"best" value for volume of a box

[PhysicsMan999]

Homework Statement

I'm given length, width and height of the box, all have uncertainties.

Homework Equations

V=lwh

The Attempt at a Solution

I've tried V=lwh, wasn't correct. Then V=lwh x (1 + relative uncertainty), wasn't correct. I just have no idea what the "best" value of a box is, my lab manual doesn't say anything about it, and I can't find anything on google either and I'm running out of tries. Appreciate any help![/B]

 

[Simon Bridge]

You need to define "best value" for this problem - what does that mean?

Please provide the complete problem statement.
If you have $l\pm\sigma_l$, $w\pm\sigma_w$ and $h\pm\sigma_h$ as independent measurements of the dimensions of a rectangular box, then the volume of the box is $V=lwh\pm\sigma_V$.

So if that is not correct (how do you know?) then there is some information missing from the problem statement, or the solution attempt. i.e. how did you evaluate $\sigma_V$?

 

[PhysicsMan999]

That's the thing..I don't know what the "best" value is, and nowhere I've looked has told me what it means. And it's for an online assignment which I have 10 tries for, so it tells me whether the answer is right or wrong.

EDIT: Never mind. I can't believe how stupid the mistake I was making was..I just wasn't paying attention to the units. Measurements were given in cm, question wanted m^3. Sorry for wasting your time.

 

[Simon Bridge]

No worries - sometimes talking to someone else is what it takes to see the obvious right under your nose :)

BTW: whenever you have an answer you know is wrong - you should say how you know.
The assessment of the problem changes whether you got an answer different from a model answer, and answer that a human marked wrong, and an answer that a computer marked wrong.

You should also always show you working and reasoning ... seeing correct calculations and a "the computer rejected the result" would have got you specific and helpful advise right away.