Friday Puzzler -- Box making

Re: Friday Puzzler -- Box making

Larry Barrett

It may be the largest you can make, but I think it is still pretty small.

Re: Friday Puzzler -- Box making

Bill Earl

One dimension rhymes with a source of wood. The other rhymes with a form of wood.

I don't know

Alex Y

That's true of the right answer, but it's also true of the wrong answer I got

Can't confirm this either

Alex Y

One of both the correct answer and my wrong one rhymes with a source of wood. But I'm not getting the rhyme with a form of wood.

Re: Can't confirm this either

Bill Earl

>But I'm not getting the rhyme with a form of wood.

Different spelling, but a Greek mythical river and a '70s rock band.

Time to put the cards on the table

Alex Y

My (wrong) answer was 3x9, with 1/2" squares cut from the corners, for a 2x8x1/2 box.

I'll post my faulty reasoning for this, but want to see your answers first.

  I don't think so

Re: Time to put the cards on the table

Larry Barrett

My answer is a 3x6 card, with 1" square cutouts, so the box is 1x4x1. I suspect this is wrong, based on your reply to Bill.

  Back to the drawing board for all of us!

Alex Y

That's what I thought Bill's answer was as well. But the 1x1x4 box has a volume of 4 in^3. Consider the box you would create by cutting 3/4" squares from your piece of card stock. The resulting box would have a volume of .75*1.5*4.5=5 1/16 in^3. (probably not maximal, but sufficient to show that the 1" squares fail that condition)

P.S., That's how I know my solution was wrong as well -- a 2/3" square gives a bigger box than the 1/2" one I used.

My flawed solution

Alex Y

Here's how I got at my answer. I'm recreating it while making this post to see if I can catch the problem with it.

Assume the card's length and width are L and W, and the sides of the squares are H. Then the box I create is (L-2H) long, (W-2H) wide, and H high.

the volume of the box is V=(L-2H)*(W-2H)*H.

We know that (L-2H) = 4*(W-2H)

Substituting, and multiplying out, V= 4W^2*H - 16W*H^2 + 16H^3

Differentiating wrt H, we find dV/dH = 4W^2 - 32 WH + 48H^2

= 4*(W^2 - 8WH + 12h^2)

=4*(W-6H)*(W-2H)

Which is zero at W=6H or 2H

So W=6H should be at least a local minimum or maximum.

If W= 6H, then L=18H, and the solution involving a prime is 3,9,1/2.

Can you help me find what is wrong with this "solution"? I think it is in the setup.

How did both of you come up with the 3,6,1 solution?

Re: My flawed solution

Larry Barrett

My "solution" was much simpler than yours. My reasoning was that if the bottom of the box was a rectangle. say, W x L, where L = 4W, (note W and L here are for the box, not the card), then the volume of the box would be max when one vertical side was a square; in this case it would be max if the cutout on the card was equal to W. So my card would have a width of 3W and a length of 6W. And the first thought I had was that 3W would be a prime number if W = 1.

I thought about this last night and approached it the same way you did, got the same derivative, and same solution you came up with. So far I do not see what is wrong with this approach.

I am in Maryland, near DC, and expecting Sandy to knock on our door at any time. We will probably lose power, so if I do not respond with any more ideas, that might be the reason (or else I don't have any more ideas).

Re: My flawed solution

Alex Y

Good luck facing Sandy! Trivia becomes really trivial in the face of something like that. We'll resume when you are back online.

Re: My flawed solution

Larry Barrett

One problem I see with your solution (and mine) is that it seems to work with any prime number. For example, you got to 6H and 18H and found that 6H = 3 would work, giving H = 1/2.

But 6H = 5, or 6H = 7, or 6H = any prime will lead to real solutions for the box. The volume (and thus the card dimensions) keeps getting larger.

But even if you stay with 6H = 3, I don't see why your solution does not lead to a max volume.

The question my "solution" answered

Alex Y

I think my faulty solution was because it answers a subtly different question.

I found the H that for a given W gives the largest box, under the constraint that the length of the box is four times its width.

What we want is the H that gives the largest box for the card dimensions, and that results in a 4x1 ratio box.

Re: The question my "solution" answered

Larry Barrett

Alex - your solution gave you a 3"x9" card, and with a 1/2" cutout gave you a box with V=1/2 x 2 x 8 = 8 cu in.

Then you said that you found that a box with cutouts of 2/3" resulted in a larger volume. I assume you mean with a card with one side still = 3".

