# stadium popcorn problem

• Nov 20th 2005, 05:21 PM
arollyson
A piece of heavy stock paper is cut into a circle with radius 4". The paper is cut from one edge to the center and shaped into a cone shaped holder. What is the maximum volume of the resulting cone?

That is the exact wording of the problem and i don't even know where to start.
• Nov 27th 2005, 02:47 PM
BubbleBrain_103
the key to these types of problems is to identify a "control variable"... something you have control over, that you can vary, and that will determine the value of the thing you are trying to maximize.

In this case I chose a certain angle to be my control variable. Draw an upside down cone, and let alpha be the angle made between the side of the cone and the vertex (the straight up and down line from the top of the cone to the bottom). Then alpha can vary anywhere between Pi/2 (if it is lying perfectly flat) and zero (if it wrapped around itself completely, and standing straight up). Our job then is to find the alpha that maximizes the volume of the cone.

So our next step is to write the volume of the cone as a function of alpha. I couldn't remember any of the formulas of the volume of the cone, so I just figure it out using integration. Let r0 be the radius of the base of the cone, and h be the height of the cone. Then verify that volume can be given as:

V = (Pi/3)*h^3*tan(alpha)^2

But since h = r0 / tan(alpha),

V = (Pi/3)*r0^3 / tan(alpha)

Finally, what is r0 in terms of alpha? Let R be the radius of the circle you started with (which in your case is 4 inches). Then notice that the side of the cone will also be of length R. Then we have also that r0 = sin(alpha)*R.
Thus:

V = (Pi/3)*(sin(alpha)*R)^3 / tan(alpha)
= (Pi/3)*R^3*cos(alpha)*sin(alpha)^2

Thus we have volume as a function of alpha. Now all thats left is to find the maximum of this function as alpha varies from 0 to Pi/2. Differentiate:

V' = -1/3*Pi*R^3*sin(alpha)^3+2/3*Pi*R^3*cos(alpha)^2*sin(alpha)

Setting V' = 0, we find three solutions, corresponding to where sin(alpha) = 0, and when tan(alpha) = +/- sqrt(2), i.e.,

alpha = 0, arctan(sqrt(2)), -arctan(sqrt(2))

only one of these, arctan(sqrt(2)) sits between 0 and Pi/2. Verify by the second derivative test that this is indeed a maximum of the function V. Thus our maximum possible volume is:

V (R = 4, alpha = arctan(sqrt(2)) ) = 128*Pi*sqrt(3)/27
• Nov 28th 2005, 04:32 AM
ticbol
Quote:

Originally Posted by arollyson
A piece of heavy stock paper is cut into a circle with radius 4". The paper is cut from one edge to the center and shaped into a cone shaped holder. What is the maximum volume of the resulting cone?

That is the exact wording of the problem and i don't even know where to start.

Here is one way.

The radius of the circular sheet, 4", will become the slanting height of the cone.

Cone.
Let x = radius
y = height

V = (1/3)(pi)(x^2)(y)

Where, by Pythagorean theorem,
4^2 = x^2 +y^2
y = sqrt(16 -x^2)
So,
V = (pi/3)(x^2)sqrt(16 -x^2) --------------(1)
V = (pi/3)sqrt(16x^4 -x^6) -----------(2)
Differentiate both sides of (2) with respect to x,
dV/dx = [(pi/6) / sqrt(16x^4 -x^6)]*(64x^3 -6x^5)
Set dV/dx to zero, for maximum V,
0 = [(pi/6) / sqrt(16x^4 -x^6)]*(64x^3 -6x^5)
0 = 64x^3 -6x^5
Divide both sides by 2x^3,
0 = 32 -3x^2
x^2 = 32/3 ------at max V.

Substitute that into (1)
max V = (pi/3)(32/3)sqrt(16 -32/3)
max V = (32pi/9)sqrt(16/3)
max V = (32pi/9)(4sqrt(1/3))
max V = (128pi/9) / sqrt(3)
max V = (128pi/27)sqrt(3) cubic inches
Or,
max V = 25.8 cu.in. -------answer.