.

I'll call the diagonal height R because l looks too much like 1.
This is the diagonal height of the cone and also the radius of the circle the cone is made from.

r will be the radius of the base and h will be the height of the cone and V its volume.

The object will be to find the maximum volume of the cone as a function of R

V = Pi*hr2/3

r = SQRT(R2-h2)

V = Pi*h(R2-h2)/3

Since r is squared it makes sense to express it as a square root.

V'(h) = Pi*R2/3 - Pi*h2 after simplification

This derivative is much simpler than with some other approaches.

Set V'(h) = 0 and solve for h

R2/3 = h2, h = R/SQRT(3) for maximum volume.

To check that this is equivalent to h/r = SQRT(2):

let h = r*SQRT(2), h2 = 2r2

h2 + r2 = R2

2r2 + r2 = R2

h = R/SQRT(3)

Q.E.D.
OnceDear is a Dear!
Wizard
Joined: Oct 14, 2009
• Posts: 25947
February 8th, 2019 at 6:57:33 PM permalink
I reworded the problem as follows:

Quote: cone problem

You have a tortilla with radius 1 and wish to form a cone. You may cut out any wedge you like from the tortilla. The point of the wedge must be at the center of the circle. After cutting out the wedge you then attach the two straight edges remaining to form a cone. What is the maximum ratio of the volume of the cone to the remaining surface area?

I welcome all comments, although, to be honest, I'm looking forward to getting past this problem and posting a new one.
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
netzer
Joined: Jan 17, 2019
• Posts: 45
February 9th, 2019 at 7:56:23 AM permalink
Quote: Wizard

I'm looking forward to getting past this problem and posting a new one.

I think there is more work to be done on this one!

The Wizard and I are not in agreement on the solution

Let r be the radius of the base, h the height of the cone, and R the slant radius, which is the radius of the circle from which the cone is cut. From my previous result, maximum volume is attained when h/r = √2, h = R/√3, r = √(2/3)R

Using the formula for the circumference of a circle and letting x be the portion of the circumference to be cut out

2 π R - x = 2 π r = 2 π √(2/3)R, x = 2 π R (1-√(2/3)

The angle of the slice to be cut out is x / R = 2 π (1-√(2/3)) = 1.152986 radians, which is about 66 degrees.

The question, however, is What is the maximum ratio of the volume of the cone to the remaining surface area? For simplicity, let us set r = 1, then this is a 1, √2, √3 right triangle.

The volume is πr2h/3 = πh/3 and the surface area is πrR = π√3 so the ratio of volume to surface area is √2/3√3 or approximately 1 to 3.67423.
Last edited by: netzer on Feb 9, 2019
OnceDear is a Dear!
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 5944
February 9th, 2019 at 8:43:34 AM permalink

Let t be the angle of the wedge, in radians
The circumference of the base of the resulting cone is t, so the base radius r = t / (2 PI)
The lateral distance is 1, so the cone height h = sqrt(1 - r2) = sqrt(1 - t2 / (4 PI2)) = sqrt(4 PI2 - t2) / (2 PI)

V = PI/3 r2 h = PI/3 * t2 / (4 PI2) * sqrt(4 PI2 - t2) / (2 PI)
= 1/(24 PI2) * t2 sqrt(4 PI2 - t2)
A = PI r d = PI t / (2 PI) = t / 2
Ratio R = V/A = 1/(12 PI2) * t sqrt(4 PI2 - t2)

R' = 1/(12 PI2) * sqrt(4 PI2 - t2) - t2 / sqrt(4 PI2 - t2)
R' = 0 when t2 = 4 PI2 - t2 -> t = sqrt(2) PI

R = 1 / (12 PI2) * PI sqrt(2) * sqrt(4 PI2 - 2 PI2)
= PI sqrt(2) sqrt(2 PI2) / (12 PI2)
= 2 PI2 / (12 PI2) = 1/6

Yes, h/r = sqrt(2) - but we still need to determine the angle of the tortilla wedge that will create a cone with h/r = sqrt(2).
The last part of your answer appears to assume that it is always true.

