Who Loves an Ice Cream Cone?

SydneyRoverP6B

SydneyRoverP6B

Well-Known Member
Staff member
Have you ever wondered while enjoying an ice cream as to whether or not the shape of the cone would influence how much ice cream it would hold? As it turns out, it most certainly does.

If you have a right circular cone of radius r and height h and the slant height of the cone is of length L, then the relationship between all three has a direct bearing on the volume of the cone. So what would be the radius r of a cone expressed in terms of the slant height L ( what fraction of L does r need to be) in order for the cone to always provide the maximum possible volume? Can you prove that your calculation is correct?

Ron.
 
Hint...The answer is not a rational number,..it involves a surd.

A question like this is an illustration of the beauty of mathematics. No numbers are involved in the question, and yet the derivation of the answer provides an outcome which is applicable to every cone of any dimension.

Ron.
 
SydneyRoverP6B said:
Hint...The answer is not a rational number,..it involves a surd.

A question like this is an illustration of the beauty of mathematics. No numbers are involved in the question, and yet the derivation of the answer provides an outcome which is applicable to every cone of any dimension.

Ron.
Click to expand...

I've got an inkling/feeling it's sqrt (2/3).
 
darth sidious said:
SydneyRoverP6B said:
Hint...The answer is not a rational number,..it involves a surd.

A question like this is an illustration of the beauty of mathematics. No numbers are involved in the question, and yet the derivation of the answer provides an outcome which is applicable to every cone of any dimension.

Ron.
Click to expand...

I've got an inkling/feeling it's sqrt (2/3).
Click to expand...

My numerical comp's seem to agree with this.

How I got there?

It's a right circular cone, i.e. it's a cone whose apex is inline with the centre point of the circular opposite end.

The slant height L is given by, from pythagoras, L^2 = r^2 + h^2

so, r^2 = L^2 - h^2

The volume of a cone is (pi * r^2 * h) /3, but since r^2 = L^2 - h^2, we can say

V = (pi * (L^2 - h^2) * h) /3 = pi/3 * (h * L^2 - h^3)

dV/dh = pi/3 * (L^2 - 3 * h^2), so dV/dh = 0 if L^2 = 3 * h^2 (or L = sqrt {3} * h)

and d^2/dh^2 {V} = -2 * pi * h (as pi and h are both positive, it must be that d^2/dh^2 {V} < 0, so maximum at the point in question)

If L^2 = 3 * h^2, then h^2 = L^2 / 3, and if if L^2 = r^2 + h^2, then r^2 = L^2 - h^2 = L^2 - ((L^2) / 3) = (2/3) * L^2

Therefore r = sqrt {(2/3) * L^2} = sqrt {(2/3)} * sqrt{ L^2} = sqrt {(2/3)} * L = ~ 0.8165 * L

r = sqrt {(2/3)] * L, or as asked, r/L = sqrt {(2/3)} = ~ 0.8165
 
Hello Darth,

Yes indeed...that is the correct answer... :D My approach was slightly different in that I chose to express h as a function of r where as you chose to make r a function of h. Your method meant that you had to substitute back in and solve for r where as that was not required in my case. Either way the answer is the same....r = Lsqrt(2/3) which is Lsqrt(6)/3 if you rationalise the denominator.

When you think of it...obtaining a number when there were none given to start with, and that number applies to all cones if you want to know what dimensions will allow the volume to be always a maximum.

The world can thank Sir Isaac Newton (1642 - 1727) and Gottfried Wilhelm Leibniz (1646 - 1716) for developing Calculus, both independently but in essence thinking the same thoughts. One of the supreme accomplishments of the human interlect. Truly amazing!!

Ron.
 
SydneyRoverP6B said:
Hello Darth,

Yes indeed...that is the correct answer... :D My approach was slightly different in that I chose to express h as a function of r where as you chose to make r a function of h. Your method meant that you had to substitute back in and solve for r where as that was not required in my case. Either way the answer is the same....r = Lsqrt(2/3) which is Lsqrt(6)/3 if you rationalise the denominator.

When you think of it...obtaining a number when there were none given to start with, and that number applies to all cones if you want to know what dimensions will allow the volume to be always a maximum.

The world can thank Sir Isaac Newton (1642 - 1727) and Gottfried Wilhelm Leibniz (1646 - 1716) for developing Calculus, both independently but in essence thinking the same thoughts. One of the supreme accomplishments of the human interlect. Truly amazing!!

Ron.
Click to expand...

I'm glad I managed to answer it!

Can I ask that you show me your method? There was a reason for me to use r as a function of h... but I want to see if my reason was justified or not...

Good problem, really opened my eyes!
 
darth sidious wrote...
Can I ask that you show me your method? There was a reason for me to use r as a function of h... but I want to see if my reason was justified or not...
Click to expand...

Hello Darth,

Sure thing,..i'll write it up and post it.

Ron.
 
