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 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 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(R2 - r2)

h = sqrt[R2 - R2(1 - q/p + q2/4p2))]

h = R sqrt(q/p - q2/4p2)

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

V = pr2h/3

V(q) = p( R(1 - q/2p))2(R sqrt(q/p - q2/4p2))/3

V(q) = 1/3(pR3)(( 2p - q)/2p2)[ 1/sqrt(q/p - q2/4p2)]

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.


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