Friday, January 15, 2010

Fox 228

2 comments:

  1. From the figure geometry, height of the isosceles trapezoid = a/2cos(λ) while the sum of parallel sides=[2b/sin(λ) + a/sin(λ)] and hence its area=(1/2)*[2b/sin(λ) + a/sin(λ)]*a/2cos(λ) =(a^2+2ab)/2sin(2λ) which is minimum when sin(2λ)=1 or when λ=45 deg. giving the minimum area as(a^2+2ab)/2
    Ajit

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  2. "Area=(a^2+2ab)/2sin(2λ) which is minimum when sin(2λ)=1"

    is so clear, you even don't need to take its derivative. Good solution!

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