Enigma 1715 - Open the box
New Scientist magazine, 15 September 2012.
By Susan Denham.
I started with a rectangular piece of
card a whole number of centimetres long
by a whole number of centimetres wide,
one of those numbers being a prime. Then
I cut out an identical small square from
each corner of the card and discarded the
four squares. I folded up the four "flaps"
on the remaining piece of card to form an
open-topped box, choosing the size of the
cut squares so that the volume of this
open box was the biggest possible.
Having made the box, I found that the
length of its rectangular base was four
times its width.
What were the dimensions of the original
piece of card?
> Enigma 1715 - Open the box
> New Scientist magazine, 15 September 2012.
> By Susan Denham.
> I started with a rectangular piece of
> card a whole number of centimetres long
> by a whole number of centimetres wide,
> one of those numbers being a prime. Then
> I cut out an identical small square from
> each corner of the card and discarded the
> four squares. I folded up the four "flaps"
> on the remaining piece of card to form an
> open-topped box, choosing the size of the
> cut squares so that the volume of this
> open box was the biggest possible.
> Having made the box, I found that the
> length of its rectangular base was four
> times its width.
> What were the dimensions of the original
> piece of card?
Let the original dimensions of the card be L x W and the fold height be H.
The base of the box so formed will be L2 = (L-2H) and W2 = (W-2H) and we are told that L2 = 4 x W2.
The volume will be L2 x W2 x H = (L-2H) x (W-2H) x H or
4(W-2H) x (W-2H) x H
After some buggering about with simultaneous equations, calculus and spreadsheets I concluded that this volume maximises when H = 2/9W. So W2 = 5/9W and L2 = 20/9W
That means L = 24/9W = 8/3W
If one side of the card is a prime number of cm then it must be 8 x 3 cm.
-- Dave Baker
>> I started with a rectangular piece of
>> card a whole number of centimetres long
>> by a whole number of centimetres wide,
>> one of those numbers being a prime. Then
>> I cut out an identical small square from
>> each corner of the card and discarded the
>> four squares. I folded up the four "flaps"
>> on the remaining piece of card to form an
>> open-topped box, choosing the size of the
>> cut squares so that the volume of this
>> open box was the biggest possible.
>> Having made the box, I found that the
>> length of its rectangular base was four
>> times its width.
>> What were the dimensions of the original
>> piece of card?
>Let the original dimensions of the card be L x W and the fold height be H.
>The base of the box so formed will be L2 = (L-2H) and W2 = (W-2H) and we are >told that L2 = 4 x W2.
>The volume will be L2 x W2 x H = (L-2H) x (W-2H) x H or
>4(W-2H) x (W-2H) x H
>After some buggering about with simultaneous equations, calculus and >spreadsheets I concluded that this volume maximises when H = 2/9W. So W2 = >5/9W and L2 = 20/9W
>That means L = 24/9W = 8/3W
>If one side of the card is a prime number of cm then it must be 8 x 3 cm.
I used the same method, but got a slightly different result:
L = 18H
W = 6H
And the simplest solution meeting the other criteria is L=9cm, W=3cm, H=0.5cm
-- Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently.
> "Dave Baker" <N...@null.com> wrote:
> I used the same method, but got a slightly different result:
> L = 18H
> W = 6H
> And the simplest solution meeting the other criteria is
> L=9cm, W=3cm, H=0.5cm
The solution W = 6H only applies when the original piece of card is square. As the card becomes progressively more rectangular i.e. L tends to infinity, then H tends to W/4. For any other rectangle the optimum value of H will be somewhere between these two limits of W/6 and W/4.
The optimum value of H if the card were 9 x 3 cm is 0.677 cm and not 0.5 cm as you can verify by calculation.
-- Dave Baker
>"alexy" <nos...@asbry.net> wrote in message >news:k6c0o3$mi5$1@dont-email.me...
>> "Dave Baker" <N...@null.com> wrote:
>> I used the same method, but got a slightly different result:
>> L = 18H
>> W = 6H
>> And the simplest solution meeting the other criteria is
>> L=9cm, W=3cm, H=0.5cm
>The solution W = 6H only applies when the original piece of card is square. >As the card becomes progressively more rectangular i.e. L tends to infinity, >then H tends to W/4. For any other rectangle the optimum value of H will be >somewhere between these two limits of W/6 and W/4.
>The optimum value of H if the card were 9 x 3 cm is 0.677 cm and not 0.5 cm >as you can verify by calculation.
Hmmm. Hard to argue with that! I better check my formulas and
calculus!
-- Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently.
>"alexy" <nos...@asbry.net> wrote in message >news:k6c0o3$mi5$1@dont-email.me...
>> "Dave Baker" <N...@null.com> wrote:
>> I used the same method, but got a slightly different result:
>> L = 18H
>> W = 6H
>> And the simplest solution meeting the other criteria is
>> L=9cm, W=3cm, H=0.5cm
>The solution W = 6H only applies when the original piece of card is square. >As the card becomes progressively more rectangular i.e. L tends to infinity, >then H tends to W/4. For any other rectangle the optimum value of H will be >somewhere between these two limits of W/6 and W/4.
