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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.

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).

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.

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.

I agree with the same formula.Quote:unJon...A =....V = ....Hmmm....so I did something wrong too I guess.

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.

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:

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.

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.

I agree with this.Quote:ThatDonGuyQuote: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.

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 |

Quote:WizardOkay, 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:

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.

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(d

^{2}- r

^{2})

Volume v = PI/3 * r

^{2}* h = PI/3 * r

^{2}sqrt(d

^{2}- r

^{2})

Lateral Surface Area a = PI * r * d

v/a = r sqrt(d

^{2}- r

^{2}) / (3d)

The first derivative of v/a with respect to r = sqrt(d

^{2}- r

^{2}) - r / sqrt(d

^{2}- r

^{2})

This is zero when d

^{2}= 2 r

^{2}-> 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.

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

Quote:WizardOkay, 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.

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.

Quote:unJonA = 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.

Quote:WizardQuote:unJonA = 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?

Quote:WizardI'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 r

^{2}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 - r

^{2}) - r / sqrt(1 - r

^{2})

d

^{2}f / dr

^{2}= -2 / sqrt(1 - r

^{2}) - (sqrt(1- r

^{2}) + r

^{2}/ sqrt(1 - r

^{2})) / (1 - r

^{2})

For r = sqrt(2)/2, r

^{2}= 1/2 -> sqrt(1 - r

^{2}) = sqrt(2)/2

Substitute x for both r and sqrt(1 - r^2):

d

^{2}f / dr

^{2}= -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

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.

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.

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 x = h/r, where r is a constant; h = rx

R = x r

^{2}/ (3 sqrt(r

^{2}+ x

^{2}r

^{2})

= x r / (3 sqrt(1 + x

^{2}))

= r / 3 * 1 / sqrt(1 + x

^{2})

dR/dx = r / 3 * (sqrt(1 + x

^{2}) - x^2 / sqrt(1 + x

^{2})) / (x

^{2}+1)

Note that 1 > 0 -> x

^{2}+ 1 > x

^{2}-> (x

^{2}+ 1) / sqrt(x

^{2}+ 1) > x

^{2}/ sqrt(x

^{2}+ 1)

-> sqrt(x

^{2}+ 1) > x

^{2}/ sqrt(x

^{2}+ 1) -> sqrt(x

^{2}+ 1) - x

^{2}/ sqrt(x

^{2}+ 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, d

^{2}is also constant

d

^{2}= h

^{2}+ r

^{2}->h = sqrt(d

^{2}- r

^{2})

R = hr / (3d) = 1 / (3d) * r sqrt(d

^{2}- r

^{2})

dR/dr = 1 / (3d) * (sqrt(d

^{2}- r

^{2}) - r

^{2}/ sqrt(d

^{2}- r

^{2}))

This is zero when sqrt(d

^{2}- r

^{2}) = r

^{2}/ sqrt(d

^{2}- r

^{2})

-> d

^{2}= 2 r

^{2}-> h

^{2}= d

^{2}- r

^{2}= r

^{2}-> 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 = r

^{2}h; since V is a constant, this is also a constant

h = d / r

^{2}

R = hr / (3 sqrt(r

^{2}+ h

^{2}) = d / r

^{2}* r / (3 sqrt(r

^{2}+ (d

^{2}/ r

^{4}))

= d / 3r * 1 / (sqrt(r

^{2}+ (d

^{2}/ r

^{4}))

= d / 3r * 1 / (sqrt(r

^{6}/ r

^{4}+ d

^{2}/ r

^{4}))

= d / 3 * r / sqrt(r

^{6}+ d

^{2})

dR/dr = d / 3 * (sqrt(r

^{6}+ d

^{2}) - r * 3 r

^{5}/ sqrt(r

^{6}+ d

^{2})) / (r

^{6}+ d

^{2})

This is zero when sqrt(r

^{6}+ d

^{2}) = 3 r

^{6}/ sqrt(r

^{6}+ d

^{2}))

-> r

^{6}+ d

^{2}= 3 r

^{6}->d

^{2}= 2 r

^{6}-> d = sqrt(2) r

^{3}

h = d / r

^{2}= 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.Quote:ThatDonGuyI 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 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 x = h/r, where r is a constant; h = rx

R = x r^{2}/ (3 sqrt(r^{2}+ x^{2}r^{2})

= x r / (3 sqrt(1 + x^{2}))

= r / 3 * 1 / sqrt(1 + x^{2})

dR/dx = r / 3 * (sqrt(1 + x^{2}) - x^2 / sqrt(1 + x^{2})) / (x^{2}+1)

Note that 1 > 0 -> x^{2}+ 1 > x^{2}-> (x^{2}+ 1) / sqrt(x^{2}+ 1) > x^{2}/ sqrt(x^{2}+ 1)

