Crap. I can't figure this out. (Math)

my head hurts, and I'm just watching...



for the record, I took Algebra one twice. the first time mostly all I did was play that paper football game where you kick field goals with your finger all year with David McElroy..

then they told me I had to take it again, because you couldn't graduate without it.

I got an A as the only sophomore in the class in high school. but I don't remember it as much as the football game.

(I didn't even like David McElroy)

Minimum number I get is 481.

477 D, 2 A, 1 B, 1 C

You should be able to fit two "A" boxes in the place of one "D" box.


That still leaves a lot of difference between a container "full" of mostly Ds and "full" of Cs, though.

Woo--this is a non trivial problem. Even for the Japanese Guy.

I also don't understand why the smallest boxes "fill" the container the least efficiently.

Me neither. I have to go back and ask some more questions.

More info, that I didn't have a few minutes ago: the container is about 29cbm.

Size A fills about 27cbm, B fills 23cbm....I haven't done the others.

I am going to have to do that chart. So when the guy calls and says I want 200 C and 15oB how much space is left and what do I need to buy to fill it up, I can answer him.

Joe( or just keep it to myself)Nation

You really need to know your container size to get the best answer.

You don't want shifting in the container.

Don't feel bad; I took college Algebra, and forgot most of it. Even went into differential calculus for a bit in an economics course.

Is the container a straight cube, or a reefer unit, or something else?

Rectangular

about 27cbm.

Joe(they say, don't worry, but hey, I'm the one the guy is asking)Nation

You know the total volume now.
And there are obviously many different combinations which will fill the space.

So just make up a little table, like this:

10 x A = (volume)
20 x A =
30 x A =
etc

then do the same for package sizes B, C and D, in adjacent columns.

Then you can pick off the numbers you want to fill the container.

You're looking for the simple arrangement

A + B + C + D = 29

Obviously you'll never get 100% usage of the space (unless you're very lucky) so you'll have to assume the wastage, say 10% or so.

Let us know how you get on, and whether you win the Employee Of The Month badge.

Yup. You've got it.

Just a lot more data to line up than I wanted....but, I love charts..

There must be a rule of sets that controls this:
If I take X of A and Y of B then the only possibilities for C and D are (m,n,o.... of C) + (d,e,f..... of D) in order to fill the space F.

Where's Paul Erdős when you need him?

Joe(swimming)Nation

My suggestion:

Create a spreadsheet with a "what if" calculation.

You need:

Total cubic volume

cubic volume of each type of box.

Then fields where you enter each type of box

And a formula that multiplies the number of each kind of box, times the volume of each kind, and subtracts it from the total volume.



You could probably set it up where you had a field for each kind of box that told you how many would fit in the volume remaining. (Just divide remaining volume by the volume of that kind of box.)

Will this help?


AVERAGE VOLUME and PIECES
Ttl vol/pcs
Pieces... Side a... Side b... Side c... Total volume.... Avg volume
1160...... 35.5...... 31.5...... 21.5...... 24042.375....... 20.726
747........41.5...... 38.5..... 23.5..... 37547.125........ 50.264
830.......41.5...... 38.5..... 22........ 35150.5......... 42.35
480.......47.5...... 45.5..... 27.5..... 59434.375....... 123.822

3217......166.......154....... 94.5....... 156174.375....... 237.162

Average volume for total pieces 48.547........ 59.29 = avg of avg

NOW WE ARE TALKING !! This is exactly the kind of thinking I needed!

Crunching numbers is not my thing... I like to think about the process but I am terrible at execution.

Joe( I should take the blindfold off)Nation

Thanks to everyone,,,,,,,Drew, Beth< CI> etc ~~~

Is this a real-world problem or a puzzle-type problem? The reason I ask is that dimensions of the container matter because this is a packing problem. Maximum counts for the small boxes are actually constraints for dimensions of the container (not just volume, but lengths of sides). A puzzle-type problem would have you trying to figure out the dimensions of the container based on the constraints you've supplied. In a real-world problem, you really should know the dimensions of your container. Also, it's strange that you're trying to minimize the number of boxes that fill the container. Usually, in a problem like this, you're trying to maximize the number. Heck, throw in as few boxes as you like, and surround them with packing material...

I think it's a real world problem for Joe, the rest of us is puzzled

It sounds like an ill-defined real-world problem.

I agree the packing matters.
I wanna see this in plan view..
and more.

The reply is meant for Joe

This reminds me of my first precept, don't give 'em what they ask fer, give 'em what they want. Geez I miss the real world, you seem to be having muchly fun.

Obtaining the dimensions of the container will allow you to minimise the number of boxes that can be stuffed while simultaneously maximising the use of the available volume of space ( presumably surface area of individual boxes costs money so you'll also need to know whether or not the individual boxes can be stuffed all ends up which you appear to have some talent for).

Have you considered that this is a comedy routine and you are the star?