Applications Of Derivatives Question 422
Question: A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Then the minimum length of the hypotenuse is
Options:
A) $ {{( {a^{\frac{3}{2}}}+{b^{\frac{3}{2}}} )}^{\frac{2}{3}}} $
B) $ {{( {a^{\frac{2}{3}}}+{b^{\frac{2}{3}}} )}^{\frac{3}{2}}} $
C) $ {{( {a^{\frac{2}{3}}}+{b^{\frac{2}{3}}} )}^{3}} $
D) $ {{( {a^{\frac{3}{2}}}+{b^{\frac{3}{2}}} )}^{3}} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] From the figure, $ PC=b\cos ec\theta $ and $ AP=asec\theta $
$ AC=PC+AP $ or $ AC=b\cos ec\theta +a\sec \theta ….(1) $
$ \therefore \frac{d(AC)}{d\theta }=-b\cos ec\theta \cot \theta +a\sec \theta \tan \theta $ For minimum length, $ \frac{d(AC)}{d\theta }=0 $ or $ a\sec \theta \tan \theta =b\cos ec\theta \cot \theta $ or $ \tan \theta ={{( \frac{b}{a} )}^{1/3}} $
$ \therefore \sin \theta =\frac{{{(b)}^{1/3}}}{\sqrt{{a^{2/3}}+{b^{2/3}}}} $ and $ \cos \theta =\frac{{{(a)}^{1/3}}}{\sqrt{{a^{2/3}}+{b^{2/3}}}} $ Also, $ \theta \in (0,\pi /2) $
$ \underset{\theta \to 0}{\mathop{\lim }},(a,\sec \theta +b,\cos ec\theta )\to \infty $
and $ \underset{\theta \to \pi /2}{\mathop{\lim }},(a,\sec \theta +b,\cos ec\theta )\to \infty $
Therefore, $ \theta ={{\tan }^{-1}}{{( \frac{b}{a} )}^{1/3}} $ is a point of minima.
For this value of $ \theta $ , $ AC=\frac{b\sqrt{{a^{2/3}}+{b^{2/3}}}}{{b^{1/3}}}+\frac{a\sqrt{{a^{2/3}}+{b^{2/3}}}}{{a^{1/3}}} $
[Using (1) and (2)]
$ =\sqrt{{a^{2/3}}+{b^{2/3}}}({b^{2/3}}+{a^{2/3}})={{({a^{2/3}}+{b^{2/3}})}^{3/2}} $
Hence, the minimum length of the hypotenuse is $ {{({a^{2/3}}+{b^{2/3}})}^{3/2}}. $
BETA
