1. ## Volume Problem

How many jam tins can fit into this box if they have a diameter of 8cm and are 11cm high?

I have worked out the volume of the box to be $21120\,\textrm{cm}^2$ and the volume of the tin to be $\pi \cdot 4^2 \cdot 11 = 552.92\,\textrm{cm}^2$.
When I divide the box by the cans, I get $\frac{21120}{552.92} = 38.20$ cans.
But the answer should be 30 cans. What have I done wrong?

2. Hello Demonstration

Welcome to Math Help Forum!
Originally Posted by Demonstration
How many jam tins can fit into this box if they have a diameter of 8cm and are 11cm high?

I have worked out the volume of the box to be $21120\,\textrm{cm}^2$ and the volume of the tin to be $\pi \cdot 4^2 \cdot 11 = 552.92\,\textrm{cm}^2$.
When I divide the box by the cans, I get $\frac{21120}{552.92} = 38.20$ cans.
But the answer should be 30 cans. What have I done wrong?
You don't tell us what the dimensions of the box are. But have you taken into account the spaces in between the tins? If they are cylindrical (as I assume they are) then you can't pack them into the box without 'wasting' some space, can you?

3. Assuming the box is specially designed for packing tins:

You must know that cylinderical things will not fit perfectly into square things.
So, using the volume formula of a cylinder is useless. Instead, assume the tins aren't round, but like a block, rectangular block.

Therefore:

$Volume = 8 \times 8 \times 11 = 704cm^3$

Now the volume of the box is 21120 $cm^3$ so...

4. Hi, sorry, I was given a diagram and I forgot to mention the dimensions.

The dimensions are $40\,\textrm{cm}\,\times 16\,\textrm{cm}\,\times 11\,\textrm{cm}$.

But what you're saying about wasted space makes perfect sense.

So if I packed 5 in a row, with 2 rows, and 3 layers, that makes 30 cans.

I get it now, thanks.