Expert: Steve Holleran - 3/19/2007

Question
Hi,
I tried to find someone on here who can answer questions about geometry, but unfortunitly the one expert that can answer geometry questions is "on vacation" so I can not ask her. I am hoping that perhaps you could help me solve this problem.
The question is this:
How many perfectly spherical scoops of ice cream could melt into this cone without overflowing? Assume the diameter of the scoops is the same as the diameter of the cone.
The diameter of the waffle cone is 3.5 in. and its height is 8 in.

My question is, how would you solve this??? I really have no idea what to do, and I would greatly appreciate it if you could tell me what I need to do in order to solve this question.
Thanks!!!!
David

Answer
Hi David,

Okay, I think I can help with this one.  Basically we are dealing with a Volume issue here.

The volume of a cone is given by V = pi/3 * r^2 * h
where r is the radius of the cone and h is the height.

Here, we have a cone with diameter = 3.5, so r = 3.5/2 or 1.75.  So, the cone's volume will be :

   V(cone) = pi/3 * 1.75^2 * 8

Don't bother to work this out--we 'll use it later.

Now, if each scoop is a sphere with diameter also 3.5, then each scoop has a radius of 1.75.  The volume of a sphere (scoop) is V = 4/3 * pi * r^3, so here we have:

 V(scoop) = 4/3 * pi * 1.75^3 .

Since we want to know how many of the scoop volumes will fit into the cone volume, just divide:

V(cone) / V(scoop) =

  [pi/3 * 1.75^2 * 8]  /  [4pi/3 * 1.75^3]

Think of each side as a fraction over 3:

  [pi * 1.75^2 * 8]/3   divided by [4 * pi * 1.75^3]/3

So when dividing fractions, invert the second and multiply:

  [pi * 1.75^2 * 8]/3  *  3 / [4 * pi* 1.75^3]

I think you'll see when you write it out that :

the 3's cancel out;  the pi's cancel out; two powers of 1.75 on the left cancel two powers on the bottom right; and the 4 will divide into the 8 twice.  That should leave you with 2 / 1.75 = 200/175 = 1 25/175  = 1  1/7

So you should be able to fit 1 1/7 scoops into the cone.

I hope this is what you needed.

Good luck.
Steve Holleran