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 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→∞