Question Number 10397 by amir last updated on 07/Feb/17

Commented bymrW1 last updated on 07/Feb/17

I think there is no maximum for the premetre of the trangle, but a minimum. Due to the symmetry it can be seen that the minimum premetre occurs when the tangent line tangents the curve at point (1, 1) or (−1,−1).

Answered by mrW1 last updated on 07/Feb/17

here the analytic solution: let B(t,s) a point on the curve xy=1. y=(1/x) s=(1/t) y′(x)=−(1/x^2 ) the slope of the tangent line at B is m_t =y′(t)=−(1/t^2 ) the equation of the tangent line is y−s=m_t (x−t) or y−(1/t)=−(1/t^2 )(x−t) the intersection points of tangent line with coordinate axes are C(0,c) and D(d,0). c−(1/t)=−(1/t^2 )(0−t) ⇒c=(2/t) 0−(1/t)=−(1/t^2 )(d−t) ⇒d=2t the premetre of triangle ΔOCD is P=OD+OC+CD=2t+(2/t)+(√((2t)^2 +((2/t))^2 )) =2[t+(1/t)+(√(t^2 +(1/t^2 )))]=2L L=t+(1/t)+(√(t^2 +(1/t^2 ))) (dL/dt)=1−(1/t^2 )+(1/(2(√(t^2 +(1/t^2 )))))×(2t−(2/t^3 )) =1−(1/t^2 )+(1/( (√(1+(1/t^4 )))))×(1−(1/t^4 )) =(1−(1/t^2 ))[1+((1+(1/t^2 ))/( (√(1+(1/t^4 )))))] (dP/dt)=0 ⇒ (dL/dt)=0 ⇒ 1−(1/t^2 )=0 ⇒t^2 =1 ⇒t=±1 i.e. the premetre P has minimum at (1,1) or (−1,−1) with t=1 we get: c=(2/t)=2 d=2t=2 P_(min) =2+2+(√(2^2 +2^2 ))=4+2(√2)

Commented byamir last updated on 07/Feb/17

thank you so much dear mrW1. your commenet is true.it is a min for premetre. but? is there any max for premeter?

Commented bymrW1 last updated on 07/Feb/17

no, there is no max for the premetre. as t→0 or t→∞, P→∞