shubhamsrg
  • shubhamsrg
what is the maximum length of rod that can pass through this L-shaped corridor :
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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shubhamsrg
  • shubhamsrg
|dw:1347549016758:dw|
shubhamsrg
  • shubhamsrg
its the same question which @siddhantsharan asked here : http://openstudy.com/study#/updates/504e2cd7e4b064acfea5f1ae there arent much people there,,and i liked this ques,,so i bring this here..
shubhamsrg
  • shubhamsrg
@mukushla @eliassaab

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anonymous
  • anonymous
|dw:1347549611645:dw| By SIMILARITY of Triangles [ three similar , right-angle triangle] we obtain (X+a)/(y+b) = x/b = a/y So ab = xy and bx + ba = yx + bx
anonymous
  • anonymous
y = ab/ x Substitrute int Length^2 = (y+b)^2 + (x+a)^2 AND Make derivation , then equate to 0 and solve for x - here is your SOLUTION !!
anonymous
  • anonymous
(ab/x + b)^2 + (x+a)^2 = a^2b^2/x^2 + b^2 + 2ab^2/x + x^2 + 2xa + a^2
shubhamsrg
  • shubhamsrg
well in the position you mentioned,,the rod cant move,so wont exit out.... |dw:1347549761719:dw|
anonymous
  • anonymous
(a^b^2/x^3) * (-2) - 2ab^2/x^2 + 2x +2a = Derivative
anonymous
  • anonymous
Now Deivative = o means the following should be solved
shubhamsrg
  • shubhamsrg
you're concept it wrong buddy,,see my last comment again,,and correct me if am wrong..
anonymous
  • anonymous
|dw:1347549859821:dw| Well - sorry to be so direct. But you ARE wrong. Rotation is NO PROBLEM
anonymous
  • anonymous
To be precise Rotatio + Simultaneous sliding of the ends
anonymous
  • anonymous
You see - the interval (lets call it "stick") that is the solution is THE SHORTEST POSSIBLE OF ALL ANGLES. Thus, a little motion provides it with MORE SPACE , NOT LESS !
anonymous
  • anonymous
Imagine a stick that fits into this space at "the most restrictive" angle. Well, it will obviously fit into positions at LES restrictive angles, wouldn't it ?
shubhamsrg
  • shubhamsrg
|dw:1347550109862:dw| for up ward movement,,downward movement should also be there,,which is not possible..
anonymous
  • anonymous
|dw:1347550236581:dw|
shubhamsrg
  • shubhamsrg
sorry didnt get this figure..
anonymous
  • anonymous
Firstly the LEFT END RISES, and the right end SLIDES into positions 3 and 3
anonymous
  • anonymous
|dw:1347550435290:dw|
anonymous
  • anonymous
|dw:1347550518272:dw|
shubhamsrg
  • shubhamsrg
it cant slide like that.. |dw:1347550476551:dw|
anonymous
  • anonymous
Look forget - drawing for a second. If Some position is the SHORTEST interval - all its small motions will create SPACE AT BOTH ENDS or AT ONE END at least !
anonymous
  • anonymous
|dw:1347550645409:dw|
anonymous
  • anonymous
|dw:1347550693270:dw|
anonymous
  • anonymous
Come on - after rotation the OLD SHORTEST STICK IS NOT FILLING THE NEW - LONGER SPACE
anonymous
  • anonymous
|dw:1347550790801:dw|
anonymous
  • anonymous
More SPACE - NO PROBLEM FITTING SHORTEST STICK !
anonymous
  • anonymous
You need a stick that will move inside the SHORTEST place. Sooo just take the SHORTEST LENGTH - some infintisimal "epsilon"
shubhamsrg
  • shubhamsrg
you mean.. |dw:1347551005335:dw| ?
anonymous
  • anonymous
This sign "- " above was MINUS
anonymous
  • anonymous
Call the minimal diagonal by M. Now take stick of length\[M - \epsilon \]
shubhamsrg
  • shubhamsrg
and since this dx is there,,it can rotate,,which will create dx upwards also ? right ?
anonymous
  • anonymous
You can take epsilon as small as you wish, BUT in some sense you are right. The reaching of the worst position and getting out of this tightest of all positions is strictly speaking impossible without some nominal "epsilon" reduction"