Volume of a box

[Acidic]

Here's an assignment I had to do when I was 14 or 15, quite easy really:
Imagine a square with a width of L.
If you cut corners of the square you could fold the paper into a box without a lid. The corners you cut out must be square in shape and all be the same size. Their length is X.
How does X relate to L in order to get the maximum volume of the box?

I'l give an example to clarify, lets say the square paper is 5x5 cm. Lets cut out 1x1 cm squares out of each corner. The paper can now be folded into a box with a base of the dimensions 3x3 cm and a highe of 1 cm. This gives a volume of 3*3*1 = 9cm^3. Cutting out a different size could result in a bigger volume.

Now, in order to get the maximum volume, how does X relate to L?

[wossname]

Isn't it something like...

let x = L / 3
so V = x^3

So the height of the box is equal to the length of a base side?

[Acidic]

no because if you have a 15x15 and take out 5 as you just said
that would give the box the dimensions 5x5x5 so Volume = 125
if you cut out corners of 4x4, the box would be
7x7x4 = 196
so, what would the optimum amount be to cut out?

[ZaidGS]

i didnt get exactly whats up,
but here us what i think:
u said there is a 15cm thing, u cut X*X cube from each corner
so
H= X
L= 15 - 2*X
W= 15 - 2*X
V = H*L*W
V= (15-2X)^2 * X
=225X - 60X^2 + 4X^3 ( correction here, n later on sure )

we want V max
then
dV/dX = 225 - 120X + 12X^2
which is equal to zero
so
use formula to get:
X=9.330 OR X=0.700

d2V/dX2 = -120 + 24X
use X=9.330 to get +ve answer
use X=0.700 to get -ve answer

so at X=0.700 is max

so ur answer is X=0.7

[Acidic]

You've got the idea but you must have misscalculated something.
I don't know where as I have lent my calculator to a freinds and I don't want to start doing this on paper. Anyway, for you volume (where X=3.064) you get:
3.604*(15-(2*3.064))^2
which is:
241.175
using 2.5 though gets:
2.5*(15-(2*2.5))^2
which is:
250

So somewhere you have gone wrong. Also, you are meant to work this out for a square with sides L.

Your method is right though.

[ZaidGS]

i discovered the error,
simple :
15 * 15 = 225
but when i expanded the V i made it:
255X + .... etc

i corrected the original post

[Acidic]

ok, well done. Now try to make the general formula for a square with sides L.

[ZaidGS]

dah !!!
y r u so lazy to do it urself ?!?!
just subsitute L in place of 15 in original formula for V
=> V= (L-2X)^2 * X
V= L2X - 4LX2 + 4X3 ( L2 is L^2 ... etc )

then differentiate,
and use formula.
note1: use minus not plus bcoz the smaller answer is the maxima
note2: use (c) and (b) of formula in terms of L, dont let that confuse you

[Acidic]

Actually, I have already done this, I was only seeing if other could do it too (read first post).

[Guv]

Volume = x*(Length - 2*x)²

Volume = 4*x³ - 4*Length*x² + x*Length²

Derivative = 12*x² - 8*Length*x + Length²

Maximum and Minimum occur at points for which the derivative is zero. Quadratic formula provides the following.

X₁ = [ 8*Length + SquareRoot(64*Length² - 48*Length²) ] / 24

X₂ = [ 8*Length - SquareRoot(64*Length² - 48*Length²) ] / 24

Some algebraic manipulation gives the following.

X₁ = Length / 2 ----- Volume = 0, which is a minimum.

X₂ = Length / 6 ------ Volume = 2*Length³/27

There is also a minimum at x = 0

[Acidic]

well done.