If t is the angle in radians, and the radius is 1, then the circumference of the wedge is 2 PI r * (t / 2 PI) = t r = t.
When this becomes the circumference of the base of the cone, the base radius = t / (2 PI), and the base height = sqrt(2) * t / (2 PI)
Solve for t such that the slant radius = 1.

netzer
Joined: Jan 17, 2019
• Posts: 45
February 9th, 2019 at 10:59:33 AM permalink
Quote: ThatDonGuy

I get what The Wizard gets

Only partly! You agree with him that the ratio of volume to surface area for the optimal cone is 1/6 but you also believe, along with me, that for the optimal cone h/r = √2. The Wizard believes that r = h = √2/2, so for him h/r = 1. He is using a square right triangle with hypotenuse 1.

Quote: ThatDonGuy

What I think is the flaw in Netzer's solution

"The last part of your answer appears to assume that it is always true." What is always true?

"If t is the angle in radians, and the radius is 1, then the circumference of the wedge is 2 PI r * (t / 2 PI) = t r = t."

No math needed! That's the definition of radian measure: the length of the arc the angle subtends divided by the radius.

"Solve for t such that the slant radius = 1." I am using a slant radius of √3 for a base radius of 1. It's a 1 √2, √3 triangle.

I have made a correction in my previous post:

The volume is πr2h/3 = πh/3 and the surface area is πrR = π√3 so the ratio of volume to surface area is √2/3√3 or approximately 1 to 3.67423

I have been going over the Wizard's development. He defines

r = radius of the base of the cone.
h = height of cone
c = circumference of the base of the cone
S = surface area of cone
V = volume of cone

Quotes from the Wizard are in italics.

He does not assign a symbol to the slant height of the cone, which is the radius of the circle from which the cone is cut. I call it R.

So, the area of the slice is π ­ rπ, thus the area of Pac Man is π ­ (π ­ rπ) = rπ
As a reminder the Pac Man shape is the cone flattened out, so S = rπ.

Actually, it is πrR, so everything that follows is tainted. A quick way to see this would be to check the units on the left and the right side of the equation. The left side is in square units but the right side is in linear units. Adding R to the right side makes it in square units.

Area of a cone, excluding the base.

Let r be the radius of the base and R be the distance from the apex to any point on the circumference of the base. Call this the slant height. The length of the circumference of the base will be 2πr.

Now cut the cone along any R line and flatten it out. It becomes a pizza with a piece missing. If it were whole its area would be πR2, however it has only 2πr of its original circumference of 2πR left and its area is reduced proportionally so it becomes

πR2(2πr/2πR) = πrR

It's as simple as that!
Last edited by: netzer on Feb 9, 2019
OnceDear is a Dear!
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 5944
February 9th, 2019 at 12:15:12 PM permalink
You can't just set a value for r and then assume it's true for all values of r.
Remember, V varies as r3, and A varies as r2, so V/A varies as r.

If r = 1, h = sqrt(2), and d = sqrt(3):
V = PI/3 r2 h = PI/3 * sqrt(2)
A = PI r d = PI sqrt(3)
V/A = sqrt(2) / (3 sqrt(3)) = sqrt(6) / 9 = about 1 / 3.67423, which is what you get

If r = 2, h = 2 sqrt(2), and d = 2 sqrt(3):
V = PI/3 r2 h = PI/3 * 4 * 2 sqrt(2) = PI * 8 sqrt(2)/3
A = PI r d = PI * 2 * 2 sqrt(3) = PI * 4 sqrt(3)
V/A = (8 sqrt(2)) / (3 * 4 sqrt(3)) = (2 sqrt(2) sqrt(3)) / (3 * 3) = 2 sqrt(6) / 9

Both are 1 - sqrt(2) - sqrt(3), but the ratios are different.

And I did find one mistake - when I said that h/r= sqrt(2). I crunched the numbers on a spreadsheet for slant length = 1, and the maximum volume/area is 1/6 when r = h = sqrt(2)/2. I did say earlier in the thread that, when the slant length is fixed (which it is), then the ratio is a maximum when r = h.

h / r = sqrt(2) is correct when either the volume or the lateral surface area is fixed, but in this case, it is the slant length that is fixed.
netzer
Joined: Jan 17, 2019
• Posts: 45
February 9th, 2019 at 12:34:00 PM permalink
Quote: ThatDonGuy

You can't just set a value for r and then assume it's true for all values of r.

