An open-empty rectangular box with negligible wall thickness is 6 feet

Similar questions to practice:
an-ant-is-clinging-to-one-corner-of-a-box-in-the-shape-of-a-135055.html
s97-184708.html
an-ant-crawls-from-one-corner-of-a-room-to-the-diagonally-134454.html

Hope it helps.

How did you arrive at your ans?? Explain and then we shall see whether this is within or beyond GMAT scope

Bunuel you rock ...you seem to have virtually every GMAT problem on earth on your finger tips..

Haha, I'm glad I actually looked at it again, anyway, I made a mistake in my calculations and it is indeed 10 in the end, damn. I'll just explain my way of thinking then:
My 2nd picture, the point C is on the line and X is a variable. Length function is \(\sqrt{36+x^2} + \sqrt{36+(2-x)^2}\).
With the help of derivatives I found that its lowest value is \(2*\sqrt{37}\) when X = 1 (although one could also apply some common sense there since if u try drawing the thing abit differently - as a double-rectangular figure omitting the folding, the shortest distance ends up being a straight line and it happens that since the rectangles have equal lengths, the line will cross the folding in its middle thus X = 1). \(2*\sqrt{37}\) > 12.
First picture is similar. Length function is \(F(x) = \sqrt{x^2+4} + \sqrt{36+(6-x)^2}\).
In order to find minimum value I went with the rule that this \(X_0\) is pretty much a root of an equation \(F'(X_0) = 0\)
\(\frac{2*x}{\sqrt{x^2+4}} - \frac{2*(6-x)}{\sqrt{36+(6-x)^2}} = 0\)
\(x* \sqrt{36+(6-x)^2} = (6-x)* \sqrt{4+x^2}\)
solving this you get x = 1,5 as the only fitting answer, y ends up being 10 which is the answer to the question

Well dear, I appreciate your insights...very thorough indeed.

However, the question can be solved using basics of Pythagoras theorem...do visit the link suggested by Bunuel at

an-ant-crawls-from-one-corner-of-a-room-to-the-diagonally-134454.html

Relevant discussions have taken place along with diagrams to make the point better. Do revert if any doubts..

Yea I did understand the approach displayed there and I find it very clever. I guess I will elaborate with the changed picture:

AB' is the shortest path the ant has to take and its the straight line, thus we got a triangle ADB' where AB' = hypotenuse and is easily found to be equal to \(\sqrt{36+64} = 10\)

Hi-
I think there is a typo in the question. I believe either width or length needs to be 2, if the answer is indeed the correct answer.
Using the data given the question:
Basically if we unfold the box, this will give us two rectangles, 1 rectangle with height 2 and base 6, 2nd rectangle with height 2 and base 6. The two rectangles have point A and B as diametrically opposite points. Now to find AB, it's sqrt( (6+6)^2 + 2^2) and that's not the correct answer.
Bunuel - I have followed the other posts that you have mentioned, but not sure what I am missing here in this question. Any inputs will be very helpful.

Thanks,
Chanakya

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.