## Poll

2 votes (28.57%) | |||

1 vote (14.28%) | |||

1 vote (14.28%) | |||

No votes (0%) | |||

1 vote (14.28%) | |||

2 votes (28.57%) | |||

1 vote (14.28%) | |||

No votes (0%) | |||

No votes (0%) | |||

3 votes (42.85%) |

**7 members have voted**

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 r

Let A = the lateral surface area = PI r sqrt(r

Let R = the desired ratio = V/A = rh / (3 sqrt(r

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

R = x r

= x r / (3 sqrt(1 + x

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

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

Note that 1 > 0 -> x

-> sqrt(x

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

d

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

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

This is zero when sqrt(d

-> d

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

h = d / r

R = hr / (3 sqrt(r

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

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

= d / 3 * r / sqrt(r

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

This is zero when sqrt(r

-> r

h = d / 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?)

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?)

February 5th, 2019 at 4:06:41 PM
permalink

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?)

The race is not always to the swift, nor the battle to the strong; but that is the way to bet.

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.

February 5th, 2019 at 6:19:16 PM
permalink

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)

February 5th, 2019 at 6:34:23 PM
permalink

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)

The race is not always to the swift, nor the battle to the strong; but that is the way to bet.

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.

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

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*hr

r = SQRT(R

V = Pi*h(R

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

V'(h) = Pi*R

This derivative is much simpler than with some other approaches.

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

R

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

let h = r*SQRT(2), h

h

2r

h = R/SQRT(3)

Q.E.D.

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.

OnceDear is a Dear!

February 8th, 2019 at 6:57:33 PM
permalink

I reworded the problem as follows:

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: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.

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

February 9th, 2019 at 7:56:23 AM
permalink

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.

Last edited by: netzer on Feb 9, 2019

OnceDear is a Dear!

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 - 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.