A right circular cone of radius ‘r’ and height ‘h’ has a slant height ‘L’. What fraction of ‘L’ does ‘r’ need to be in order for the cone to always contain a maximum volume?

The volume of a cone is given by V = (1/3)(PIr^2h)

From Pythagoras, L = (r^2 + h^2)^1/2 thus

L^2 = r^2 + h^2. Now express ‘h’ as a function of ‘r’.

h(r) = (r^2 – L^2)^1/2

Substitute h(r) into V to form V(r)

V(r) = (1/3)PIr^2(r^2 – L^2)^1/2

Now differentiate V(r) explicitly as a function of ‘r’…dV/dr to obtain V’(r)

V’(r) = (2/3)PIr(r^2 – L^2)^1/2 + (1/3)PIr^3(r^2 – L^2)^-1/2

Equate V’(r) to zero and solve to obtain the critical points. In this case there is only one.

r = (1/3)L(6)^1/2 or sqrt(6)/3 using an alternative terminology. To validate the result I used the first derivative test. Substitute values either side of sqrt(6)/3 back into V’(r) and test the gradient either side of the critical point. You could also choose a value for L and test all three.

How do you find my method Darth? What was your concern about making ‘h’ a function of ‘r’?

I’ll post up another question that will be slightly harder still….a gradual process of complexity... :D

Ron.
 
My reason for using using r instead of h was because the formula for volume uses r^2, which meant not having to use square roots. I found that easier than carrying square roots all over the place!

I have tried your method also, I get it to work too. Only one nit-pick...

SydneyRoverP6B said:
From Pythagoras, L = (r^2 + h^2)^1/2 thus

L^2 = r^2 + h^2. Now express ‘h’ as a function of ‘r’.

h(r) = (r^2 – L^2)^1/2
Click to expand...

Don't you mean h(r) = (L^2 – r^2)^1/2 ?
 
Hello Darth,

Yes indeed....a typo on my part. If that were correct you would have the square root of a negative and that brings in the wonderful world of complex numbers.. :D

Ron.
 
LMAO yes!

Mind you, I used a lot of them in my university days

Z = R + j(wL - 1/wC) ohms, etc

Engineers use j instead of i!

Awesome stuff, stuff that would make younger generations' hair turn white! :p
 
Hello Darth,

Impedance is equal to resistance + reactance....inductive and capacitive. What is wrong with keyboard manufacturers,...leaving off an omega key..still w is not too far off.. :LOL:

It is all great stuff.!! and Engineers certainly do use j.

Rectangular or polar coordinates including complex exponential form for the latter..Ae^j0 (i'll use 0 in place of phi)

Ron.
 
SydneyRoverP6B said:
Hello Darth,

Impedance is equal to resistance + reactance....inductive and capacitive. What is wrong with keyboard manufacturers,...leaving off an omega key..still w is not too far off.. :LOL:

It is all great stuff.!! and Engineers certainly do use j.

Rectangular or polar coordinates including complex exponential form for the latter..Ae^j0 (i'll use 0 in place of phi)

Ron.
Click to expand...

Ooh yes, good stuff! :)

Some older books use cis(theta), I can't remember if it was/is mathematically correct to use degrees with cis. It's not mathematically correct to use degrees with e^j{theta}, but many engineers DO! (Myself included)
 
Hello Darth,

I'll start another thread on complex numbers and post on all this great stuff over there.

Ron.
 
SydneyRoverP6B said:
marty wrote,..
All this just for an ice cream!!
Click to expand...

The mathematics is applicable to all cones for any purpose,..the ice cream is just a treat.

Ron.
Click to expand...

I seem to recall there was a method of evaluating the area of a doughnut too! Maybe Marty would like that instead? :p
 
darth sidious wrote,..
I seem to recall there was a method of evaluating the area of a doughnut too!
Click to expand...

Hello Darth,

That will be a Torus. To find the surface area you can revolve a circle such as (x - b)^2 + z^2 = a^2 (a < b) in the xz plane around the z axis. Will be a double integral in polar co ordinates,..0 to 2PI the limits for both integrals, integrating with respect to theta and phi. So the integrand will be a(b + acos(phi))d(theta) d(phi). This will give a surface area of 4PI^2ab. I have not used parentheses as ab are not part of the exponent.

Ron.
 
SydneyRoverP6B said:
darth sidious wrote,..
I seem to recall there was a method of evaluating the area of a doughnut too!
Click to expand...

Hello Darth,

That will be a Torus. To find the surface area you can revolve a circle such as (x - b)^2 + z^2 = a^2 (a < b) in the xz plane around the z axis. Will be a double integral in polar co ordinates,..0 to 2PI the limits for both integrals, integrating with respect to theta and phi. So the integrand will be a(b + acos(phi))d(theta) d(phi). This will give a surface area of 4PI^2ab. I have not used parentheses as ab are not part of the exponent.

Ron.
Click to expand...

Thanks! :)
 
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