Finding the maximum volume of a cone

[physjeff12]

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.

 

[Mark44]

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.

 

[physjeff12]

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)

 

[Mark44]

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.

 

[physjeff12]

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)

http://z.about.com/d/math/1/5/t/L/conerr.jpg

 

[Mark44]

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?

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.

 

[physjeff12]

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 irrelevant because I'm trying to find volume, not area?

 

[Mark44]

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.

 

[physjeff12]

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]

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.

 

[dingman08]

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 each other
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!