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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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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}}$

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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 $l=a\sec\theta+bcosec\theta$
$\qquad\qquad\qquad=\large\frac{a.\sqrt{a^{\Large\frac{2}{3}}+b^{\Large\frac{2}{3}}}}{a^{\Large\frac{1}{3}}}+\frac{b.\sqrt{a^{\Large\frac{2}{3}}+b^{\Large\frac{2}{3}}}}{b^{\Large\frac{1}{3}}}$
$\qquad\qquad\qquad=\sqrt{a^{\Large\frac{2}{3}}+b^{\Large\frac{2}{3}}}(a^{\Large\frac{2}{3}}+b^{\Large\frac{2}{3}})$
$\qquad\qquad\qquad=\big(a^{\Large\frac{2}{3}}+b^{\Large\frac{2}{3}}\big)^{\Large\frac{3}{2}}$
answered Aug 12, 2013 by sreemathi.v
edited Sep 2, 2013 by sharmaaparna1
 

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