EMAT 6600

Problem Solving

Maximum Volume of a Cone

A **circular disc** of radius R is used to make a **Cone**
by removing a sector with angle ø and then joining the edges where the sector
was removed. What is the **maximum volume** that such a cone can attain?
That is, what angle ø for a disc of radius R would you remove to make a cone of
maximum volume?

Because the circumference will miss a sector the circumference C = 2pR - qR. Now to find the radius r of the cone when the ends of the sectors are joined, we have

2pR - qR = 2pr

R - qR/2p = r

R(1 - q/2p) = r

In order to find the volume of the cone, we will need the height of the cone.

Let us construct the height of the cone, h, which is perpendicular to r.

h = sqrt(R^{2}
- r^{2})

h = sqrt[R^{2}
- R^{2}(1 - q/p + q^{2}/4p^{2}))]

h = R sqrt(q/p - q^{2}/4p^{2})

Now, we write the volume V as function of q, while R is fixed.

V = pr^{2}h/3

V(q) = p( R(1
- q/2p))^{2}(R
sqrt(q/p - q^{2}/4p^{2}))/3

V’(q) =
1/3(pR^{3})((
2p
- q)/2p^{2})[
1/sqrt(q/p - q^{2}/4p^{2})]

Let V’(q) = 0 and we have

q = 2p, at which point the volume is zero. To find the optimum volume let us compare our answers in a spreadsheet.