Find the Magnitude of Theta to Maximize Volume of a Cone

[vr0nvr0n]

Homework Statement

A sector with central angle θ is cut from a circle of radius R = 6 inches, and the edges of the sector are brought together to form a cone. Find the magnitude of θ such that the volume of the cone is a maximum.

Homework Equations

Volume of a Cone = ⅓ * π * r² * h

Area of a Sector of a Circle = R²/2 * Θ

The Attempt at a Solution

So, this problem is a lot like other cone/volume problems I have done before, but I have never had to find the magnitude of Theta before.

Normally, I would use the Pythagorean Theorem to determine what to sub in for "r" in the Volume of a Cone equation (in this case, that would be 36-h² = r which can be determined by using the Pythagorean theorem as R² = r² + h². In this problem "R" always = 6).

I know I can use this information to determine the critical numbers of "h," but I can't seem to determine when the Area of a Sector of a Circle would come into play. It is related to "R," which is always 6 in this problem, so I determined that this equation could be simplified to 18*Θ.

I attempted a solution by finding the critical numbers as I mentioned above and plugged those into Θ, but that doesn't really make sense to me. Any guidance would be appreciated. I don't want to be given the answer, obviously -- This isn't the forum for that. But, any hint or clue would be amazing.

Thank you so much.

 

[mfb]

To find theta with your optimal value of r (or h), think about the circumference of the cone base. Where does it come from? How is the length of this related to the original circle with a section cut out?

 

[vr0nvr0n]

To find theta with your optimal value of r (or h), think about the circumference of the cone base. Where does it come from? How is the length of this related to the original circle with a section cut out?

Is r = sin(Θ½)? I had thought of that, but I think I confused myself by solving for 18Θ early on.

 

[mfb]

Is r = sin(Θ½)? I had thought of that, but I think I confused myself by solving for 18Θ early on.

r cannot be the sine of anything, that wouldn't work in terms of dimensions - there is no reason why r should be smaller than 1.
I have no idea where 18Θ comes from.

 

[vr0nvr0n]

r cannot be the sine of anything, that wouldn't work in terms of dimensions - there is no reason why r should be smaller than 1.
I have no idea where 18Θ comes from.

The 18Θ came from the Area of a Sector of a Circle formula (R²*½*Θ).

 

[mfb]

That area is related to the cone surface area, which you don't need.

 

[Ray Vickson]

Homework Statement

A sector with central angle θ is cut from a circle of radius R = 6 inches, and the edges of the sector are brought together to form a cone. Find the magnitude of θ such that the volume of the cone is a maximum.

Homework Equations

Volume of a Cone = ⅓ * π * r² * h

Area of a Sector of a Circle = R²/2 * ΘThank you so much.

Avoid plugging in $R = 6$ until the last minute; that way, you can tell where the various numbers are coming from. Not only that, you get a more general formula.

Anyway, if you cut out a sector of angle $\theta$ from a circle of radius $R$, then fold it up into a cone, what is the radius of the base of the cone (call it $r$)?

Now you have the cone's base $r$ and slanted-side length $R$, so you can determine its height $h$ and thus find its volume.

 

[vr0nvr0n]

Avoid plugging in $R = 6$ until the last minute; that way, you can tell where the various numbers are coming from. not only that, you get a more general formula.

Anyway, if you cut out a sector of angle $\theta$ from a circle of radius $R$, then fold it up into a cone, what is the radius of the base of the cone (call it $r$)?

Now you have the cone's base $r$ and slanted-side length $R$, so you can determine its height $h$ and thus find its volume.

Hmm... 2π - $\theta$?

So, am I better off approaching this in Pythagorean form using $r$(2π - $\theta$); $R$; and $h$² = $R$² - $r$²?

 

[haruspex]

Hmm... 2π - θ?

That cannot be a radius.

The circumference of the original full circle is 2πR, right?
After the sector is removed, what length of that remains?
When formed into a cone, what happens to that part-circle?

 

[vr0nvr0n]

That cannot be a radius.

The circumference of the original full circle is 2πR, right?
After the sector is removed, what length of that remains?
When formed into a cone, what happens to that part-circle?

Sorry, I wrote the correct term further down in my post $r$(2π - $\theta$).

So, if that's correct, I should be working with: ⅓(2π$r$ - $\theta$ $r$)²π√($R$² - $r$²)

 

[haruspex]

wrote the correct term further down in my post r(2π - θ).

What is that the answer to?
Ray defined r as the radius after forming the cone. You want an equation of the form r= some function of R and θ.

 

[mfb]

Sorry, I wrote the correct term further down in my post $r(2\pi - \theta)$.

That is the wrong radius.

 

[vr0nvr0n]

What is that the answer to?
Ray defined r as the radius after forming the cone. You want an equation of the form r= some function of R and θ.

Ray wrote: "Anyway, if you cut out a sector of angle θ from a circle of radius R, then fold it up into a cone, what is the radius of the base of the cone (call it r)?"

The radius of the base of the cone would definitely be $r$² = $R$² - $h$². To make this relate to the angle of $\theta$, the circumference of the base of the cone would be 2π$r$ - $\theta$ $r$, so am I plugging in √($R$² - $h$²) for $r$?

 

[Ray Vickson]

Ray wrote: "Anyway, if you cut out a sector of angle θ from a circle of radius R, then fold it up into a cone, what is the radius of the base of the cone (call it r)?"

The radius of the base of the cone would definitely be $r$² = $R$² - $h$². To make this relate to the angle of $\theta$, the circumference of the base of the cone would be 2π$r$ - $\theta$ $r$, so am I plugging in √($R$² - $h$²) for $r$?

OK, if you know $h$ you can figure out $r$. However, who tells you the value of $h$? All you know is $\theta$.

 

[vr0nvr0n]

OK, if you know $h$ you can figure out $r$. However, who tells you the value of $h$? All you know is $\theta$.

I don't know $h$ or $r$ (except in the formulaic terms mentioned above), I am looking for $\theta$, and all I am given is $R$ (the radius of the circle from which the sector is cut).

 

[Ray Vickson]

I don't know $h$ or $r$ (except in the formulaic terms mentioned above), I am looking for $\theta$, and all I am given is $R$ (the radius of the circle from which the sector is cut).

Exactly; but you can work out a formula for $r$ in terms of $R$ and $\theta$. (You tried that before, but made some errors.)

In the end, you will get a formula for the volume $V$ in terms of $R$ and $\theta$, and since $R$ is just a constant input parameter, the only variable involved is $\theta$.

 

[haruspex]

circumference of the base of the cone would be 2πr- θr

Eh? You have a circle radius r. What is its circumference?

 

[mfb]

@vr0nvr0n: You are confusing yourself by trying too many steps at once.

What is the circumference of the base of the cone - a full circle with radius r?

Going back to the original circle with radius R with a section cut out: How large was the original circumference, how large is the remaining part of the circumference?