-> sqrt(x^{2}+ 1) > x^{2}/ sqrt(x^{2}+ 1) -> sqrt(x^{2}+ 1) - x^{2}/ sqrt(x^{2}+ 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, d^{2}is also constant

d^{2}= h^{2}+ r^{2}->h = sqrt(d^{2}- r^{2})

R = hr / (3d) = 1 / (3d) * r sqrt(d^{2}- r^{2})

dR/dr = 1 / (3d) * (sqrt(d^{2}- r^{2}) - r^{2}/ sqrt(d^{2}- r^{2}))

This is zero when sqrt(d^{2}- r^{2}) = r^{2}/ sqrt(d^{2}- r^{2})

-> d^{2}= 2 r^{2}-> h^{2}= d^{2}- r^{2}= r^{2}-> 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 = r^{2}h; since V is a constant, this is also a constant

h = d / r^{2}

R = hr / (3 sqrt(r^{2}+ h^{2}) = d / r^{2}* r / (3 sqrt(r^{2}+ (d^{2}/ r^{4}))

= d / 3r * 1 / (sqrt(r^{2}+ (d^{2}/ r^{4}))

= d / 3r * 1 / (sqrt(r^{6}/ r^{4}+ d^{2}/ r^{4}))

= d / 3 * r / sqrt(r^{6}+ d^{2})

dR/dr = d / 3 * (sqrt(r^{6}+ d^{2}) - r * 3 r^{5}/ sqrt(r^{6}+ d^{2})) / (r^{6}+ d^{2})

This is zero when sqrt(r^{6}+ d^{2}) = 3 r^{6}/ sqrt(r^{6}+ d^{2}))

-> r^{6}+ d^{2}= 3 r^{6}->d^{2}= 2 r^{6}-> d = sqrt(2) r^{3}

h = d / r^{2}= 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?)

Quote:unJonI 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.Quote:ThatDonGuyComing 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?)

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!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)

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*hr

^{2}/3

r = SQRT(R

^{2}-h

^{2})

V = Pi*h(R

^{2}-h

^{2})/3

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

V'(h) = Pi*R

^{2}/3 - Pi*h

^{2}after simplification

This derivative is much simpler than with some other approaches.

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

R

^{2}/3 = h

^{2}, h = R/SQRT(3) for maximum volume.

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

let h = r*SQRT(2), h

^{2}= 2r

^{2}

h

^{2}+ r

^{2}= R

^{2}

2r

^{2}+ r

^{2}= R

^{2}

h = R/SQRT(3)

Q.E.D.

Quote:cone problemYou 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?

Here is my answer (PDF)

I welcome all comments, although, to be honest, I'm looking forward to getting past this problem and posting a new one.

Quote:WizardI'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 πr

^{2}h/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.

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 - r

^{2}) = sqrt(1 - t

^{2}/ (4 PI

^{2})) = sqrt(4 PI

^{2}- t

^{2}) / (2 PI)

V = PI/3 r

^{2}h = PI/3 * t

^{2}/ (4 PI

^{2}) * sqrt(4 PI

^{2}- t

^{2}) / (2 PI)

= 1/(24 PI

^{2}) * t

^{2}sqrt(4 PI

^{2}- t

^{2})

A = PI r d = PI t / (2 PI) = t / 2

Ratio R = V/A = 1/(12 PI

^{2}) * t sqrt(4 PI

^{2}- t

^{2})

R' = 1/(12 PI

^{2}) * sqrt(4 PI

^{2}- t

^{2}) - t

^{2}/ sqrt(4 PI

^{2}- t

^{2})

R' = 0 when t

^{2}= 4 PI

^{2}- t

^{2}-> t = sqrt(2) PI

R = 1 / (12 PI

^{2}) * PI sqrt(2) * sqrt(4 PI

^{2}- 2 PI

^{2})

= PI sqrt(2) sqrt(2 PI

^{2}) / (12 PI

^{2})

= 2 PI

^{2}/ (12 PI

^{2}) = 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.

Quote:ThatDonGuyI 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:ThatDonGuyWhat 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 πr

^{2}h/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 πR

^{2}, however it has only 2πr of its original circumference of 2πR left and its area is reduced proportionally so it becomes

πR

^{2}(2πr/2πR) = πrR

It's as simple as that!

Remember, V varies as r

^{3}, and A varies as r

^{2}, so V/A varies as r.

If r = 1, h = sqrt(2), and d = sqrt(3):

V = PI/3 r

^{2}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 r

^{2}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.

Quote:ThatDonGuyYou 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 r^{3}, and A varies as r^{2}, 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 r

^{2}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?

Quote:netzerQuote:ThatDonGuyYou 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 r^{3}, and A varies as r^{2}, 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.

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.

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?

Quote:netzerWhat 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.

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?