Question

An open box is made from a `30"cm" xx 30"cm"` sheet of metal by removing a square from each corner and folding up the sides. To the nearest `"cm"` what is the size of the removed squares to make a box of volume `1000"cm"^3` ?

Answer 1

The dimensions of each square to the nearest centimetre could be `1"cm" xx 1"cm"` or `10"cm" xx 10"cm"`

Explanation:

It's easy to see that the squares could be `10"cm" xx 10"cm"`, which would result in a box all of whose sides would be `10"cm"`, but is this the only solution?

Let the side of each of the removed squares be `t` cm

Then the base of the box is `(30-2t)"cm" xx (30-2t)"cm"` and each of the rectangular sides is `t "cm" xx (30-2t)"cm"`, resulting in a volume in `"cm"^3` of `t(30-2t)(30-2t) = 4t^3-120t^2+900t`

So `4t^3-120t^2+900t = 1000`

Subtract `1000` from both sides and divide by `4` to get:

`t^3-30t^2+225t-250 = 0`

Divide by `t-10` to find:

`t^3-30t^2+225t-250 = (t-10)(t^2-20t+25)`

So `t=10` or `t = (20+-sqrt(300))/2 = 10+-5sqrt(3)`

`(10-5sqrt(3))"cm" ~= 1.34"cm"` so about `1"cm" xx 1"cm"` to the nearest `"cm"`.

`(10+5sqrt(3))"cm" ~= 18.66"cm"` so about `19"cm" xx 19"cm"` to the nearest `"cm"`. Discard this 'solution' as `30-2t < 0` - so the cut outs would overlap and the base would be a square of negative lengths.

Here's `(t^3-30t^2+225t-250)-:100` ...

graph{0.01*(x^3-30x^2+225x-250) [-11.42, 28.58, -10.08, 9.92]}