Statement 1:

Remember, V varies as r3, and A varies as r2, so V/A varies as r.

Statement 2:

V/A = sqrt(2) / (3 sqrt(3)) = sqrt(6) / 9 = about 1 / 3.67423, which is what you get

I appreciate the amount of thought you are giving this and I'm glad we're coming closer together, but Statement 1 contradicts Statement 2, does it not? Is V/A a function of r or is it a constant?

V varies as r2 and A varies as rR, and since R varies as r, V/R doesn't vary.

While you were writing I added some observations on the Wizard's solution. If you agree he made a mistake we should tell him before he posts it on his puzzle site. Draw straws?
Last edited by: netzer on Feb 9, 2019
OnceDear is a Dear!
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 5944
February 9th, 2019 at 12:48:15 PM permalink
Quote: netzer

Quote: ThatDonGuy

You can't just set a value for r and then assume it's true for all values of r.

Statement 1:

Remember, V varies as r3, and A varies as r2, so V/A varies as r.

Statement 2:

V/A = sqrt(2) / (3 sqrt(3)) = sqrt(6) / 9 = about 1 / 3.67423, which is what you get

I appreciate the amount of thought you are giving this and I'm glad we're coming closer together, but Statement 1 contradicts Statement 2, does it not? Is V/A a function of r or is it a constant?

While you were writing I added some observations on the Wizard's solution. If you agree he made a mistake we should tell him before he posts it on his puzzle site. Draw straws?

You left out "statement 3":
V/A = (8 sqrt(2)) / (3 * 4 sqrt(3)) = (2 sqrt(2) sqrt(3)) / (3 * 3) = 2 sqrt(6) / 9
This is in line with statement 1 - that V/A varies as r, and is not a constant

And I don't see what mistake Wizard made; as I said when I corrected myself, the V/A ratio is a maximum for a given slant length when h = r, which is what he has (and you agreed with this) in his solution.

However, I think there is an error in his solution, as I get PI (2 - sqrt(2)) instead of PI/2 as the angle to cut out of the tortilla.
netzer
Joined: Jan 17, 2019
• Posts: 45
February 9th, 2019 at 1:22:01 PM permalink
Quote: ThatDonGuy

And I don't see what mistake Wizard made; as I said when I corrected myself, the V/A ratio is a maximum for a given slant length when h = r, which is what he has (and you agreed with this) in his solution.

I agree that that is what he wrote but I do not agree that it is correct. Also, I think the Wizard will see the error immediately.

Quote: ThatDonGuy

However, I think there is an error in his solution, as I get PI (2 - sqrt(2)) instead of PI/2 as the angle to cut out of the tortilla.

That's 1.84030 radians , or 105.4414 degrees. A little wide, I think.
OnceDear is a Dear!
netzer
Joined: Jan 17, 2019
• Posts: 45
February 11th, 2019 at 6:07:19 AM permalink
I'd like to see if we can develop a consensus on a formula as a necessary first step to getting the right answer. You don't have to derive it; just look it up in any reference source on geometric figures.

What is the formula for the surface area of a right circular cone, not including the base?
Let S be the surface area, r be radius of the base, and R the slant height: the distance from any point on the perimeter of the base all the way up the side to the apex. The area can be expressed in terms of r and R. You don't need to know the height of the cone.

S = ? Anybody?
OnceDear is a Dear!
Wizard
Joined: Oct 14, 2009
• Posts: 25947
February 11th, 2019 at 10:24:08 AM permalink
Quote: netzer

What is the formula for the surface area of a right circular cone, not including the base?

Solving for the surface area of a cone is surprisingly tricky.

Not including the base, the formula is pi*radius*slant height.

Without using that formula, what is the surface area of a cone, not including the base, for a cone of height of 12 and radius of 5?
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan