Applications Of Derivatives Question 70
Question: If $ A>0,B>0 $ and $ A+B=\pi /3, $ then the maximum value of tan A tan B is
Options:
A) $ \frac{1}{\sqrt{3}} $
B) $ \frac{1}{3} $
C) $ 3 $
D) $ \sqrt{3} $
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Answer:
Correct Answer: B
Solution:
[b] We have, $ A+B=\frac{\pi }{3} $
$ \therefore B=\frac{\pi }{3}-A\Rightarrow \tan B=\frac{\sqrt{3}-\tan A}{1+\sqrt{3}\tan A} $ Let $ Z=\tan A.\tan B $ . Then, $ Z=\tan A.\frac{\sqrt{3}-\tan A}{1+\sqrt{3}\tan A}=\frac{\sqrt{3}\tan A-{{\tan }^{2}}A}{1+\sqrt{3}\tan A} $
$ \Rightarrow Z=\frac{\sqrt{3}x-x^{2}}{1+\sqrt{3}x}, $ where $ x=\tan A $
$ \Rightarrow \frac{dZ}{dx}=-\frac{(x+\sqrt{3})(\sqrt{3}x-1)}{{{(1+\sqrt{3}x)}^{2}}} $ For max $ Z,\frac{dZ}{dx}=0\Rightarrow x=\frac{1}{\sqrt{3}},-\sqrt{3} $ . $ x\ne -\sqrt{3} $ because $ A+B=\pi /3 $ which implies that $ x=\tan A>0 $ . It can be easily checked that $ \frac{d^{2}Z}{dx^{2}}<0 $ for $ x=\frac{1}{\sqrt{3}} $ .
Hence, Z is maximum for $ x=\frac{1}{\sqrt{3}}i.e.,\tan A=\frac{1}{\sqrt{3}}orA=\pi /6 $ . For this value of $ x,Z=\frac{1}{3}. $
BETA
