# Finding the maximum volume of a cone

## Homework Statement

In England, you can purchase fish and chips for a reasonable price. The reason it is so reasonable is because they give you no silverware, nor a plate. They just roll up a piece of paper in a cone and toss your food in. The vendors need to roll the cone in a perfect size, not too fat nor too skinny. Find out how to optimize the volume of the cone. For modeling purposes, assume that the piece of paper is a circle of radius 5 inches, and that we are cutting a wedge out of it whose central angle is Θ. Find the maximum volume of this cone.

## Homework Equations

A = 2πr

i need to find out a formula to subtract the cut out wedge of the circle from the rest of the circle.

## The Attempt at a Solution

I am stuck trying to find a main formula. I would think that after i get that formula, i find the derivative and then set it equal to zero, and solve.

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Mark44
Mentor
Your relevant equation is irrelevant. That's not the area of a circle; it's the circumference. Some actually relevant equations that you should look up are the area of a circle, the volume of a cone, and the area of a sector of a circle. See if you can find those in your textbook.

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ok, i found out the volume forumula
i divided the cone into 2 right triangles, using 5 as the hypotenuse, and r as the radius of the base of the cone.

so i have

V = (1/3)(pi)(square root of 25 - x squared)(x squared)

i am working on finding the derivative but there is so much work with the product rule, chain rule, and quotient rule, i think i may have screwed up somewhere

so far i am at,

dV/dx = [(1/3)(pi)][(1/2)(25-x^2)^(-1/2) *-2x] * 2x + x^2 * (derivative of first part)

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Mark44
Mentor
ok, i found out the volume forumula
i divided the cone into 2 right triangles, using 5 as the hypotenuse, and r as the radius of the base of the cone.
This really doesn't make any sense. A cone is a 3D object. How can you divide it up into two triangles, which are 2D objects.
so i have

V = (1/3)(pi)(square root of 25 - x squared)(x squared)

i am working on finding the derivative but there is so much work with the product rule, chain rule, and quotient rule, i think i may have screwed up somewhere
You're jumping into this problem without setting yourself up with the formulas you need. The volume of a right circular cone (what you're dealing with) is V = 1/3 * pi * r^2 * h. That's one of your relevant formulas.

Another one is the area of a sector of a circle, sort of like a wedge of pie of any size. This formula is A = (1/2)$\theta r^2$, where $\theta$ is measured in radians.

V = 1/3 * pi * r^2 * h

thats what i had, but i substituted in h in terms of x but using pythagorean.

x^2 + h^2 = 5 ^2

h^2 = 25 - x^2
h = square root of (25 - x^2)

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Mark44
Mentor
Even with the picture, this formula still doesn't make any sense. You used 5 for the radius. What does x represent? What does sqrt(25 - x^2) represent?
physjeff12 said:
V = (1/3)(pi)(square root of 25 - x squared)(x squared)
You're also not taking into account that the circular piece of paper has a wedge cut out of it.

Even with the picture, this formula still doesn't make any sense. You used 5 for the radius. What does x represent? What does sqrt(25 - x^2) represent?

You're also not taking into account that the circular piece of paper has a wedge cut out of it.
the sqrt(25 - x^2) represented h. x represented the radius (r in the picture). 5 represented the S on the picture.

i thought the A = (1/2)theta r^2 would be irrelevent because i'm trying to find volume, not area?

Mark44
Mentor
OK, now it makes sense. It would have helped me understand better if you had explained what your were doing to get the formula you showed.

The 5 dimension in the drawing is what is called the slant height of the cone. Here's your formula again, this time with r instead of x. r matches your diagram and suggests what it represents, the radius of the cone.

$$V(r)~=~(1/3)\pi r^2(25 - r^2)^{1/2}$$

Now you want to take the derivative. You'll need to use the product rule and then the chain rule, in that order.

ok cool.

so v'(r) = (1/3)pi(r^2)*(-1/(2*sqrt(25 - x^2)) + [(25-r^2)^(1/2)]*[(2/3)*pi*r]

correct?

Mark44
Mentor
No. In the chain rule, you forgot the factor that comes from d/dr(25 - r^2). Also, since V is now a function of r, then V'(r) shouldn't have any x's floating around in it.

Ok so if you don't want to use calculus you can AM-GM it.
So the volume of the cone would be pi(r^2)h/3.
Arc length=circumference of cone=theta/360*10pi
so the radius of the cone=(theta/360*10pi)/2pi = theta/72
the slant height of the cone is 5 so the height is sqrt(25-(theta/72)^2)
so the volume of the cone is pi(theta/72)^2*sqrt(25-(theta/72)^2)/3
now all we have to do is find the maximum of this equation
we can get rid of the /3 and pi because those are constant
if we let x=(theta/72)^2 then the formula becomes (x)sqrt(25-x)
much simpler, eh?
this formula is maximized when (x^2)(25-x) is maximized (for obvious reasons)
so now we can use AM-GM on the formula (x^2)(50-2x), which is maximized when the three terms are closest to eachother
so we let x=x=50-2x and 50/3
going back to what we defined x to be, x=50/3=(theta/72)^2
theta=293.939 degrees and you're done!