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Another super easy problem, If an ant wants to crawl over the rectangular block of dimensions \( 6\times5\times4 \) from one vertex to a diagonally opposite vertex, what is the shortest distance it would need to travel?

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Is it a flying ant?

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Other answers:

No, no :)
4 + sqrt(61)
sqr 74
No No,
Read the question carefully Parth.
sqr 117
Bingo ninhi5! \(\sqrt{117} \) is the right answer.
The ant needs to travel one side, and one diagonal. We have three cases: side '4' + diagonal of (6 and 5) = 4 + sqrt61= 11.something side '6' + diagonal of (4 and 5) = 6 + sqrt41= 12.something side '5' + diagonal of (4 and 6) = 5 + sqrt52 =12.something hence shortest = 4 + sqrt61
Now where am I wrong?
|dw:1338100316725:dw|\[\sqrt{x^2 + 5^2} + \sqrt{(6-x)^2 + 4^2}\] By the pythagorean theorem, the sum of those two diagonals is: To minimize this distance, derive and set equal to 0.
dont make it over complicated :)
Oh damn! I completely forgot that !! :/ Damn my soul
There are only 3 possible roots, consistent to Smooth's diagram.
11.7? lol
Oh No :(
@ninhi5: Can you post the explanation?
I let Alpha do the derivation and optimization for me.
i just use pythogorean theorem
So x = 10/3, giving this distance:
My new question guys
can antonia travel around the outside of the box?

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