Question

A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle.Show that the maximum length of the hypotenuse is ⎛⎜⎝a23+b23⎞⎟⎠32.

Open in App
Solution

Let P be a point on the hypotenuse AC of right △ABC such thatPL⊥AB=aPM⊥BC=bLet ∠APL=∠ACB=θAP=asecθ and PC=bcosecθLet l be the length of the hypotenuse, then l=AP+PC⇒asecθ+bcosecθ, 0<θ<π2Now differentiate l with respect to θ, we getdldθ=asecθ.tanθ−bcosecθ.cotθFor maxima and minima dldθ=0Thus asecθ.tanθ=bcosecθ.cotθ⇒asin3θ=bcos3θab=cos3θsin3θThus ba=tan3θ⇒tanθ=[ba]13Now d2ldθ2=a(secθ.sec2θ+tanθ.secθ.tanθ)=−b(cosecθ(−cosec2θ)+cotθ(−cosecθ.cotθ))=asecθ(sec2θ+tan2θ)+bcosecθ×bcosecθ+cot2θSince 0<θ<π2, so all t ratios of θ are positive.Also a>0 and b<0Therefore, d2ldθ2 is positive.⇒ l is least when tanθ=[ba]13Least value of l =asecθ+bcosecθ=a.√a23+b23a13+b.√a23+b23b13=√a23+b23(a23+b23)=(a23+b23)32Hence, proved.

Suggest Corrections
0
Related Videos
Parametric Differentiation
MATHEMATICS
Watch in App