## Poll

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**7 members have voted**

February 1st, 2019 at 7:00:37 AM
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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.

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.

It's not whether you win or lose; it's whether or not you had a good bet.

February 1st, 2019 at 9:01:10 AM
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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).

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).

February 1st, 2019 at 9:14:20 AM
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10:10?

100% risk of ruin

February 1st, 2019 at 11:14:26 AM
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I think I may have asked the problem badly. Give me a moment.

It's not whether you win or lose; it's whether or not you had a good bet.

February 1st, 2019 at 3:26:29 PM
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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.

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.

It's not whether you win or lose; it's whether or not you had a good bet.

February 1st, 2019 at 4:00:48 PM
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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.

February 1st, 2019 at 4:20:22 PM
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I agree with the same formula.Quote:unJon...A =....V = ....Hmmm....so I did something wrong too I guess.

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.

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.

February 1st, 2019 at 4:41:40 PM
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Quote:charliepatrickI 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.

February 1st, 2019 at 5:32:08 PM
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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:

How about the other extreme of a radius of 0.99?

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

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.

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

lateral length | 1 |

radius | 0.01 |

height | 0.999949999 |

volume | 0.000104715 |

lateral surface area | 0.031415927 |

Ratio volume to surface area | 0.003333167 |

How about the other extreme of a radius of 0.99?

lateral length | 1 |

radius | 0.99 |

height | 0.14106736 |

volume | 0.144785658 |

lateral surface area | 3.110176727 |

Ratio volume to surface area | 0.046552229 |

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

lateral length | 1 |

radius | 0.5 |

height | 0.866025404 |

volume | 0.226724921 |

lateral surface area | 1.570796327 |

Ratio volume to surface area | 0.144337567 |

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.

It's not whether you win or lose; it's whether or not you had a good bet.

February 1st, 2019 at 5:56:27 PM
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Quote:WizardSo 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.