shubhamsrg
  • shubhamsrg
what is the maximum length of rod that can pass through this L-shaped corridor :
Mathematics
katieb
  • katieb
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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

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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| ?