Thread: Nail in Can

Hi All,
A similar question to the ladder one i posted earlier. Now that i have an idea from what i was shown earlier, i've tried to attempt it.
The Question says: A 20cm nail fits inside a cylindrical can. Find the maximum radius of 3 balls that will fit exactly inside the can?

I have drawn the diag to the best of my ability. At first i thought the nail stood upright in the can.
So i the length of the can is 20cm and to fit 3 balls in =

20/6 = radius = 3.33cm (wrong ans)

But, using the help i got from the ladder question, I then thought the nail is probably lying diagonally (AB) and then i tried splitting the diag to make right angel triangles.

*h = Radius

(AO)^2 + h^2 = 10^2 > (AO)^2 + h^2 = 100

(2AO)^2 + (2h)^2 = 20^2 > (4AO)^2 + (4h^2) = 400

Subtracting the two doesn't work..
I tried making (AO^2) the subject = AO^2 = 100 - h^2
and sub it into the second equation...still nothing :-(

Hello, BobBali!

Actually, this is not a max-min calculus problem at all!

A 20 cm nail fits inside a cylindrical can.
Find the maximum radius of 3 identical balls that will fit exactly inside the can?

I have drawn the diagram to the best of my ability.

I thought the nail is lying diagonally (AB) and I split the diag to make right triangles.

*h = Radius

. These two equations are equivalent!

Consider the largest right triangle.

The height is

The base is

The hypotenuse is


Therefore: .

. . . . . . . . .

Hi Soroban, Thank you! I substituted the above value of [tex]sqrt{10}=\pm 3.16[\math]

into the equation constructed to cross check if i get the correct hypotenuse of 20cm and i obtained 19.98[tex]\simeq[\math] 20.

However, in the textbook the ans is 3.03 and when that is substituted for h = 19.16 , which is a large margin of error, must be an error in the txtbook...
Thank you ;-)