Tuesday, December 22, 2009

Fox 218


  1. If we assume that θ to be the angle between BO and the negative y-axis & if x=sin(θ) then the perimeter,P=1/(1-x^2)+1/x+(4/x^2+1/(1-x^2))^(1/2)
    I tried differentiating this to put dP/dx=0 but it just becomes too complicated. By trial and error it was found that the derivative is close to 0 at sin(θ)=0.77159061 at which value the minimum P = 6.79881.
    Not at all sure if this is correct. Can someone please verify this or find the correct solution?

  2. The expression should've read:
    P= 1/(1-x^2))^(1/2)+ 2/x + (4/x^2+1/(1-x^2))^(1/2) Sorry abt the typo. Now actually the derivative turns out to be;
    2x (4/x^2+ 1/(1-x^2))^(1/2)/(1-x^2)^(3/2)+2x/(1-x^2)^2 -8/x^3-(4/x^2)(4/x^2+1/(1-x^2))^(1/2) and this when equated to 0 gives us: x=sin(θ)=0.79951
    which in turn gives the least P as 7.1713
    If thisn't right then I give up and leave the question open to someone else

  3. When inclanation is 45 degrees, perimeter comes out to be: 3V2 + V10 = 7.404918347

    I couldn't confirm it yet, but your answer looks reasonable.

  4. I am sure that your answer is right Joe
    I have a solution but yours is better