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? Ajit

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

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)

ReplyDeleteI 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?

Ajit

The expression should've read:

ReplyDeleteP= 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

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

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

I am sure that your answer is right Joe

ReplyDeleteP=7.171284668:)

I have a solution but yours is better