>The optimum value of H if the card were 9 x 3 cm is 0.677 cm and not 0.5 cm >as you can verify by calculation.
Aha! I see the fault in my solution. I was maximizing volume for a
L=4W box, while the maximization should have been first, and the 4x1
ratio a result. -- Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently.
Dave: A little help please. I recognize from the specific example you
gave that my solution is wrong, but I can't figure out why. Please
point out the problem in my solution below:
"Dave Baker" <N...@null.com> wrote:
>Let the original dimensions of the card be L x W and the fold height be H.
>The base of the box so formed will be L2 = (L-2H) and W2 = (W-2H) and we are >told that L2 = 4 x W2.
>The volume will be L2 x W2 x H = (L-2H) x (W-2H) x H or
>4(W-2H) x (W-2H) x H
I'm with you to this point, but here we diverge. Multiplying, I got
V = 4W^2H - 16WH^2 + 16H^3
Differentiating wrt H, I get DV/DH = 4W^2 - 32WH + 48H^2 =4*(W-2H)*(W-6H)
From which I concluded that W=6H.
Your specific example shows that I am wrong, but WHY? I can't see the
breakdown in the argument above, although I'm sure I will give myself
a dope-slap when you point it out! <g>
>After some buggering about with simultaneous equations, calculus and >spreadsheets I concluded that this volume maximises when H = 2/9W. So W2 = >5/9W and L2 = 20/9W
>That means L = 24/9W = 8/3W
>If one side of the card is a prime number of cm then it must be 8 x 3 cm.
-- Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently.
> I started with a rectangular piece of card a whole number of
> centimetres long by a whole number of centimetres wide, one of those
> numbers being a prime. Then I cut out an identical small square from
> each corner of the card and discarded the four squares. I folded up
> the four "flaps" on the remaining piece of card to form an
> open-topped box, choosing the size of the cut squares so that the
> volume of this open box was the biggest possible.
Let l be the length of the card; let w be the width.
Let d be the depth of the box.
Then the volume of the box as a function of d is
v(d) = (l-2d)(w-2d)(d)
= 4d^3 - 2(l+w)d^2 + lwd
Its derivative is
v'(d) = 12d^2 - 4(l+w)d + lw
The maximum volume of the box is attained when the derivative equals
zero, and the quadratic formula gives us:
Did someone say that I did some handwaving about whether this is a
minimum or maximum? And about whether either of the two values for d
is reasonable when compared to l and w? Nobody? Great. So let's
stop here for now.
> Having made the box, I found that the length of its rectangular base
> was four times its width.
We notice that
l - 2d = 4(w - 2d)
l - 2d = 4w - 8d
4w - l = 6d
Hey, we have a nice formula for 6d!
4w - l = l + w +- sqrt(l^2 - lw + w^2)
3w - 2l = +- sqrt(l^2 - lw + w^2)
(3w - 2l)^2 = l^2 - lw + w^2
9w^2 - 12wl + 4l^2 = l^2 - lw + w^2
8w^2 - 11wl + 3l^2 = 0
(w - l) (8w - 3l) = 0
> What were the dimensions of the original piece of card?
If w == l, then the original card is a square, but that doesn't work.
So 8w == 3l. Since the card's dimensions (in cm) are integers and one
is prime, our answer must be "3cm by 8cm".
On Friday, October 26, 2012 2:49:02 PM UTC-4, alexy wrote:
> Dave: A little help please. I recognize from the specific example you
> gave that my solution is wrong, but I can't figure out why. Please
> point out the problem in my solution below:
> "Dave Baker" <N...@null.com> wrote:
> >Let the original dimensions of the card be L x W and the fold height be H.
> >The base of the box so formed will be L2 = (L-2H) and W2 = (W-2H) and we are > >told that L2 = 4 x W2.
> >The volume will be L2 x W2 x H = (L-2H) x (W-2H) x H or
> >4(W-2H) x (W-2H) x H
The box is the maximum volume for the given card (not yet subject to the constraint that the box length is four times the box width). You are substituting too soon.
Jonathan Dushoff <jdush...@gmail.com> wrote:
>On Friday, October 26, 2012 2:49:02 PM UTC-4, alexy wrote:
>> Dave: A little help please. I recognize from the specific example you
>> gave that my solution is wrong, but I can't figure out why. Please
>> >Let the original dimensions of the card be L x W and the fold height be H.
>> >The base of the box so formed will be L2 = (L-2H) and W2 = (W-2H) and we are
>> >told that L2 = 4 x W2.
>> >The volume will be L2 x W2 x H = (L-2H) x (W-2H) x H or
>> >4(W-2H) x (W-2H) x H
>The box is the maximum volume for the given card (not yet subject to the constraint that the box length is four times the box width). You are substituting too soon.
>JD
thanks. That is the conclusion that I had [finally] come to.
-- Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently.
