IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

bumba

one corner of a rectangle sheet of a paper of width a cm is folded over so as to reach the opposite edge of the sheet.find the minimum length of the crease.

kindly explain the answer.

him26.89

i got it [{(3)^3/2}/4]a

avinash.sharma

The lower corner A of the rectangle page OABC ( which has OA=a) is folded over so as it touches the inner edge OC at R. By folding pattern we get that

(i) Length PR = PA

(ii) Ð APQ = Ð QPR ( = q )

Let the length of the crease is x. In triangle OPR, we have OP = PR cos (p- 2q)

Þ OP = PA cos (p-2q) = -PA Cos2q (1)

In triangle APQ we have AP/PQ = cos q

Þ AP = PQ Cosq = x Cosq (2)

using it in (1) we get OP = -x Cosq Cos2q

Now a =OP+AP so

a= x Cosq-x Cosq Cos2q = x (Cosq- Cosq Cos2q)

Þ a/x = (Cosq- Cosq Cos2q) in this situation when x is minimum a/x ( let it is y) will be maximum.

Þ y = (Cosq- Cosq Cos2q)

For minimum length of crease (PQ=x) we should have maximum y and so dy / dq = 0 and d²y/dq² should be negative.

Now there is no trick left try to solve it and you will get the maximum y at sinq = [2/3]^(1/2) and length of the crease = (3Ö3 a)/4

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