# 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 $(a^{\large\frac{2}{3}} + $$b^{\large\frac{2}{3}})^{\Large\frac{3}{2}} ## 1 Answer Toolbox: • \large\frac{d}{dx}$$(\sin \theta)=\cos\theta$
• $\large\frac{d}{dx}$$(\cos \theta)=-\sin\theta Step 1: Let P be a point on the hypotenuse AC of right \Delta ABC such that PL\perp (AB)=a PM\perp (BC)=b Let \angle APL=\angle ACB=\theta AP=a\sec\theta PC=b cosec\theta Let l be the length of the hypotenuse then l=AP+PC \Rightarrow a\sec\theta+b cosec\theta \;0 < \theta <\large\frac{\pi}{2} Step 2: Differentiating l with respect to \theta l=a\sec\theta+bcosec\theta \large\frac{dl}{d\theta}=$$a\sec\theta.\tan\theta-bcosec\theta.\cot\theta$
For maxima & minima $\large\frac{dl}{d\theta}$$=0 a\sec\theta+\tan\theta=b cosec\theta\cot\theta=0 \Rightarrow a\sec\theta\tan\theta=b cosec\theta.\cot\theta \Rightarrow a.\large\frac{1}{\cos\theta}.\frac{\sin\theta}{\cos\theta}=$$b.\large\frac{1}{\sin\theta}.\frac{\cos\theta}{\sin\theta}$
$\Rightarrow a\sin^3\theta=b\cos^3\theta$
$\large\frac{a}{b}=\frac{\cos^3\theta}{\sin^3\theta}$
$\large\frac{b}{a}=\frac{\sin^3\theta}{\cos^3\theta}$
$\large\frac{b}{a}=$$\tan^3\theta \tan\theta=\big[\large\frac{b}{a}\big]^{\Large\frac{1}{3}} Step 3: Now \large\frac{d^2l}{d\theta^2}$$=a(\sec\theta.\sec^2\theta+\tan\theta.\sec\theta\tan\theta)$
$\qquad\quad\;\;=-b[cosec\theta(-cosec^2\theta)+\cot\theta(-cosec \theta\cot\theta)$
$\qquad\quad\;\;=a\sec\theta(\sec^2\theta+\tan^2\theta)+bcosec\theta\times cosec\theta+\cot^2\theta)$
Since $0< \theta<\large\frac{\pi}{2}$ so all $t$-ratios of $\theta$ are +ve.
Also $a>0$ and $b>0$
Therefore $\large\frac{d^2l}{d\theta^2}$ is +ve.
$\Rightarrow l$ is least when $\tan\theta=\big[\large\frac{b}{a}\big]^{\Large\frac{1}{3}}$
Step 4:
Least value of $l=a\sec\theta+bcosec\theta$
$\qquad\qquad\qquad=\large\frac{a.\sqrt{a^{\Large\frac{2}{3}}+b^{\Large\frac{2}{3}}}}{a^{\Large\frac{1}{3}}}+\frac{b.\sqrt{a^{\Large\frac{2}{3}}+b^{\Large\frac{2}{3}}}}{b^{\Large\frac{1}{3}}}$
$\qquad\qquad\qquad=\sqrt{a^{\Large\frac{2}{3}}+b^{\Large\frac{2}{3}}}(a^{\Large\frac{2}{3}}+b^{\Large\frac{2}{3}})$
$\qquad\qquad\qquad=\big(a^{\Large\frac{2}{3}}+b^{\Large\frac{2}{3}}\big)^{\Large\frac{3}{2}}$
edited Sep 2, 2013