If Iam doing the arithmetic correctly, the box width will be 5/3", the length will then be 20/3" and the volume will be 2/3 x 5/3 x 20/3 = 200/27 = 7.4 cu in, which is less than the 8 cu in for your original solution. This card will be 3 x 8, as opposed to your original 3 x 9. What am I missing?

I just read your new ideas, but have not absorbed them yet.

PS, Sandy has passed, we lost power for the night and part of next day, but no where near the damage as those in NJ and NY received.

Re: The question my "solution" answered

Alex Y

Alex - your solution gave you a 3"x9" card, and with a 1/2" cutout gave you a box with V=1/2 x 2 x 8 = 8 cu in.

Then you said that you found that a box with cutouts of 2/3" resulted in a larger volume. I assume you mean with a card with one side still = 3".

No. That's what my solution provided, and where I went wrong. The original problem states that the cutouts create the largest box possible from my card, NOT the largest possible with the width of my card and a 4:1 ratio. From my 3x9 piece of card stock, I could have cut out 2/3" squares and created a box that was 2/3 x 5/3 x 23/3 = 230/27 = 8.5 cu in. It will not meet the 4:1 ratio test, but the maximum volume condition is not limited to boxes of that ratio.

If Iam doing the arithmetic correctly, the box width will be 5/3", the length will then be 20/3" and the volume will be 2/3 x 5/3 x 20/3 = 200/27 = 7.4 cu in, which is less than the 8 cu in for your original solution. This card will be 3 x 8, as opposed to your original 3 x 9. What am I missing?

I think that is the right answer (to the wrong question ). Similarly, if I used a 3x10 card, with a 1/3 cutout, I would have 1/3 x 7/3 x 28/3 = 196/3, again smaller than my box. This supports the idea that my "solution" successfully answers the question "what relationship between W and H gives you the maximum volume where the resulting box has a base of 1x4 shape?"

I just read your new ideas, but have not absorbed them yet.

PS, Sandy has passed, we lost power for the night and part of next day, but no where near the damage as those in NJ and NY received.

Glad to hear it.

Anyone?

Alex Y

Still working on this, and want me to hold off on the answer? If not, I'll post it.

Answer

Alex Y

The card was 3"x8", and I cut a 2/3" square from each corner.

The resulting box is 1 2/3" x 6 2/3" x 2/3"

The base has a 1:4 ratio, and the volume, at 200/27 ~= 7.407 cu. in. is the largest that can be made from that card.

Short of calculus, this volume max (at least a local max) can be shown (or at least supported) by looking at slightly smaller and slightly larger squares taken from the corners.

.65": .65*1.7*6.7 = 7.404

.7": .7*1.6*6.6 = 7.392

Question

Larry Barrett

Alex - did you need calculus to find this solution, or some other method?

Solution

Alex Y

Here's how to get to the correct answer:

Let the card have dimensions L and W, and cut out squares that are H on a side.

Condition 1: The resulting box has a base that is 4:1

L-2H = 4*(W-2H)

L-2H = 4W - 8H

6H = 4W -L

Condition 2: The resulting box is the largest that can be made from the given card

First get a formula for Volume:

V = H * (L-2H) * (W-2H)

V = 4*H^3 -2*(L+W)*H^2+L*W*H

Then to find the H that gives you the largest volume (independent of condition 1--where I went wrong), differentiate wrt to H

DV/DH = 12*H^2 - 4*(L+W)*H + L*W

Setting that equal to zero and solving the quadratic gives us

H = (4*(L+W) +/- sqrt(16*(L+W)^2-48*L*W))/24

removing 4 from the numerator and denominator:

H = ((L+W) +/- sqrt((L+W)^2-3*L*W))/6

or

6H = (L+W) +/- sqrt((L+W)^2-3*L*W)

Combining conditions 1 and 2, get

4W-L = (L+W) +/- sqrt((L+W)^2-3*L*W)

3W - 2L = +/- sqrt((L+W)^2-3*L*W)

Squaring both sides,

9W^2 - 12WL + 4L^2 = (L+W)^2-3LW

9W^2 - 12WL + 4L^2 = L^2+2WL+W^2-3WL

Combining terms:

8W^2 - 11WL + 3L^2 = 0

Factoring:

(8W-3L)*(W-L) = 0

So either L=W (a square, obviously not the answer) or L=8/3 * W

And from the integer and prime conditions, we get W=3, L=8

Whew! It's been 40 years since I did so much calculus and algebra!

Re: Solution

Larry Barrett

Thanks. I started down that path, but gave up (or ran out of room on the paper I was scribbling on) too soon. Good puzzle.